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OSMANIA UNIVERSITY LIBRARY
Call No. 3/7 Sy /V? / Accession No. // S ^ 2
Author
Title
This book should be ret,urnd orvor befoe the date last marked below.
LECTURES
ON
THE THEORY OF FUNCTIONS OF
HEAL VARIABLES
VOLUME II
BY
JAMES PIERl'ONT, LL.D.
PKOFESSOK OF MATHEMATICS IN YALE UNIVERSITY
GINN AND COMPANY
BOSTON NEW YORK CHICACiO LONDON
COPYRIGHT, 1912, BY
JAMES P1ERPONT
ALL RIGHTS RESERVED
PRINTED IN THK UNITED STATUS OF AMERICA
926.1
Cbe fltbcnicure grc
GINN AND COMPANY PRO
PRIETORS BOSTON U.S A.
TO
ANDREW W. PHILLIPS
THESE LECTURES
ARE INSCRIBED
WITH AFFECTION AND ESTEEM
PREFACE
THE present volume has been written in the same spirit that
animated the first. The author has not intended to write a
treatise or a manual ; he has aimed rather to reproduce his uni
versity lectures with necessary modifications, hoping that the
freedom in the choice of subjects and in the manner of presenta
tion allowable in a lecture room may prove helpful and stimulating
to a larger audience.
A distinctive feature of these Lectures is an attempt to develop
the theory of functions with reference to a general domain of
definition. The first functions to be considered were simple
combinations of the elementary functions. Kiemann in his great
paper of 1854, " Ueber die Darstellbarkeit einer Function durch
eine trigonometrische Reihe," was the first to consider seriously
functions whose singularities ceased to be intuitional. The re
searches of later mathematicians have brought to light a collection
of such functions, whose existence so long unsuspected lias revolu
tionized the older notion of a function and made imperative the
creation of finer tools of research. But while minute attention
was paid to the singular character of these functions, practically
none was accorded to the domain over which a function may be
defined. After the epochmaking discoveries inaugurated in 1874
by G. Cantor in the theory of point sets, it was no longer neces
sary to consider a function of one variable as defined in an in
terval, a function of two variables as defined over a field bounded
by one or more simple curves, etc. The first to make use of this
new freedom was C. Jordan in his classic paper of 1892. He
has had, however, but few imitators. In the present Lectures the
author has endeavored to develop this broader view of Jordan,
persuaded that in so doing he is merely carrying a step farther
the ideas of Dirichlet and Riemann.
Often such an endeavor leads to nothing new, a mere statement
for any n of what is true for n = 1, or 2. A similar condition
v
vi PREFACE
prevails in the theory of determinants. One may prefer to treat
only two and three rowed determinants, but he surely has no
ground of complaint if another prefers to state his theorems and
demonstrations for general n. On the other hand, the general
case may present unexpected and serious problems. For example,
Jordan has introduced the notion of functions of a single variable
having limited variation. How is this notion to be extended to
two or more variables ? An answer is far from simple. One was
given by the author in Volume I ; its serviceableness has since
been shown by B. Camp. Another has been essayed by Lebesgue.
The reader must be warned, however, against expecting to find
the development always extended to the general case. This,
in the first place, would be quite impracticable without greatly
increasing the size of the present work. Secondly, it would often
be quite beyond the author's ability.
Another feature of the present work to which the author would
call attention is the novel theory of integration developed in
Chapter XVI of Volume 1 and Chapters I and II of Volume II.
It rests on the notion of a cell and the division of space, or in fact
any set, into unmixed partial sets. The definition of improper
multiple integrals leads to results more general in some respects
than yet obtained with Riemann integrals.
Still another feature is a new presentation of the theory of
measure. The demonstrations which the author has seen leave
much to be desired in the way of completeness, not to say rigor.
In attempting to find a general and rigorous treatment, he was
at last led to adopt the form given in Chapter XI.
The author also claims as original the theory of Lebesgue
integrals developed in Chapter XII. Lebesgue himself considers
functions such that the points e at which a <f(x) < 6, for all a, b
form a measurable set. His integral he defines as
where l m <f(x)<l m+l i n e m whose measure is e f m9 and each
lm+i Z m = 0, as n = oo. The author has chosen a definition which
occurred to him many years ago, and which to him seems far
more natural. In Volume I it is shown that if the metric field 21
PREFACE vii
be divided into a finite number of metric sets S v S a of norm <2,
then
f / = M ax Zm& , f / = Min 2 M&
/5!l *^2(
where m t , Jf t are the minimum and maximum of/ in 8 t . What
then is more natural than to ask what will happen if the cells
&i ^2"* are infiftit 6 instead of tinite in number? From this
apparently trivial question results a theory of ^integrals which
contains the Lebesgue integrals as a special case, and which,
furthermore, has the great advantage that riot only is the relation
of the new integrals to the ordinary or Riemannian integrals
perfectly obvious, but also the form of reasoning employed in
Riemaim's theory may be taken over to develop the properties
of the new integrals.
Finally the author would call attention to the treatment of
the area of a curved surface given at the end of this volume.
Though the above are the main features of novelty, it is hoped
that the experienced reader will discover some minor points, not
lacking in originality, but not of sufficient importance to em
phasize here.
It is now the author's pleasant duty to acknowledge the in
valuable assistance derived from his colleague and former pupil,
Dr. W. A. Wilson. He has read the entire manuscript and
proof with great care, corrected many errors and oversights in
the demonstrations, besides contributing the substance of 372,
373, 401406, 414424.
Unstinted praise is also due to the house of Ginn and Com
pany, who have met the author's wishes with unvarying liberality,
and have given the utmost care to the press work.
JAMES PIERPONT
NEW HAVEN; December, 1911
CONTENTS
CHAPTER I
POINT SETS AND PROPER INTEGRALS
ARTICLES
110. Miscellaneous Theorems .
1115. Iterable Fields ....
1625. Union and Divisor of Point Sets
PAGE
1
14
22
CHAPTER II
IMPROPER MULTIPLE INTEGRALS
2628. Classical Definition
29 Definition of de la Valle'e Poussin
. 30
31
30. Author's Definition .........
3101. General Theory
. 32
. 32
59
70 78. Iterated Integrals
. 63
CHAPTER III
SERIES
7980. Preliminary Definitions and Theorems
81. Geometric, General Harmonic, Alternating, and Telescopic Series
82. Dini's Series
83. Abel's Series
84. Trigonometric Series
85. Power Series
86. Cauchy's Theorem on the Interval of Convergence .
8791. Tests of Convergence. Examples
92. Standard Series of Comparison ......
9398. Further Tests of Convergence ........
99. The Binomial Series
100. The Hypergeornetric Series
101108. Pringsheim's Theory
109113 Arithmetic Operations on Series
114115. Twoway Series
77
81
80
.87
88
89
90
91
101
104
110
112
113
125
133
CONTENTS
CHAPTER IV
MULTIPLE SERIES
ABTIOLIS
116126, General Theory
126188. Iterated Series .
PAGE
187
148
CHAPTER V
SERIES Of FUNCTIONS
184145. General Theory. Uniform Convergence .
146. The MooreOsgood Theorem .
J47149. Continuity of a Series ....
150152. Term wise Integration ....
158156. Termwise Differentiation ....
156
170
178
177
181
CHAPTER VI
POWER SERIES
157168. Termwise Differentiation and Integration 187
169. Development of log (1 + #), arcsin x. arctan x, e*, sin ac, coax . 188
160. Equality of two Power Series 191
161162. Development of a Power Series whose Terms are Power Series . 192
163. Multiplication and Division of Power Series 196
164^166, Undetermined Coefficients 197
166167. Development of a Series whose Terms are Power Series . . . 200
168. Inversion of a Power Series 203
169171. Taylor's Development 206
172. Forms of the Remainder 208
173. Development of (1 4 x)n 210
174. Development of log (1fx), etc 212
176181. Criticism of Current Errors 214
182. Pringsheiin's Necessary and Sufficient Condition .... 220
183. Circular Functions ".222
184. Hyperbolic Functions 228
186192. Hypergeometric Function 229
193. Bessel Functions 238
CHAPTER VII
INFINITE PRODUCTS
195202. General Theory
203206. Arithmetical Operations
207212. Uniform Convergence
213218. Circular Functions .
242
250
254
257
CONTENTS
xi
ARTIOLBS PAGE
219. Bernouillian Numbers ......... 266
220*228. B aud T Functions .......... 267
CHAPTER VIII
AGGREGATES
229230. Equivalence ........... 276
231. Cardinal Numbers .......... 278
232241. Enumerable Sets .......... 280
242. Some Space Transformations ........ 286
243260. The Cardinal c .......... 287
251261. Arithmetic Operations with Cardinals ...... 292
262264. Numbers of Liouville ......... 299
CHAPTER IX
ORDINAL NUMBERS
266267. Ordered Sets ........... 302
268270. EutacticSets ........... 304
271279. Sections ............ 807
280284. Ordinal Numbers .......... 310
285288. Limitary Numbers .......... 814
289300. Classes of Ordinals .......... 318
CHAPTER X
POINT SETS
301312. Pantaxis ........... 324
813320. Transfinite Derivatives ......... 380
321333. Complete Sets ........... 337
CHAPTER XI
MEASURE
334343. Upper Measure .
344368. Lower Measure
369870. Associate Sets .
371376. Separated Sets .
343
348
365
366
CHAPTER XII
LEBESGUE INTEGRALS
377402. General Theory
403400. Integrand Sets .
371
385
Xll
CONTENTS
ARTICLES PAGE
407409. Measurable Functions . . 388
410. Quasi and Semi Divisors 390
411413. Limit Functions 392
414424. Iterated Integrals 394
IMPROPER ITINTEGRALS
425428. Upper and Lower Integrals 402
429431. 7,Integrals 405
432435. Iterated Integrals 409
CHAPTER XIII
FOURIER'S SERIES
436437. Preliminary Remarks 415
438. Summation of Fourier's Series ........ 420
439442. Validity of Fourier's Development 424
443446. Limited Variation 429
447448. Other Criteria . 437
449456. Uniqueness of Fourier's Development 438
CHAPTER XIV
DISCONTINUOUS FUNCTIONS
467462. Properties of Continuous Functions .
403404. Pointwise and Total Discontinuity .
405473. Examples of Discontinuous Functions
474489. Functions of Class 1
490497. Semicontinuous Functions ....
452
467
459
468
485
CHAPTER XV
DERIVATKS, EXTREMES, VARIATION
498518. Derivates
519525. Maxima and Minima
526534. Variation
535537. Nonintuitional Curves
538539. Pompeiu Curves
540542. Faber Curves
493
521
531
537
542
546
CHAPTER XVI
SUB AND INFRAUNIFORM CONVERGENCE
Continuity
643650.
551556. Integrability ,
557561 . Differentiability
665
662
670
CONTENTS
xin
CHAPTER XVII
GEOMETRIC NOTIONS
ARTICLES PA6X
662663. Properties of Intuitional Plane Curves 678
664. Motion 679
666. Curve as Intersection of Two Surfaces 679
<3^66. Continuity of a Curve . 680
667. Tangents 680
668672. Length 681
673. Spacefilling Curves 688
674. Hilbert's Curve . 690
676. Equations of a Curve 693
676580. Closed Curves 694
581. Area 699
682. Osgood's Curve 600
583. Resume" 603
584585. Detached and Connected Sets 603
586591. Images 605
692597. Side Lights on Jordan Curves 610
598600. Brouwer's Proof of Jordan's Theorem . 614
601. Dimensional Invariance 619
602. Schonfliess* Theorem 621
603608. Area of Curved Surfaces 623
Index 639
List of Symbols 644
FUNCTION THEORY OF REAL
VARIABLES
CHAPTER I
POINT SETS AND PROPER INTEGRALS
1. In this short chapter we wish to complete our treatment of
proper multiple integrals and give a few theorems on point sets
which we shall either need now or in the next chapter where we
take up the important subject of improper multiple integrals.
In Volume I, 702, we have said that a limited point set whose
upper and lower contents are the same is measurable. It seems
best to reserve this term for another nofcion which has come into
great prominence of late. We shall therefore in the future call
sets whose upper and lower contents are equal, metric sets. When
a set 31 is metric, either symbol
or
expresses its content. In the following it will be often con
venient to denote the content of 21 by
This notation will serve to keep in mind that 21 is metric, when
we are reasoning with sets some of which are metric, and some
are not.
The frontier of a set as 21, may be denoted by
Front .
2. 1. In I, 713 we have introduced the very general notion of
cell, division of space into cells, etc. The definition as there
1
2 POINT SETS AND PROPER INTEGRALS
given requires each cell to be metric. For many purposes this
is not necessary ; it suffices that the cells form an unmixed divi
sion of the given set 91. Such divisions we shall call unmixed di
visions of norm S. [I, 711.] Under these circumstances we have
now theorems analogous to I, 714, 722, 723, viz :
2. Let 33 contain the limited point set 21. Let A denote an un
mixed division of 33 of norm S. Let 2l fi denote those cells of 33 con 
taining points of 21. Then
lim H a = H.
5=0
The proof is entirely analogous to I, 714.
8. Let 33 contain the limited point set 21. Let f(x^ # m ) be
limited in 2(. Let A be an unmixed division of 33 of norm 8 info
cells S v S 2 , . Let 2ft t , tn t be respectively the maximum and mini
mum of f in S t . Then
lim S* = lira 22ftA = f /<% (1
6=0 6=0 c/2l
lim S = lim 2mA = f fdft. (2
6=0 6=0 J%
Let us prove 1) ; the relation 2) may be demonstrated in a similar
manner. In the first place we show in a manner entirely analo
gous to I, 722, that
(3
The only modifications necessary are to replace S t , S[, S l/e , by their
upper contents, and to make use of the fact that A is unmixed, to
establish 5).
To prove the other relation
(4
we shall modify the proof as follows. Let U be a cubical division
of space of norm e < e . We may take e so small that
(5
PROPER INTEGRALS 3
The cells of E containing points of 21 fall into two classes.
1 the cells e M containing points of the cell S t but of no other cell
of A ; 2 the cells e{ containing points of two or more cells of A.
Thus we have _.
where M^ M{, are the maxima of / in e itt , e{. Then as above we
have __
4' o
if e is taken sufficiently small.
On the other hand, we have
Now we may suppose S , e are taken so small that
diflfer from 21 by as little as we choose. We have therefore for
properly chosen S , ,
This with 6) gives
which with 5) proves 4).
4. Let f(xi  . XM) be limited in the limited field 21. Let A be
an unmixed division oftyof norm 8, into cells S v 8 2 . Let
where as usual ra t , Jf t are the minimum and maximum of f in 8 t .
~
Max >S A , Cfd* = Min A^ A .
J%
The proof is entirely similar to I, 723, replacing the theorem
there used by 2, 8.
5. In connection with 4 and the theorem I, 696, 723 it may be
well to caution the reader against an error which students are apt
to make. The theorems I, 696, 1, 2 are not necessarily true if /
4 POINT SETS AND PROPER INTEGRALS
has both signs in 21. For example, consider a unit square 8
whose center call O. Let us effect a division E of S into 100
equal squares and let 21 be formed of the lower lefthand square
and of 0. Let us define / as follows :
/= 1 within *
=  100 at a
For the division JE?,
^=l + Tta
Hence, Min ^ < _
On the other hand, lim
The theorems I, 723, and its analogue 4 are not necessarily true
for unmixed divisions of space. The division A employed must
be unmixed divisions of the field of integration 21. That this is
so, is shown by the example just given.
6. In certain cases the field 21 may contain no points at all.
In such a case we define
7. From 4 we have at once :
Let A be an unmixed division of 21 into cells 8j, 8 2 , Then
S = Min 2S t ,
with respect to the class of all divisions A.
8. We also have the following :
Let D be an unmixed division of space. Let d v d v denote those
cells containing points of 21. Then
with respect to the class of the divisions D.
For if we denote by S t the points of 81 in d t we have obviously
Also by I, 696, 21 = Miu
PROPER INTEGRALS 5
with respect to the class of rectangular division of space JP= je t j.
But the class E is a subclass of the class D.
Thus
Min 2^ < Min 2<? t <Min 22 t .
A D JS
Here the two end terms have the value 21.
3. Let/^j # m ), ^(zj # m ) be limited in the limited field
We have then the following theorems :
1. Letf s g in 21 except possibly at the points of a discrete set
T 7
//=//
For let  / , \g\< M. Let D be a cubical division of norm d.
Let MI, NI denote the maximum of/, g in the cell d^ Let A de
note the cells containing points of J)> while A may denote the
other cells of 2l/>.
Then, 2 M& = 2 W 4
Hence,
and the term on the right =^ as d = 0.
2. Letf > g in 21 except possibly at the points of a discrete set
Then
For let 91 = A +
/>/>
But in 4, />^, hence
The theorem now follows at once.
POINT SETS AND PROPER INTEGRALS
fc..fr t
, fc.fr.
For in any cell d,
Max of = c Max/; Min </ = c Min/
when c > ; while
Max <?/=<? Min/; Min <?/=<? Max/
when <? < 0.
4. If g is integrable in 21,
For from
Max/+Min#<Max
we have
fc + &*& + ^fc + fc' (2
But g being integrable,
Hence 2) gives
which is the first half of 1). The other half follows from the
relation
Min / + Min g < Min (/ + #) < Min / + Max g.
5. The integrands f, g being limited,
For in any cell d,
Min (/ + g} < Min / + Max g < Max (/ + g).
PROPER INTEGRALS
6. Letf*=g + h, \h\<H a constant, in 21. Then,
For
Then by 2 and 4
f ff +fff<ff<fff
*s9f *^9f '91 *^ 9f
or  r r r"
ff+J ^<J /<J y
4. Letf(x v # m ) 6e limited in limited 21. 2%ew,
J)^ / 1 < Jlf, ^e have also,
~3t
0
(2
(3
(4
(5
Let us effect a cubical division of space of norm 8.
To prove 1) let JV;=sMax/ in the cell rf t . Then using the
customary notation,
Hence
<
Letting 8=0, this gives
which is 1).
POINT SETS AND PROPER INTEGRALS
To prove 3), we use the relation
i/i </<
Hence
from which 3) follows on using 3, 3.
The demonstration of 4) is similar.
To prove 5), we observe that
5. 1. Let f> be limited in the limited fields 39, g. Let 51 be
the aggregate formed of the points in either 35 or . Then
This is obvious since the sums
may have terms in common. Such terms are therefore counted
twice on the right of 1) and only once on the left, before passing
to the limit.
Remark. The relation 1) may not hold when /is not > 0.
Example. Let 21 = (0, 1), 35 = rational points, and (= irra
tional points in 31 Let/= 1 in 35, and 1 in . Then
and 1) does not now hold.
2. Let 21 be an unmixed partial aggregate of the limited field
Let S  33  21. If
ff=*f in
=s in 6,
then
PROPER INTEGRALS
But ~ f
V = J/ by8 ' 1 '
and obviously
3. The reader should note that the above theorem need not be
true if 91 is not an unmixed part of $3.
Example. Let 21 denote the rational points in the unit square
Then
4. jLetf 21 fte a part of the limited field 33 Letf>. 6e limited in
91. i^ #=/ tw 21 awci = in g = S  21 Then
 (2
For let J^, N L be the maxima of/, g in the cell d t . Then
Passing to the limit we get 1).
To prove 2) we note that in any cell containing a point of 31
Min/> Min#.
6. 1. Letf(xi a? w ) be limited in the limited field 21. Let 4B M
be an unmixed part of 21 awcA Aa ) = 21 as u === 0.
f/=li,nf/.
*1 u=0
10 POINT SETS AND PROPER INTEGRALS
For let / <Jf in . Let <5 M = 21  S u . Then
But
by 4, 1), 5).
Hence passing to the limit u = in 2) we get 1).
2. We note that 1 may be incorrect if the $d u are not unmixed.
For let 51 be the unit square. Let SS U be the rational points in a
concentric square whose side is 1 u. Let/= 1 for the rational
points of 21 and = 2 for the other points. Then
7. In I, 716 we have given a uniform convergence theorem
when each 33 M <21. A similar theorem exists when each M .>21,
viz. :
Let $8 U < 95 U , if u< u f . Let 91 be a part of each $Q U . Let <B M =
21 as u = 0. Then for each e> 0, there exists a pair u^ d Q such that
For SS Uo < 21 + ^, UQ sufficiently small.
Also for any division D of norm d < some d Q .
But
,/>< MO ,^ if
Hence
8. 1. Let 21 be a point set in m = r + a way space. Let us set
certain coordinates as # r +i a? m= i n ea h point of 21. The
resulting points $8 we call a projection of 21. The points of 21
PROPER INTEGRALS 11
belonging to a given point 6 of 33, we denote by ( 6 or more shortly
by S. We write
a = 33 e,
and call 33* & components of a.
We note that the fundamental relations of I, 733
hold not only for the components y, ^, etc., as there given, but
also for the general components 31, 53.
In what follows we shall often give a proof for two dimensions
for the sake of clearness, but in such cases the form of proof will
admit an easy generalization. In such cases 33 will be taken as
the ^projection or component of a*
2. If a = 33 & is limited and 33 is discrete, 91 is also discrete.
For let 91 lie within a cube of edge ^ (7> 1 in m = r + s way
space. Then for any d < some c# ,
Then a/> < C' D < e.
3. That the converse of 2 is not necessarily true is shown by
the two following examples, which we shall use later :
Example 1. Let 91 denote the points #, y in the unit square
determined thus :
For
Tfl
= i 7i=l, 2, 3, , 7/1 odd and < 2 n ,
let
Here a is discrete, while 48 = 1, where 33 denotes the projection
of a on the #axis.
4. ISxample 2. Let a denote the points #, y in the unit square
determined thus :
12 POINT SETS AND PROPER INTEGRALS
For
2 * _, nt, n relatively prime,
n
let 1
o<y<l
Then, $ denoting the projection of 81 on the tfaxis, we have
f = 0, # = 1.
9. 1. Let 21= $ & 60 a limited point set. Then
<. a
For let/=l in 21. Let ^ = 1 at each point of 21 and at the
other points of a cube A = B  containing 21, let g = 0. Then
Byl ' 733 '
But by 5, 4,
Thus
21
which gives 1), since
2. Jw case 21 * metric we have
iJt*. (2
andf S w aw integrable function over $
This follows at once from 1).
PROPER INTEGRALS 13
3. In this connection we should note, however, that the converse
of 2 is not always true, i.e. if is integrable, then 21 has content
and 2, 2) holds. This is shown by the following :
Example. In the unit square we define the points #, ^of 21 thus :
For rational a?,
For irrational #,
Then = i for every x in 33. Hence
/"*
But 91 = 0, l.
10. 1. Letf(x^ " # m ) be limited in the limited field 81 = 33
(2
,,
Let us first prove 1). Let 91, S3, ( lie in the spaces 9t m , $ 9J
r f g = m. Then any cubical division D divides these spaces into
cubical cells rf,, d(, d" of volumes rf, c?', rf" respectively. Ob
viously d = d'd". J? also divides 33 and eacA S into unmixed cells
S', 8". Let M. = Max/ in one of the cells d a while M'J = Max/
in the corresponding cell S". Then by 2, 4,
since M MJ > 0. Hence
Letting the norm of D converge to zero, we get 1). We get
2) by similar reasoning or by using 3, 3 and 1).
14 POINT SETS AND PROPER INTEGRALS
2. To illustrate the necessity of making/ > in 1}, let us take
31 to be the Pringsheim set of I, 740, 2, while / shall = 1 in 21.
Then
On the other hand
Hence
and the relation 1) does not hold here.
Iterable Fields
11. 1. There is a large class of limited point sets which do not
have content and yet _ ._
21= f . (1
%/jjg
Any limited point set satisfying the relation 1) we call iterable^
or more specifically iterable with respect to $$.
Example 1. Let 21 consist of the rational points in the unit
square. Obviously __  ^
21=  6=  = 1,
j/33 */(
so that 21 is iterable both with respect to 48 and (.
Example 2. Let 21 consist of the points x, y in the unit square
defined thus :
For rational x let < y < J.
For irrational # let < y < 1.
Here 21 = 1.
Thus 21 is iterable with respect to S but not with respect to 53,
ITERABLE FIELDS 15
Example 8. Let 91 consist of the points in the unit square de
fined thus: ,. 11^ A ^ ^o
For rational a? let < y < .
For irrational a? let \ < y < 1.
Here 91 = 1, while
Hence 91 is iterable with respect to & but not with respect to $8.
Example 4. Let 21 consist of the sides of the unit square and
the rational points within the square.
Here 91 = 1, while
and similar relations for &. Thus 21 is not iterable with respect
to either 33 or (.
Example 5. Let 21 be the Pringsheim set of I, 740, 2.
Here 21 = 1, while
Hence 21 is not iterable with respect to either 33 or @.
2. Every limited metric point set is iterable with respect to any of
its projections.
This follows at once from the definition and 9, 2.
12. 1. Although 21 is not iterable it may become so on remov
ing a properly chosen discrete set J).
Example. In Example 4 of 11, the points on the sides of the
unit square form a discrete set 5 ; on removing these, the deleted
set 21* is iterable with respect to either S3 or .
2. The reader is cautioned not to fall into the error of suppos
ing that if 2l x and 2^ are unmixed iterable sets, then 21 = 2l x + 21 2
is also iterable. That this is not so is shown by the Example in 1.
For let 2t x = 21*, 91 2 = > in that example. Then $) being dis
crete has content and is thus iterable. But 21 = 2lj + 21 2 is
iterable with respect to either $Q or <.
16 POINT SETS AND PROPER INTEGRALS
13. 1. Let 21 be a limited point set lying in the m dimensional
space 9t m . Let 48, & be components of 31 in 9t r , 9?,, r + s = m.
A cubical division D of norm 8 divides 9t m into cells of volume
d and W r and 9t, into cells of volume d^ c?,, where d = d r rf r Let
b be any point of 48, lying in a cell d r . Let 2rf, denote the sum
b
of all the cells d# containing points of 31 whose projection is b.
Let 2,d s denote the sum of all the cells containing points of 31
dr
whose projection falls in d r , not counting two d t cells twice.
We have now the following theorem :
is iterable with respect to 48,
For
Hence
<
8 *
Let now 8 0. The first and third members = 31, using I, 699,
since 31 i iterable. Thus, the second and third members have
the same limit, and this gives 1).
2. If 31 i* iterable with respect to 48,
lim Zd^d, = I.
=o 33 b
This follows at once from 1).
3. Let 31 =as g be a limited point set, iterable with respect to $&.
Then any unmixed part Q of 31 is also iterable with respect to the
SBcomponent ofQ*.
For let b = a point of 48 ; 6' points of 31 not in @ ; C b = points
of S 6 in g, (7^ = points of ( 6 in g'. Then for each >0 there
exist a pair of points, b v 6 2 , distinct or coincident in any cell d f
such that as 6 ranges over this cell,
ITERABLE FIELDS 17
Let $ denote, as in 13, l, the cells of 2cZ, which contain points of (',
and F the cells containing points of both (, (' whose projections
fall in d r . Then from
ii t
we have dr
<E 6 <C*+Cl< Min G b + 0' + S< Max <? 6 4 "
Multiplying by d r and summing over 33 we have,
(i
Passing to the limit, we have
f tf + y + <*, (2
*/
the limit of the last term vanishing since (g, (' are unmixed parts
of 31. Here r/, 77" are as small as vye please on taking /9 sufficiently
small. From 2) we now have
4. Let 31 = 55 fo iterable with respect to SJ. i^ S be a part
of $8 and A all those points of 31 whose projection falls on B. Then
A is iterable with respect to B.
For let D be a cubical division of space of norm d. Then
= lim
2d r d,\, (1
r,* >
where the sum on the right extends over those cells containing
no point of A. Also
(2
where the second sum on the right extends over those cells d,
containing no point of JB.
Subtracting 1), 2) gives
= lim { A D  2rf r S } 4 lim
dU <> M > d0
18 POINT SETS AND PROPER INTEGRALS
As each of the braces is > we have
S.
JB
14. We can now generalize the fundamental inequalities of I,
733 as follows :
Let f(x l XM) be limited in the limited field 21 = 33 S, iterable
with respect to SQ. Then
For let us choose the positive constants A, B such that
f+A>0, /^<0, in .
Let us effect a cubical division of the space of 9t m of norm 8 into
cells d. As in 13, this divides 9^, 91 ,, into cells which we denote,
as well as their contents, by c? r , d s . Let b denote any point of 48.
As usual let m, M denote the minimum and maximum of / in the
cell d containing a point of 21. Let m' , M 1 be the corresponding
extremes of /when we consider only those points of 21 in d whose
projection is b. Let / < F in 21.
Then for any 6, we have by I, 696,
or __
4 6)+ 2md s < \J\ (2
snce
In a similar manner
? <M3, + 4(2d,<). (3
Thus for any 5 in S3, 2), 3) give
 JB(2df,6)+2wrf,< Cf<2Md s + 4(2<Z.<). (4
ft 6 .( 6 6
Let /8 > be small at pleasure. There exist two points b v 6 2 dis
tinct or coincident in the cell rf r , for which
ITERABLE FIELDS 19
where  ff l ,  &  < and & v and (^ stand for S v S 6j , and finally
where __
yMinf/ f J"
*/(j
for all points 6 in d r .
Let c = Min & in c? r , then 4) gives
5  c) + 2mrf, < y + ft < </+ /3 2 < SMi, h ^(2^,  c)
11 22
where the indices 1, 2 indicate that in 2 we have replaced b by
Multiplying by d r and summing over all the cells d r containing
points of 53, the last relation gives
~ c) 4 ^
1 SB 1
5  c). (5
SB SB SB 2 SB 2
2rf r 2rf 4 = 2l, by 13, 2.
SB 2
s= ) 6 = S, since 51 is iterable.
^
Thus the first and last sums in 5) are evanescent with 8. On
the other hand
2<2,m  2(2 4 w) I <
d, 1  SB efc 1
= as 8=0, by 13, l,
Thus
(6
(7
Hence passing to the limit 8=0 in 5) we get 1), since 2 1 d r ,
2/8 2 d r have limits numerically <>8& which may be taken as small
as we please as (3 is arbitrarily small.
20 POINT SETS AND PROPER INTEGRALS
2. If 91 is not iterable with respect to 33, let it be so on remov
ing the discrete set ). Let the resulting field A have the com
ponents B, 0. Then 1 gives
snce
3. The reader should guard against supposing 1) is correct if
only 21 is iterable on removing a discrete set J). For consider
the following :
Example. Let the points of 51 = 9lj + T) lie in the unit square.
Let 9lj consist of all the points lying on the irrational ordinates.
Let !) lie on the rational ordinates such that, when
tyy*
x~~, m,n relatively prime,
n
.
n
Let us define/ over 31 thus :
/=! in 9l r
/=0 in >.
The relation 1) is false in this case. For
/=!,
.*
while
/
.*
15. 1. Let /(tfj x m ) be limited in the limited point set 91.
Let D denote the rectangular division of norm d. All the points
of 9l/> except possibly those on its surface are inner points / of 91.
[I, 702.]
The limits H m Cf , u m Cf (1
d=0 */ rf=0 %/W
7) J)
exist and will be denoted by
JTV , fV , (2
^/a ^a
and are called the inner, lower and upper integrals respectively.
1TERABLE FIELDS 21
To show that 1) exist we need only to show that for each c >
there exists a d Q such that for any rectangular divisions 2>', D" of
norms < d
A:
To this end, we denote by E the division formed by superimpos
ing D" on D 1 . Then E is a rectangular division of norm < d .
Let %s ~ la* = 4', *  />" = ^"
If d ft is sufficiently small, A , An
VI , A <^ ??,
an arbitrarily small positive number. Then
I//T 7* \ //*" 7*\l
A= r  u ~r < +
if i; is taken small enough.
2. The integrals

, r/
/
heretofore considered may be called the outer, lower and upper in
tegrals, in contradistinction.
3. Let f be limited in the limited metric field 31. Then the inner
and outer lower (upper) integrals are equal.
For y( D is an unmixed part of 31 such that
Cont !, = , asd = 0.
Then by 6, l, r r
limj /=)/.
d=0^/, ^
But the limit on the left is by definition
4. JTAew 21 has no inner points,
22 POINT SETS AND PROPER INTEGRALS
For each fl/, = 0, ar^d hence each
Point Sets
16. Let 21 = 33 + be metric. Then
For let D be a cubical division of space of norm d. The cells
of 31^ fall into three classes : 1, cells containing only points of 33;
these form $8 D . 2, cells containing points of (; these' form (/>.
3, cells containing frontier points of 33, not already included in 1
or 2. Call these fo. Then
/> = / , + 6 1> +L. ( 2
Let now d = 0. As 91 is metric, \ D = 0, since f^ is a part of
Front 21 and this is discrete. Thus 2) gives 1).
17. 1. Let 31, , 6  (1
be point sets, limited or not, and finite or infinite in number.
The aggregate formed of the points present in at least one of the
sets 1) is called their union^ and may be denoted by
or more shortly by
g _
If 31 is a general symbol for the sets 1), the union of these sets
may also be denoted by
or even more briefly by ,>
If no two of the sets 1) have a point in common, their union
may be called their sum, and this may be denoted by
The set formed of the points common to all the sets 1) we call
their divisor and denote by
POINT SETS 23
or fe y Dv{%\,
if 21 is a general symbol as before.
2. Examples.
Let 31 be the interval (0, 2); SS the interval (1, oo). Then
, 8) = (0, oo), J(a, 8) = (1, 2).
Let Sit = (0,1), *, = (!, 2)
Then 91 2 ) = (0, oo),
a 9 .o=o.
Let
Then {jA)i gf ...^ = (o*
Then
3. Let a^a^aa^a,^ (i
Let = D(a, a r a,, )
Let 2l = 2
Then 2l = S + S 1 + S 2 + 
Let us first exclude the = sign in 1). Then every element of
21 which is not in ) is in some 9l n but not in 2l n+1 . It is therefore
in S B+ i but not in ( n+3 , & n + 3 , The rest now follows easily.
4. Some writers call the union of two sets 91, 8 their sum,
whether a, $ have a point in common or not. We have not done
this because the associative property of sums, viz. :
does not hold in general for unions.
24 POINT SETS AND PROPER INTEGRALS
Example. Let SI = rectangle (123 4),
SQ = (5 6 7 8),
Then
and (Z7(, #)), (2
are different.
Thus if we write + for U, 1), 2) give
* + ()*(* +).
18. 1. Let Slt^^^Sls be a set of limited complete point
aggregates. Then
/>(*!. V )><>
Moreover 33 is complete.
Let a n be a point of 2l n , n = 1, 2, and 21 = & v a& a$
Any limiting point a of 31 is in every 2l n . For it is a limit
ing point of
#im #m+H #m+2i "
But all these points lie in 2l m , which is complete. Hence a lies in
3l m , and therefore in every Sl^ 2l 2 i Hence a lies in 33, and
>0.
48 is complete. For let $ be one of its limiting points. Let
As each 6 m is in each 8l n , and 3l rt is complete, /9 is in 2l n . Hence /3
is in #.
2. i< 91 6e a limited point set of the second species. Then
', ",'", )><>,
and is complete.
For < n) is complete and > 0. Also w >. (<H>I) .
19. f Slj, 31 2 ... ?ie tn ; /et 21= Z7{2l n S. X^ A n be the com
plement of 2l n with respect to 53, so that A n + 2l n = 53 i^
A and 81 arc complementary, so that A 4 21 = 53*
POINT SETS 25
For each point 6 of 53 lies in some 2l n , or it lies in no 2l n , and
hence in every A n . In the first case b lies in 31, in the second in
A. Moreover it cannot lie in both A and 21
20. 1. Let 2l a <21 2 <21 8 (1
i
be an infinite sequence of point sets whose union call 21. This
fact may be more briefly indicated by the notation
Obviously when 21 is limited,
n . (2
That the inequality may hold as well as the equality in 2) is
shown by the following examples.
Example 1. Let 2l n = the segment f, 1 J
Then = (0*, 1).
n
Example 2. Let a n denote the points in the unit interval whose
abscissae are given by
x = , m < n = 1, 2, 3, m, n relatively prime.
n
Let . = a 1 + ... + o..
Here 21 = Z7{2U
is the totality of rational numbers in (0*, 1*).
A a _ _
91 = 1 and 21 B = 0, we see
t > lim I B .
2. Let i>l2> (3
Let SQ be their divisor. This we may denote briefly by
Obviously when SB 1 is limited,
< lim ..
26 POINT SETS AND PROPER INTEGRALS
Example 1. Let $ = the segment f 0, V
Then = Dv\% n \ =(0), the origin.
Here fg = 0. lim B = lim  = 0,
n
and = i. ^r
S3 = lim 4Bn
Example 2. Let 2l n be as in 1, Example 2. Let b n = 31 3l n .
Let g* ("\ 2^ 4 b
Here ^ = the segment (j^ 2) and *g n = 2.
Hence ^ < Um =
3. i^ 33 j < $3 2 < 6e unmixed parts of 31. ie < n = S.
= U { 33 n } . Then S = 21  & discrete.
For let 31 = 33 W + S n 5 th en Sn is an unmixed part of 31. Hence
Passing to the limit n = oo, this gives
lim g n = 0.
Hence is discrete by 2.
4. We may obviously apply the terms monotone increasing,
monotone decreasing sequences, etc. [Cf. 1, 108, 211] to sequences
of the type 1), 3).
21. Let 6 = 31 + S. If 31, S are complete,
For 8=Dist(3l,
since 91, SB are complete and have no point in common. Let D be
a cubical division of space of norm d. If d is taken sufficiently
small 210, $8 D have no cells in common. Hence
Letting d = we get 1).
POINT SETS 27
22. 1. If 21, 33 are complete, so are also
g=(2l, 33), ) = !>* (21, 33).
Let us first show that is complete. Let c be a limiting point
of (. Let <?j, c 2 , be points of ( which = <?. Let us separate
the c n into two classes, according as they belong to 21, or do not.
One of these classes must embrace an infinite number of points
which = c. As both 51 and 33 are complete, c lies in either 21 or
33 Hence it lies in g.
To show that J) is complete. Let d v d^ be points of ) which
= d. As each d n is in both 21 and 33, their limiting point d is in
21 and $Q, since these are complete. Hence d is in ).
2. If 21, 93 are metric so are
= (21,33) $D = Dt>(, 33).
For the points of Front (5 lie either in Front 21 or in Front 33i
while the points of Front ) < Front 21 and also < Front 33. But
Front 21 and Front 33 are discrete since 21, 33 are metric.
23. Let the complete set 21 have a complete part 33. Then how
ever small e > is taken, there exists a complete set in 21, having no
point in common with 33 such that
Moreover there exists no complete set S, having no point in common
with 33 such that _ _ _
<>.
The second part of the theorem follows from 21. To prove 1)
let D be a cubical division such that
,= ! + ', * = + ", 0<',"<. (2
Since 33 is complete, no point of 33 lies on the frontier of 33^
Let denote the points of 21 lying in cells containing no point of
33. Since 21 is complete so is &, and 33> & have no point in common.
Thus _
z>. (3
28 POINT SETS AND PROPER INTEGRALS
But the cells of (/> may be subdivided, forming a new division A,
which does not change the cells of 8^, so that JBz> = 33*1 but so that
S A = S + "', <'"<. (4
Thus 2), 3> 4) give
r l=t#~
>!
24. Ze 31, 8 i complete. Let
Then _____
I + CU + fc. (1
Forlet 11 = 31 + A
Then ^4 contains complete sets (7, such that
>U3l6, (2
but no complete set such that
<7>UI, (8
by 23. On the other hand,
Hence A contains complete sets (7, such that
>>, (4
but no complete set such that
<?>$>. (5
From 2), 3), and 4), 5) we have 1), since e is arbitrarily small.
8 Let
each 31 B being complete and tuck that 31 B > some constant k,
Then
POINT SETS 29
For suppose Z = tl)>0.
Let
Then by 23 there exists in 91, a complete set S r having no point
in common with $) such that
or as 2L > A, such that
j v
Let 6 a = JD V (91 2 , (, ), U
Then by 24, 8,4C^H +
Thus
Thus 21 3 contains the nonvanishing complete set S a having no
point in common with 2). In this way we may continue. Thus
Slj, 21 2 , contain a nonvanishing complete component not in D,
which is absurd.
Corollary. Let 21 = ( Slj < ?I 2 < ) be complete . ?%<w l n = t
This follows easily from 23, 25.
CHAPTER II
IMPROPER MULTIPLE INTEGRALS
26. Up to the present we have considered only proper multiple
integrals. We take up now the case when the integrand f(x l 'x m )
is not limited. Such integrals are called improper. When m = 1,
we get the integrals treated in Vol. I, Chapter 14. An important
application of the theory we are now to develop is the inversion
of the order of integration in iterated improper integrals. The
treatment of this question given in Vol. I may be simplified and
generalized by making use of the properties of improper multiple
integrals.
27. Let 91 be a limited point set in wway space 9J m . At each
point of 91 let f(x l # m ) have a definite value assigned to it.
The points of infinite discontinuity of f which lie in 21 we shall
denote by Q. In general $ is discrete, and this case is by far the
most important. But it is not necessary. We shall call $ the
singular points.
Vfli
Example. Let 51 be the unit square. At the point # = ,
r n
y = , these fractions being irreducible, let f=ns. At the other
s
points of 21 let /= 1. Here every point of 21 is a point of infinite
discontinuity and hence $ = 21.
Several types of definition of improper integrals have been
proposed. We shall mention only three.
28. Type I. Let us effect a division A of norm S of 9t m into
cells, such that each cell is complete. Such divisions may be
called complete. Let 2ls denote the cells containing points of 21,
but no point of $, while 2t 6 ' may denote the cells containing a
point of Q. Since A is complete, / is limited in 2ls. Hence /
admits an upper and a lower proper integral in 21$. The limits,
when they exist, ~
lim f /, lim f /, (1
ao J% 8 =o s/a g
30
GENERAL THEORY 31
for all possible complete divisions A of norm S, are called the
lower and upper integrals of / in 21, and are denoted by
(2
*3l %
or more shortly by
When the limits 1) are finite, the corresponding integrals 2)
are convergent. We also say/ admits a lower or an upper improper
integral in 21. When the two integrals 2) are equal, we say that
/ is integrable in 21 and denote their common value by
ffd* or by ff. (3
*/Sl */2l
We call 3) the improper integral of f in 21 ; we also say that
f admits an improper integral in 2( and that the integral 3) is
convergent.
The definition of an improper integral just given is an extension
of that given in Vol. I, Chapter 14. It is the natural develop
ment of the idea of an improper integral which goes back to the
beginnings of the calculus.
It is convenient to speak of the symbols 2) as upper and lower
integrals, even when the limits 1) do not exist. A similar remark
applies to the symbol 3).
Let us replace /by / in one of the symbols %), 3). The
resulting symbol is called the adjoint of the integral in question.
We write
(4
When the adjoint of one of the integrals 2), 3) is convergent,
the first integral is said to be absolutely convergent. Thus if 4) is
convergent, the second integral in 2) is absolutely convergent, etc.
29. Type II. Let X, /A>0. We introduce a truncated func
tion/^ defined as follows :
f^ ~f(?i  Zm) when  X </< p
= X when/< X
= fi when / > fji.
32 IMPROPER MULTIPLE INTEGRALS
We define now the lower integral as
A similar definition holds for the upper integral. The other
terms introduced in 28 apply here without change.
This definition of an improper integral is due to de la ValUe
Poussin. It has been employed by him and ft. Gr. D. Richardson
with great success.
30. Type IIL Let , /9>0. Let 2l a8 denote the points of 21
at which

We define now
f/ = lim f f ; f/ = Urn f /. (1
J* ,*/./ *V a, 0=~ Al a / V
The other terms introduced in 28 apply here without change.
This type of definition originated with the author and has been
developed in his lectures.
31. When the points of infinite discontinuity $ are discrete
and the upper integrals are absolutely convergent, all three defini
tions lead to the same result, as we shall show.
When this condition is not satisfied, the results may be quite
different.
Example. Let 21 be the unit square. Let 2l a , 2^ denote respec
tively the upper and lower halves. At the rational* points S3,
x=* n \ y = , in 2l r let/= m. At the other points of 2l p let
H\j 8
/==! In2l 2 let/=0.
1 Definition. Here $ = Sl r
Hence
2 Definition. Here
' i, f/=+oo.
* Here as in all following examples of this sort, fractions are supposed to be
irreducible.
GENERAL THEORY 33
3 Definition. Here 51^ embraces all the points of 8 a , S and a
finite number of points of $B for a > 2, ft arbitrarily large. Hence
//!, //It
i ^
and thus
32. In the following we shall adopt the third type of definition,
as it seems to lead to more general results when treating the im
portant subject of inversion of the order of integration in iterated
integrals.
We note that if /is limited in 91,
lim jf = the proper integral J /.
For a, y8 being sufficiently large, 9l aj8 == 91
Also, if 91 is discrete,
fr
For 9l a8 is discrete, and hence
Hence the limit of these integrals is 0.
33. Let w=Min/ , JlfMax/ in 91.
Then
lim I /= lim I /, m finite.
lim I /=lim f /, M finite.
a, 0=00*61^ =^ tt ,ir
For these limits depend only on large values of , & and when
m is finite. ftf ~ . ^
 > foralla>w.
Similarly, when J!f is finite
5l ft<8 ^j/ , forall/8>Jf.
34 IMPROPER MULTIPLE INTEGRALS
Thus in these cases we may simplify our notation by replacing
2L, M i 3lm0
by 2l_. , 21, ,
respectively.
2. Thus we have:
/=limj /, when Min / is finite.
J% p=*J%p
J / = lim J / , when Max/ is finite.
3. Sometimes we have to deal with several functions/, #, 
In this case the notation 2l aj8 is ambiguous. To make it clear we
let 3l/,a,8 denote the points of 91 where
Similarly, 2l g , at ^ denotes the points where
a<_g<_ft, etc.
34. I f is a monotone decreasing function of a for each ft.
J%afl
I f is a monotone increasing function of ft for each a.
J ^
If Max / 1 finite
Xf are monotone decreasing functions of a.
J
If Min / is finite
\ f are monotone increasing functions of ft.
J&p
Let us prove the first statement. Let a f > a.
Let J!) be a cubical division of space of norm d.
Then ft being fixed,
X/=lim2 m&, (1
_~0 9,0
f /=lim2m;<, (2
Jxa'p *= n a <p
using the notation so often employed before.
GENERAL THEORY 85
But each cell d, of 3L0 lies among the cells dj of 31. /. Thus we
can break up the sum 2), getting
Here the second term on the right is summed over those cells
not containing points of 2l a/3 . It is thus < 0. In the first term
on the right m^ <m t . It is thus less than the sum in 1). Hence
Thus r r
I < I , '>.
J^iJ^
In a similar manner we may prove the second statement ; let
us turn to the third.
We need only to show that
( / is monotone decreasing.
'a'.
Let '>. Then *
I == lim 2jf t <* t . (3
J%_ a d^ ^ a
f ^limSJK'rf/. (4
J%_ a , d =o %_ a>
As before ^M(d( = 2Jf (d{ + I.M'Jd'J. (5
But in the cells d t , MJ = M,. Hence the first term of 5) is
ihe same as 2 in 3). The second term of 5) is < 0. The proof
? ollows now as before.
35. If Max / is finite and I fare limited, ( f is convergent and
J& a !
If Min / is finite and \ are limited, ( f is convergent and
/2l0 */$!
f/< f /
/M JVL
36 IMPROPER MULTIPLE INTEGRALS
For by 34
/../ 4'
are limited monotone functions. Their limits exist by I, 277, 8.
36. If M = Max / is finite, and \ f is convergent, the correspond
ing upper integral is convergent and
where f >; a in 2L a .
Similarly, if w= Min/ is finite and I f is convergent, the corre
sponding lower integral is convergent and
Let us prove the first half of the theorem.
We have * *
I /= lim ( .
*/2l a=J8J a
Now C C C
U<J <L <^3la
*L2l ^_ a ^21a
We have now only to pass to the limit.
37, ffyf w convergent, and 53 < 81,
/
does not need to converge. Similarly
does not need to converge, although \ f does.
Example. Let 21 be the unit square ; let S3 denote the points
for which x is rational.
GENERAL THEORY 37
e /as 1 when x is irrational
=  when x is rational
y
Then r
A/ 1 ;
On the other hand,
Hence x /
I = lim I =lim Iogy8= foo
*/$ 0. ^p
is divergent.
38. 1. In the future it will be convenient to let ^J denote the
points of 21 where /> 0, and 9Z the points where/ <. 0. We may
call them the positive and negative components of 21.
2. If\f converges, so do I f.
If \ f converges, so do \ f.
For let us effect a cubical division of space of norm d. Let
ft f > . Let e denote those cells containing a point of ^ ; e r
those cells containing a point of ^ but no point of ^ ; S those
cells containing a point of 2l a/3 but none of typ,.
Then .
f = limjSJf.  e + SJtf"., *' + 2JK,  S{.
/ d=o
Obviously lf.>M. ,
Hence ~ ~
\  f =li
^a' / a <*
38 IMPROPER MULTIPLE INTEGRALS
We find similarly
f  f Um{2(JT t Jf.>
*/^/ */fp 5 <*=:<)
Now
for a sufficiently large a, and for any & ' > .
Hence the same is true of the left side of 1>
As corollaries we have :
3. If the upper integral of f is convergent in 31, then
If the lower integral off is convergent in 31,
f < f < f etc.
J p pJyfi "/v
4. Iff> 0wrf I /i convergent, so is
*/$(
J/ ' <
Moreover the second integral is < the first.
This follows at once from 3, as 31 = ^.
39. If J f and (f converge, so do J /.
We show thatj /converges; a similar proof holds forj . To
this end we have only to show that
>0; , /3>0;
; a< '
GENERAL THEORY
3?
Let D be a cubical division of space of norm d. Let ^ , %r
denote cells containing at least one point of 2l a '0 f , 2l a '^ M at which
/>0. Let n a ' > tla" denote cells containing only points of St a '/3'
2l a 0" at which/< 0. We have
9U"/3"
Subtracting,
(2
Let Jf[ == Max / for points of S J? in d,. Then since / has one
sign in 9t,
ISJtfftZJtW, <2JfJ(l t 2af!dJ. (3
"a' "a" "~ 9^a' ^a"
Letting d = 0, 2) and 3) give
Now if /3 is taken sufficiently large, the first term on the right is
< e/2. On the other hand, since I / is convergent, so is I / by
J$t ^W
36. Hence for a sufficiently large, the last term on the right is
<6/2. Thus 4) gives 1).
40. Iff is intec/rable in 21, it is in any 53 < 21.
Let us first show it is integrable in any 2t a/ g.
Let
Let D be a cubical division of space of norm d.
Then A aft = lim 2o> t d t , o> t = Osc/ in d..
*0 %afi
Let ' > , ff > 0. Then
X'^' ~ A^ = Km {2(^(
40 IMPROPER MULTIPLE INTEGRALS
Now any cell d t of 2l a/3 is a cell of 2t a ' /8 ', and in d^ &[ > ( t .
Hence A a >p > A aft . Thus A^ is a monotone increasing function
of a, . On the other hand
lim A a p =s 0,
by hypothesis. Hence A a p= and thus/ is integrable in 2l a/3 .
Next let / be limited in , then /<some 7 in 48. Then
< 3l y , y But / being integrable in 2l y , y , it is in $8 by I, 700, 3.
Let us now consider the general case. Since/ is integrable in 21
both converge by 38. Let now P, N be the points of $, 91 lying
in. Then
both converge. Hence by 39,
both converge. But if $8 a ^ b denote the points of SS at which
a<f<b,
f/= lim f /
/S , &=>*/5B a6
by definition.
But as just seen,
and /is integrable in $8.
41. As a corollary of 40 we have :
1 . Iff is integrable in 21, it admits a proper integral in any part
o/2l in which f is limited.
2. Iff is integrable in any part of 21 in which f is limited, and if
either the lower or upper integral off in 21 is convergent, f is Integra
ble in 21.
GENERAL THEORY 41
For let
7f ~
J/limJ/
^a a^oo^SU
exist. Since
necessarily
exists and 1), 2) are equal.
42. 1. In studying the function/ it is sometimes convenient to
introduce two auxiliary functions defined as follows :
#=/ where/ >0,
= where /<0.
A=/ where/<0,
= where />0.
Thus g, h are both > and
We call them the associated nonnegative functions.
2. As usual let 3l a/3 denote the points of 21 at which ;</^$.
Let Sip denote the points where g</3, and 2l a the points where h<a.
Then
f A = lim f A. (2
^sr a^aw^a^
For
by5 ' 4 '
Letting , y8 = oo, this last gives 1).
A similar demonstration establishes 2).
3. We cannot say always
X^aslim f g ; fA = lim f
aooi si 8oo^
_*
as the following example shows.
42 IMPROPER MULTIPLE INTEGRALS
Let /= 1 at the irrational points in 21 = (0, 1),
s w, for # = in 21.
n
Then / /
= , J <? = 1.
X
Again let / = 1 for the irrational points in 21,
= n for the rational points x =
w
Then
/=!.
!t .0
43. 1.
; (2
3) JTA= r/, r/.<  r/, (4
' / ^ ^l ~ J W
provided the integral on either side of the equations converges, or
provided the integrals on the right aide of the inequalities converge.
Let us prove 1); the others are similarly established. Effecting
a cubical division of space of norm d, we have for a iixed /3,
f ^lim
^8 </'>
= limS^f t rf t = f /. (5
*= Va J ^
Thus if either integral in 1) is convergent, the passage to the
limit /3 = oo in 5), gives 1).
2. If \ fis convergent, \ g converge.
If I f is convergent^ I h converge.
*/^( /5(
This follows from 1 and from 88.
GENERAL THEORY 43
3. If I / is convergent, we cannot say that I / is always con
/$r Aft
imilar remark holds for the lower integra
/ = 1 at the rational points of 21 = (0, 1)
Then
vergent. A similar remark holds for the lower integral.
For let
J = A
= at the irrational points.
x
4. That the inequality sign in 2) or 4) may be necessary is
shown thus :
Let j
/== for rational x in 21 = (0, 1)
\Jx
= for irrational x.
Then r r
J^ = , J/=2.
44. 1. ff^fg lim f h, (1
j/ = lim f 9fy ( 2
provided, 1 the integral on the left exists, or 2 the integral and the
limit on the right exist.
For let us effect a cubical division of norm d. The cells con
taining points of 21 fall into two classes :
a) those in which /is always <0,
6) those in which /is >0 for at least one point.
In the cells a), since /= g A,
Max/ = Max (g  A) = Max g  Min A, (3
as Max g = 0. In the cells 6) this relation also holds as Min A = 0.
Thus 3) gives  
f=] a\ * (^
*/9f 'M
44 IMPROPER MULTIPLE INTEGRALS
Let now a, /3 = oo. If the integral on the left of 1) is conver
gent, the integral on the right of 1) is convergent by 43, 2. Hence
the limit on the right of 1) exists. Using now 42, 2, we get 1).
Let us now look at the 2 hypothesis. By 42, 2,
r r
Thus passing to the limit in 4), we get 1).
2. A relation of the type
X'X'X*
does not always hold as the following shows.
ra
Example. Let/ = n at the points x =
92n
ra
,
= 1 at the other points of SI = (0, 1).
Then f/=l f 17 = f h = 0.
J% JK J.X
45. 7/ ' I f is convergent, it is in any unmixed part 53 0/'2l
J%
Let us consider the upper integral first. By 48, 2,
X'
exists. Hence a fortiori,
X'
exists. Since 21 = 53 f is an unmixed division,
f h= f h+ f h.
J%afi !#<# J&ap
Hence I h < ) h.
*^9T
GENERAL THEORY 45
As the limit of the right side exists, that of the left exists also.
From this fact, and because 1) exists,
exists by 44, 1.
A similar demonstration holds for the lower integral over
46. J/2lx, 21 2 " ^mform an unmixed division 0/21, then
f/ = f/+ ... + f /, (1
S& Jl thm
provided the integral on the left exists or all the integrals on the
right exist.
For if 2l m$ a/3 denote the points of 2l a3 in 2l m , we have
 r + ... + r . (2
f J*
Now if the integral on the left of 1) is convergent, the integrals
on the right of 1) all converge by 45. Passing to the limit in 2)
gives 1). On the other hypothesis, the integrals on the right of
1) existing, a passage to the limit in 2) shows that 1) holds in
this case also.
47. If \ f and \ f converge, so does I I/I, and
Jy Jw ^21
f/< /*//"/ (1
*/ *A A
(2
For let ^.3 denote the points of 21 where
0</
Then since , /. ,
I/I =
<f ff +Ch (3
c/9f *s9I
<ffff by 43,1. (4
xflj *s 9ft
Passing to the limit in 3), 4), we get 1), 2).
4(> IMPROPER MULTIPLE INTEGRALS
48. 1. If I \f\ converges^ both I /converge.
J% j/a
For as usual let ty denote the points of 21 where/>0.
Then
is convergent by 38, 3, since J / is convergent.
*^*i
Similarly, 
)(/)= //
An /<
is convergent. The theorem follows now by 39.
2. If I  / converges, so do
*
For by 1,
both converge. The theorem now follows by 43, '2.
3. For
ff (2
V2f
/>//^ ^ converge it is necessary and sufficient that
i convergent.
For if 3) converges, the integrals 2) both converge by 1.
( hi the other hand if both the integrals 2) converge,
converge by 38, 2. Hence 3) converges by 47.
4. Iffis integrable in 31, so is \f\.
For let Aft denote the points of 21 where < / < /3. Then
and the limit on the right exists by 3.
GKNKUAL THKORY 47
But by 41, l,/ is integrable in A ft . Hence / is integrable in
A ft by I, 720. Thus
49. From the above it follows that if both integrals
converge, they converge absolutely. Thus, in particular, if
converges, it is absolutely convergent.
We must, however, guard the reader against the error of sup
posing that only absolutely convergent upper and lower integrals
exist.
Example. At the rational points of 21 =(0, 1) let
At the irrational points let
/< = 
x
Here
(/=oo.
Jy
Thus, / admits an upper, but not a lower integral. On the
other hand the upper integral of / does not converge absolutely.
For obviously
50. We have just noted that if
f /(*!
J .
is convergent, it is absolutely convergent. For m = 1, this result
apparently stands in contradiction with the theory developed in
Vol. I, where we often dealt with convergent integrals which do
not converge absolutely.
48 IMPROPER MULTIPLE INTEGRALS
Let us consider, for example,
sin
If we set x = , we get
u
U
which converges by I, 667, but is not absolutely convergent by
I, 646.
This apparent discrepancy at once disappears when we observe
that according to the definition laid down in Vol. I,
J = R lim I fdx,
a=0 J a
while in the present chapter
7= lim I fdx.
a, /3 = ao ^a/
Now it is easy to see that, taking a large at pleasure but fixed,
I fdx ==oo as fi = QO,
^a/3
so that 7does not converge according to our present definition.
In the theory of integration as ordinarily developed in works
on the calculus a similar phenomenon occurs, viz. only absolutely
convergent integrals exist when m > 1.
51. 1. If J / is convergent,
Jyn
Iff < f I/I (1
\J% J%
For 9l tt/3 denoting as usual the points of 21 where </<y8
we have ' ~
i/.
.
Passing to the limit, we get 1).
GENERAL THEORY 49
2. If I / 1 is convergent, ( f are convergent for any JB<91.
*^2l Ji33
For j I/ 1 is convergent by 38, 4.
*/<B
Hence 
S*
~sd
converge by 48, 3.
3. If, 1, f / 1 is convergent and Minf is finite, or if, 2, I f is
J% 5/K
convergent and Max / i finite, then
zs convergent.
This follows by 36 and 48, 3.
52. ie/>0 in 21. Xe^ A0 integral
converge. If
then for any unmixed part 33 < 91,
f/=f/+"', (2
./$ */%
<<*'< (3
For let 21 = + S. Then ^=SS ft + ^ f is an unmixed division,
Also
f + by 1)
^
f + f +.
j^ e
f><) IMPROPER MUI/riPLK INTEGRALS
Hence
u . (4
j93 & *$0 :<$,fi
From 2)
/*
= {//! by 4)
l/( J/(B '
which establishes 3).
53. If the, integral \ \ f \
8, then
> 0, a > 0,
for any 33 < 31 situh tfi
< (2
Let us suppose first that/>0. If the theorem is not true,
there exists, however small o>0 is taken, a 33 satisfying 3) such
that
Then there exists a cubical division of space such that those
points of 31, call them (5, which lie in cells containing a point of
3J, are such that (5<cr also. Moreover S is an unmixed part of 21.
Then from 4) follows, as/>0, that
f f>6 (5
c/(" V
also.
Let us now take ft so that
%/r = L +
Then
and 0<'<a
by 52. But f
1<#S,<*
GENERAL THEORY 51
Let now @<r < e, then
which contradicts 5).
Let us now make no restrictions on the sign off. We have
But since 1) converges, the present case is reduced to the pre
ceding.
54. 1 . Let I f I converge
J<% '
Let as usual 3( a8 denote the points of 21 at which # </ < /3. Let
A ab be such that each 2l a /3 lie* in some A ab in which latter f is limited.
Let 3X/3 = A ab 2la0 and let a, b ~ ao with a, f$. Then
lim IXp = 0.
a, 0=n
For if not, let
Inn > a = /, I > 0.
a, 0= x,
Then for any < X < /, there exists a monotone sequence Ja n , /3 n 
such that
)a n j3 n > ^ f r w > some m.
Let ^n=Min(a n , &), then / > ^ n in S) an ^ n , and /i n =oo.
Hence r
J
^an
But !) an p n being a part of
by 38, 3. This contradicts 1).
2. Definition. We say ^4 fli b is conjugate to 2l a /s with respect to/.
55. 1. J.8 ?/*maZ Zef a<f<j3in 3l aj3 . Ze*
Z^^ ^4^ 6^ conjugate to 2l a/3 w;zYA reference to f; and A b conjugate
to 2(0 with respect to /.
52 IMPROPER MULTIPLE INTEGRALS
If, 1,
or if, 2,
lim J
=00 ^
77 T?
lim I /= I /
a, 6=00 *Z^4 a, 6 ^1^1
For, if 2 holds, 1 holds also, since
Thus case 2 is reduced to 1. Let then the 1 limit exist.
We have
as 4) in 44, l shows. Let now
3X 3 =
Then,
L 9<\ 4 ff<L 9+ I 9
J *o# JA >> ^ *^>/S
(3
But 5DB = as , /3 = oo, by 54. Let us now pass to the limit
, /8 = oo in 3). Since the limit of the last term is by 53, 54, we
get ^
lim f # = lim J g. (4
a, /3co ^St^ 0,6 = 00*^^0,6
Similarly, * ~
lim I A = lim I A. (5
a, /3GO il^a/3 a, 6=< *L A ab
Passing to the limit in 2), we get, using 4), 5),
f /= lim ( f g f h }
J * ,* \ J *< *** f
 lim
a> ;>nao
In a similar manner we may establish 1) for the lower integrals.
GENERAL THEORY 53
2. The following example is instructive as showing that when
the conditions imposed in 1 are not fulfilled, the relation 1) may
not hold.
Example. Since / a ?
I =+oo,
/o x
there exists, for any b n > 0, a < b n+1 < 6 n , such that if we set
then n ^ n ^ ^
Cr 1 <Cr 2 < =00,
as b n = 0. Let now
/ = 1 for the rational points in 21 = (0, 1),
=  for the irrational.
x
Then
Let
Let ^4 n denote the points of 21 in (6 n , 1) and the irrational points
in (6 n +r *)
Then J[>^+.
ii^n
But obviously the set A H is conjugate to 210. On the other hand,
while r
lim I
noo i^4n
56. Jf ^Ae integral
converges, then
<r>
(2
for any unmixed part SB of 21
54 IMPROPER MULTIPLE INTEGRALS
Let us establish the theorem for the upper integral; similar
reasoning may be used for the lower. Since 1) is convergent,
fff (3
JK
and \ = lim f h (4
a,0r*/H aj8
exist by 44, l. Since 3) exists, we have by 53,
<' (5
for any 53 < 21 such that 33 < some <r'.
Since 4) exists, there exists a pair of values a, b such that
4'
since the integral on the right side of 4) is a monotone increasing
function of a, b.
Since 31 = 93 f S is an unmixed division of 31,
XC C
h== J k + J<. ^
Since h > 0, and the limit 4) exists, the above shows that
ft= lim I h , i>= lim I 7t
a, /3=ao /SB a a, /3=^ ^afl
exist and that
Then a, 6 being the same as in 6),
/i= f A + y, (8
^Be>
and we show that
O^V<iy (9
as in 52. Let now c > a, 6 ; then
if we take g <.
GENERAL THEORY 55
Thu8 ' <. (11
by 44, 1. Thus 2) follows on using 5), 11) and taking <r <<r', a".
57. If the integral I / converges and $Q U is an unmixed part of
J.W
2i such that $Q U = 31 as u = 0, then
lim f /= f/. (1
w=oc/ tt *?
For if we set 31 = 33,, f S M , the last set is an unmixed part of 91
and S M = 0. Now
Js M
Passing to the limit, we get 1) on using 56.
58. 1. Let ^
//, 1, Ae u/>/?t?r contents of
== 8 a, =ao ,
^/, 2, A0 upper integrals off, g,f + g are convergent, then
If\ holds, and if, 3, the lower integrals off, g,f\g are conver
gent, then
(3
Let us prove 2); the relation 3) is similarly established. Let
Z> a . be a cubical division of space. Let S a denote the points of
'Dap lying in cells of D^, containing no point of the sets 1). Let
56 IMPROPER MULTIPLE INTEGRALS
Then D aft may be chosen so that g a/3 = 0.
Now 7? C
J*fi a/ ""^aP
since the fields are unmixed. By 56, the second integral on the
right = as a, ft =00 . Hence
lim f /= lim f f.
a, //, a/3 a, /3=cc '<g a/8
Similar reasoning applies to # and/f 9
Again,
Thus, letting a, /3=ao we get 2).
2. TFAen ^A singular points of f, g are discrete, the condition 1
holds.
3. If g is integrable and the conditions 1, 2, 3 are satisfied,
4. If f> g are integrable and condition 1 is satisfied^ f + g is i
tegrable and
5.
provided the integral off in question converges or is definitely infinite.
For
Also _ _
lim 3X0 = lim ^
where 2l a/J refers to/.
6. When condition 1 is not satisfied, the relations 2) or 3)
may not hold.
GENERAL THEORY 57
Example. Let 91 consist of the rational points in (0, 1).
Let
at the point x = . Then
n
Now
/=! + n , g = 1 n
Then
/ + g == 2 in 91.
embrace only a finite number of points for a given a, {3. On the
other hand,
21,^ = 91 for/3>2.
Thus the upper content of the last set in 1) does not == as
, f3 = oo and condition 1 is not fulfilled. Also relation 2) does
not hold in this case. For
59. Ifc>Q, thenj^cfef^f, (1
, (2
provided the integral on either side is convergent.
For
f ifc>0 (3
/ if<?<0. (4
c/,a^
Let c > 0. Since
n
therefore ^
in this set.
Hence any point of 9l f/ a/3 , is a point of 51 A ?. ? and conversely.
Thus **=, s.fi whenc>0.
58 IMPROPER MULTIPLE INTEGRALS
Similarly ^ = a , . whenc<0 .
6 C
Thus 3), 4) give
f < 0.
^ * J
c' c
We now need only to pass to the limit a, /3 = oo
60. Let one of the integrals
converge. Iff = g, except at a discrete set !j) in 31, both integrals
converge and are equal. A similar theorem holds for the lower
integrals.
For let us suppose the first integral in 1) converges. Let
_
l/= }/+)/= )/. (2
^81 ^y) ^ ^,1
stt' 7 = Um Jar 9 = liln J ^
a a,p=
then
Now
/, a/3
Thus the second integral in 1) converges, and 2), 3) show that
the integrals in 1) are equal.
61. 1. Let f /, fff (1
fa ^H
converge. Letf>^g except possibly at a discrete set. Let
$.* = /><,, #,,,) ; f = a/,.pS>i ; 9^ = ^,^^
If 
f^ = 0, 9 ap =0, as , ^= oo,
then 
I /'> I y ( 2
/a y  /a f/ v
RELATION BETWEEN THE INTEGRALS OF TYPES I, II, III 59
For let @ aj8 be defined as in 58, l. Then
Let , /9 = QO, we get 2) by the same style of reasoning as in
58.
2. If the integrals 1) converge, and their singular points are dis
crete, the relation 2) holds.
This follows by 58, 2.
8. If the conditions of 1 do not hold, the relation 2) may not
be true.
Example. Let 21 denote the rational points in (0*, 1*). Let
f=n at 3 =  in 21.
Then f>y in 21.
But
r r
l/=0 I
/a %/H*
Relation between the Integrals of Type* /, //, ///
62. Let us denote these integrals over the limited field 21 by
<? , Vk , P
respectively. The upper and lower integrals may be denoted by
putting a dash above and below them. When no ambiguity arises,
we may omit the subscript 21. The singular points of the inte
grand/, we denote as usual by $.
63. Tf one of the integrals P is convergent, and Q is discrete, the
corresponding O integral Converges, and both are equal.
L^ _ _
PH = Pa a + Aty using the notation of 28,
= <?,+ Fj.
Now = as 8 = by 56.
60 IMPROPER MULTIPLE INTEGRALS
Hence
50
= Cki by definition.
64. If O is convergent, we cannot say that P converges. A
similar remark holds for the lower integrals.
Example. For the rational points in 31 = (0, 1) let
f (<r^
J \f)
for the irrational points let
Then
On the other hand, _ 
P H = lim j /
a^ft/K a/3
does not exist. For however large y8 is taken and then fixed,
I f^= oo as = 00.
J **p
65. If C is absolutely convergent and $ & discrete, then both P
converge and are equal to the corresponding C integrals.
For let D be any complete division of 21 of norm S. Then
J*afi A aM A' aM
using the notation of 28. Now since
Ck / converges, C%' 8 1/ = as S = 0.
But
(2
Again, D being fixed, if are sufficiently large,
f f^C^f > , /S>/3 .
^6
RELATION BETWEEN THE INTEGRALS OF TYPES I, II, III 61
Hence 1), 2) give
f /= CK* + *' K < 1 for any 8 < some S .
^a3 *
On the other hand, if S is sufficiently small,
tf=tfa s + e" "<! forS<S .
Hence f /ff + e'" e"'<.
X
Passing to the limit a, /3 = oo, we get
66. .// P r si/ is absolutely convergent, the singular points Q are
discrete.
For suppose $ > 0 ^ et 33 denote the points of 21 where
>yS. Then $8 > <J for any ^8. Hence
as /8 == oo unless ^ = 0.
67. Jf F^/ is absolutely convergent, so is O.
For let D be a cubical division of space of norm d.
Then
/ 1 < some /8 in 2l d .
Hence Jj/S/J/, < r./.
Hence (7 is absolutely convergent.
68. Letf>Q. If V%f is convergent, there exists for each >0,
a a >
/or any _
. (2
<]2 IMPROPER MULTIPLE INTEGRALS
For
for X sufficiently large. Let X be so taken, then
Al8 ' X/A<XS<, (4
if a is taken sufficiently small in 2).
From 3), 4) follows 1).
69. If V<&f is absolutely convergent, both converge and are
equal to the corresponding \ r integrals.
For by 67, O is absolutely convergent. Hence C converge by 65.
Thus ~ C e
OW/= I f 4 a , la < for some d.
*/H ( / 8
Also  r
V^f = I ,/A^t f P , \ P \ <  tor some X, JJL.
Hence
Now
But
and 7 <  if d is sufficiently small, and for any X, /A, by 68.
o
Taking a division of space having this norm, we then take X,
so large that
/AM=/ i Sid
Then Q
i) = a p 7,
and hence , ,
 7?  < .
From this and 1) the theorem now follows at once.
ITERATED INTEGRALS 83
Iterated Integrals
70. 1. We consider now the relations which exist between the
integrals
and
(2
where 21 = 93 S lies in a space 9t m , m = p + q, and 83 is a projection
of 21 in the space 9? p .
It is sometimes convenient to denote the last q coordinates of a
point x = (x l x p Xp+i x p+q ) by y l y q . Thus the coordinates
#! x p refer to 93 and y l  y g to (. The section of 21 correspond
ing to the point x in S3 may be denoted by g x when it is desirable
to indicate which of the sections S is meant.
2. Let us set
then the integral 2) is
It is important to note at once that although the integrand / is
defined for each point in 21, the integrand < in 4) may not be.
Example. Let 21 consist of the points (#, y) in the unit square :
n n
Then 31 is discrete. At the point (#, y)'m 21, let
~y
Then f/ = by 32.
c/21
On the other hand ~
for each point of 93. Thus the integrals 2) are not defined.
64 IMPROPER MULTIPLE INTEGRALS
To provide for the case that <f> may not be defined for certain
points of 5) we give the symbol 2) the following definition.
f f/ = Urn f f/, (5
&'' a^oo^Jr
where F = S when the integral 3) is convergent, or in the con
trary case F is such a part of S that
and such that the integral in 6) is numerically as large as 6) will
permit.
Sometimes it is convenient to denote F more specifically by r oj8 .
The points 93 aj8 are the points of S3 at which 6) holds. It will
be noticed that each 93 a/3 in 5) contains all the points of 93 where
the integral 3) is not convergent. Thus
Hence when 93 is complete or metric,
lim 93 a p=. (7
a,/3~co
Before going farther it will aid the reader to consider a few
examples.
71. ^Example 1. Let 31 be as in the example in 70, 2, while/ = ri*
at x = . We see that
n
J>=
On the other hand 93 aj 8 contains but a finite number of points
for any a, ft. Thus
Thus the two integrals 1), 2) exist and are equal.
Example 2. The fact that the integrals in Ex. 1 vanish may
lead the reader to depreciate the value of an example of this kind,
This would be unfortunate, as it is easy to modify the function so
that these integrals do not vanish.
ITERATED INTEGRALS 65
Let 21 denote all the points of the unit square. Let us denote
the discrete point set used in Ex. 1 by >. We define /now as
follows : /shall have in $) the values assigned to it at these points
in Ex. 1. At the other points A = 21 J),/ shall have the value 1.
Then ///*/*
1=1 + 1=1=1. (3
/ JA J JA ^
On the other hand 33 aj3 consists of the irrational points in 93
a finite number of other points. Thus
Hence again the two 3), 4) exist and are equal.
Let us look at the results we get if we use integrals of types I
and II. We will denote them by and V as in 62.
We see at once that
(7 a== ra=P = l.
Let us now calculate the iterated integrals
Cfetffc, (5
and F$ F<. (6
We observe that
C(i = 1 for x irrational
= + QO for x rational.
Thus the integral 5) either is not defined at all since the field
$85 does not exist, or if we interpret the definition as liberally as
possible, its value is 0. In neither case is
Let us now look at the integral 0). We see at once that
does not exist, as V& = 1 for rational x, and = +00 for irrational
x. On the other hand
Hence in this case
f>6 IMPROPER MULTIPLE INTEGRALS
Example 3. Let 31 be the unit square.
Let
f=: n for x = n even
n
= n for # = ~ M odd.
n
At the other points of 31 let/= 1.
Then
Here every point of 31 is a point of infinite discontinuity and
thus ^ = 31.
Here Cfo is not defined, as 31 5 does not exist; or giving the
definition its most liberal interpretation,
The same remarks hold for C^Ots
O V&*
On the other hand Tr
V* = 400,
while v^
does not exist, since rr m
K ff = n tor x =
* ft
= 1 for irrational x.
Moreover T , Tr fr rr ,
58^ =  r 8 r c = + Qo.
Example 4. Let 31 denote the unit square. Let
/=n 2 for z = , neven, 0<y<i
W 71
= n a for ^ = ~ , n odd, < y <  .
~~
At the other points of 31 let/= 1.
Then *
r/=i
^
ITERATED INTEGRALS 67
Let us look at the corresponding C and V integrals.
We see at once that
Again the integral O^Cg does not exist, or on a liberal interpre
tation it has the value 0. Also in this example
do not exist or on a liberal interpretation, they = 0.
Turning to the F' integrals we see that
while V% Fjg does not exist finite or infinite.
Example 5. Let our field of integration 21 consist of the unit
square considered in Ex. 4, let us call it @, and another similar
square gf> lying to its right. Let / be defined over ( as it was
defined in Ex. 4, and let/= 1 in .
Then r r
f/= f f=2
*/2l J&J&
Also __ v __ 9
C 5l  K 2l = ^
Then 1
33S"" A
while V^V^ does not exist,
and Tr Tr 77 fr
72. 1. In the following sections we shall restrict ourselves as
follows :
1 21 shall be limited and iterable with respect to SB.
2 S3 shall be complete or metric.
3 The singular points $ of the integrand /shall be discrete.
2. Let us effect a sequence of superposed cubical divisions of
space
A* JV
whose norms d n = 0.
68 IMPROPER MULTIPLE INTEGRALS
Let 2l n = SS n  S n denote the points of 21 lying in cells of D n
which contain no point of $. We observe that we may always
take without loss of generality
=*.
For let us adjoin to 31 a discrete set ) lying at some distance
from 21 such that the projection of on 9? p is precisely $Q.
Let 4 = 2l + ) = S(7 , <?=<5+c , c = 0.
We now set .  
<p = / in VI
= in 3\
A nen * * 7* 7*
+ 1 ^,= j ^
ife
Similarly
Hence
8. The set S n being as in 2, we shall write
73. Let B^ n denote the points of 33 at which c n > <r. Then if 21
is iterable, with respect to S3,
lim S a n = 0. (1
n=*
For since 21 is iterable,
21 = I S by definition.
*/33
Hence S considered as a function of # is an integrable function
in.
Similarly
x, _
X s "
and S n is an integrable function in $Q.
ITERATED INTEGRALS 69
We have now ~ 7? ,  ~ ^> A
6 = Sn+C n , C n >0
as S n , c n are unmixed. Hence c n is an integrable function in SB*
But
_ r
:cw I Src
9* = L(Sn'
/S
As the left side = as n = oo ,
lim ( c n = 0. (2
*/9^ V
But
As the left side == 0, we have for a given cr
lim B n = 0,
which is 1).
74. ie 21 = 93 fo iterable. Let the integral
ff , />0
convergent and limited in complete 33. S n denote the points
of S3 a which
< 2
lim g n = SB. (3
7i=='
For let . . . A
<r 1 ><7 2 > =0.
Since fft n === as w === oo by 48, we may take i>j so large, and
then a cubical division of 9t p of norm so small that those cells con
taining points of B fflVv have a content <^/2. Let the points of
93 lying in these cells be called B v and let SQj = 93 J? r Then
S v 53j form an unmixed division of 93 and
j is complete since SB is.
70 IMPROPER MULTIPLE INTEGRALS
We may now reason on S8 t as we did on 93, replacing ij/Z by i?/2 2 .
We get a complete set $ 2 .<_$Bi such that
Continuing we get ^ > ,
Thus ^ 
Let now b = Dt;{8 ll j.
Then r, (4
by 25.
Let b n denote those points of b for which 2) does hold. Then
b = 5b n j. For let ft be any point of b. Since 1) is convergent,
there exists a o\ such that
at ft,
for any c such that c <<r t . Thus ft is a point of bi/ t and hence of
jb n j. Thus b n = b as b is complete. But S n > b n .
Hence
which with 4), gives 3).
75. Let 21 = 93 S ie iterable. Let the integral
^j convergent arid limited in complete 93
TA0W. /* 7i
lim j ( /=0. (1
n=oo & ^ n V
For let D be a cubical division of 9? p of norm d.
Then
/ r ^"
(f I /=lim2rf 4 Min I / = lim
1^ /C n (/^Q t ilCn r/~0
Let df[ denote those cells of D containing a point of S n where (5 n
is defined as in 74.
ITERATED INTEGRALS 71
Let d" denote the other cells containing points of 33. Then
where 
0< \f<M.
J&
Hence
4
Letting d = 0, we get
f/<
^SB Cn
Letting now n = oo and using 3) of 74, we get 1), since e is
small at pleasure.
76. Let 21 = 33 &e iterable with respect to 33, w/m?A / is com
or metric. Let the singular points $ of f be discrete. Then
if, f>& , f/< (* f/< f / (1
J ' ^ J& J "*/ . ~ /
if, /<^ , r/<r r/< c/ (2
47 t/ "  */2l' J<%Jss. J Jyi J v
any on o/ /if? members in 1) may ft^ infinite. Then all
that follow are also infinite. A similar remark applies to 2).
Let us first suppose :
/> , 53 is complete , 1 1 / is convergent.
We have by 14, ~ ~ ~. *
I /< I I /< I /.
*. ""iZr</ "~ ^n
Passing to the limit gives
r/<limf f /. (3
Jsr ~ J&J&n J
and also ^ ^ =
lim I I / <. I / , finite or infinite. (4
JBfe ^a
Now e > being small at pleasure, there exists a # such that
72 IMPROPER MULTIPLE INTEGRALS
But for a fixed n ^
I is limited in 93.
/ (
Hence for Gr sufficiently large,
f f< f/ , at each point of S3, 6? < 6^. (6
*cSw ^r
Then
where F n , y n are points ot F in ( n , c w .
Hence
J$(t*Ln
Now S3^ ; may not be complete ; if not let B tt be completed 93#.
As S3 is complete,
n ~JBoJ^ '
We may therefore write 8), using 5)
+//V /+/ f^ff+S f
J%J& /,/^ii "LBffJy* ^*/n JUff^Vn
By 75, the last term 011 the right = as 71 = oo. Thus passing
to the limit,
n/<lim f f/, (9
__/=/& V
since > is small at pleasure.
On the other hand, passing to the limit Q = oo in 7), and then
w=oo, we get
lira f f < f f . (10
^=00 /.Z(S ?/ V
Thus 3), 10), 9), and 4) give 1).
M$ now suppose that the middle term of 1) is divergent.
We have as before
f f<]mif f < f/.
i/a < ; ^/r n^^ ^ ^^
Hence the integral on the right of 1) is divergent.
ITERATED INTEGRALS 73
Let us now suppose 93 is metric. We effect a cubical division
of 9tp of norm d, and denote by B d those cells containing only
points of 93. Then B d is complete and
Let A d denote those points of 91 whose projections fall on JS d .
Then A d is iterable with respect to B d by 13, 3, and we have as
i Y fVi^i i~iTir<iri i M rp r>ocjo
in the preceding case
If the middle integral in 11) is divergent, j is divergent and 1)
holds, also if the last integral in 11) is divergent, 1) holds. Sup
pose then that the two last integrals in 11) are convergent.
Then by 57
limf f=f f.
d=0*//W< //(
limf = f.
d=^Aa ^21
Thus passing to the limit d = in 11) we get 1).
Let us now suppose f > (?, Q > 0.
Then
and we can apply 1) to the new function g.
Thus
ffl= f fj/< (V (12
?3i <33 J& *'yi
Now
by 58, 6, since $ is discrete. Also by the same theorem,
C g = Cf + a iim s y = f / + or, (14
/S /^ y  /g
denoting by S y the points of S where
and setting
T = lim <
y==X
74 IMPROPER MULTIPLE INTEGRALS
Now for any n
Hence #H = lim C f # = <?lim fg n ,
x /33 c/g^ noo 4/93
or (15
Now for a fixed /&, 7 may be taken so large that for all points
of S,
Hence _
6 > Urn
yes oo
Hence >
f
33
Hence (16
and thus F is integrable in S3.
This result in 14) gives, on using 58, 3,
c r<7= c f
*Z ?Ze 5/ ?^s
From 12), 13), and 17) follows 1).
77. As corollaries of the last theorem we have, supposing 21 to
be as in 76,
1. Iff is integrable in 31 andf> 6r, then
J 21 Jjd J<
2. ?// > ff and  is divergent, then
J%
^ divergent.
ITERATED INTEGRALS 75
3. If f > Gr and one of the integrals I I f is convergent, then
/<*/(
's convergent.
78. Let 21 = S3 S be iterable with respect to 33, which last is com
plete or metric. Let the singular points $ be discrete. If
f/' (1
r IT/, (2
^93 ^(S
>oA converge, they are equal.
For let J>!, D 2 be a sequence of superimposed cubical divisions
is in 72, 2. We may suppose as before that each 93 n = 33
Since 1) is convergent
Since /is limited in 3t n , which latter is iterable,
This shows that
s an integrable function in $8, and hence in any part of S3
From 3), 4) we have
I f  f f <^ n>n<r (6
 Ja J^J&n 2
We wish now to show that
When this is done, 6) and 7) prove the theorem.
To establish 7) we begin by observing that
iim r r.
,/i^^^r
76
IMPROPER MULTIPLE INTEGRALS
Now for a fixed n, , y8 may be taken so that F shall embrace all
the points of < for every point of SB Let us set
r =
Then
(8
As
lira f f = f f by I, 724.
,0v a 3~X *AB*X
On the other hand,
<
i/i<
1  7 ' 
Thus 7) is established when we show that
To this end we note that / is integrable in 31 by 48, 4. Hence
by 77, i,
Also by I, 734,
From 10), 11) we have for n > j
Ar ' Ar ' J<y\*/(s
2
( 12
since the left side == 0.
But as in 8)
Passing to the limit 6?= oo gives
n!f= f f /+ f f I/I.
  J SB J ^ ' JsaJc* '
This in 12) gives 9).
CHAPTER III
SERIES
Preliminary Definitions and Theorems
79. Let }, a<p a%  be an infinite sequence of numbers.
The symbol A = a 1 + <*t + <%+. (1
is called an infinite series. Let
yl n = a 1 + ^+ .. + a n . (2
If lim A H (3
=oo
is finite, we say the series 1) is convergent. If the limit 3) is infi
nite or does not exist, we say 1) is divergent. When 1) is conver
gent, the limit 3) is called the sum of the series. It is customary
to represent a series and its sum by the same letter, when no con
fusion will arise. Whenever practicable we shall adopt the fol
lowing uniform notation. The terms of a series will be designated
by small Roman letters, the series and its sum will be denoted by
the corresponding capital letter. The sum of the first n terms of a
series as A will be denoted by A n . The infinite series formed by
removing the first n terms, as for example,
will be denoted by A n , and will be called the remainder after n
terms.
The series formed by replacing each term of a series by its nu
merical value is called the adjoint series. We shall designate it
by replacing the Roman letters by the corresponding Greek or
German letters. Thus the adjoint of 1) would be denoted by
A = j f 2 4 3 4 = Adj A (5
Where n=K.
77
78 SERIES
If all the terms of of a series are >0, it is identical with its
adjoint.
A sum of p consecutive terms as
 n+p
we denote by A n ^ p .
B=a^ + a^ + a^\  ,
be the series obtained from A by omitting all its terms that vanish.
Then A and B converge or diverge simultaneously, and when conver
gent they have the same sum.
Thus if the limit on either side exists, the limit of the other side
exists and both are equal.
This shows that in an infinite series we may omit its zero terms
without affecting its character or value. We shall suppose this
done unless the contrary is stated.
A series whose terms are all > we shall call a positive term
series; similarly if its terms are all < 0, we call it a negative term
series. If a n > 0, n>m we shall say the series is essentially a pos
itive term series. Similarly if a n < 0, n>m we call it an essen
tially negative term series.
If A is an essentially positive term series and divergent,
lim A n = f <x> ; if it is an essentially negative term series and di
vergent, lim A n = QO.
When lim A tl = oo, we sometimes say A is 00.
80. 1. For A to converge, it is necessary and sufficient that
e>0, m, A n , p <e, n>m, p = l, 2, ... (1
For the necessary and sufficient condition that
lim A n
n=w
exists is A A A ,
> 0, m, \ A v A n \ < e, v, n > m. (2
But if v = n + p
Thus 2) is identical with 1).
PRELIMINARY DEFINITIONS AND THEOREMS 79
2. The two series A, A a converge and diverge simultaneously.
When convergent, __
A = A. + A.. (3
For obviously if either series satisfies theorem 1, the other
must, since the first terms of a series do not enter the relation 1).
On the other hand, A __ A A
^+P == A t { A 9 ^ p .
Letting p==<x> we get 3).
3. If A is convergent, A n = 0.
For lim A n = lira (A  A n )
= A lim A n = A A
= 0.
For A to converge it is necessary that a n == 0.
For in 1) take p = 1 ; it becomes
I +! I < c n > m
We cannot infer conversely because a n = 0, therefore A is con
vergent. For as we shall see in 81, 2,
l + i + + 
is divergent, yet lirn a n = 0.
4. The positive term series A is convergent if A n is limited.
For then lim A n exists by I, 109.
5. A series whose adjoint converges is convergent.
For the adjoint A of A being convergent,
>0, m, A n , p <e, n>m, p =1, 2, 3
But
Thus
and A is convergent.
Definition. A series whose adjoint is convergent is called
absolutely convergent.
80 SERIES
Series which do not converge absolutely may be called, when
necessary to emphasize this fact, simply convergent.
6. Let A = a l f 2 +
be absolutely convergent.
Let B = a tj f a l2 + ; 4 1 <* 2 <
fo any series whose terms are taken from A, preserving their relative
order. Then B is absolutely convergent and
\B\<A.
For  J BJ<B ra <A n <A, (1
choosing n so large that A n contains every term in B m . Moreover
for m > some m 1 , A n B m > some term of A. Thus passing to the
limit in 1), the theorem is proved.
7. Let A a l + a%+ The series B = ka l + ka% +
converges or diverges simultaneously with A. When convergent,
We have now only to pass to the limit.
From this we see that a negative or an essentially negative term
series can be converted into a positive or an essentially positive
term series by multiplying its terms by k= 1.
8. If A is simply convergent, the series B formed of its positive
terms taken in the order they occur in A, and the series C formed of the
negative terms, also taken in the order they occur in A, are both
divergent.
If B and are convergent, so are B, F. Now
A n = B ni 4 r 2 , n = ^ + n r
Hence A would be convergent, which is contrary to hypothesis.
If only one of the series B, is convergent, the relation
shows that A would be divergent, which is contrary to hypothesis.
PRKLIMfXAKY DEFINITIONS AND THRORKMS 81
9. The following theorem often affords a convenient means of
estimating the remainder of an absolutely convergent series.
Let A = a l f a 2 + be an absolutely convergent series. Let
_Z? = 6j f > 2 f be a positive term convergent series ivhose sum is
known either exactly or approximately. Then if \ a n \ < 6 n , n > m
\A n \<B n <.
<B n <B.
Letting p == a> gives the theorem.
EXAMPLES
81. 1. The geometric series is defined by
The geometric series is absolutely convergent when \g\< 1 and di
vergent when </>1. When convergent,
ft 1 (f>
u . 4
.
Hence ,
a  1 _ ff
^^n ^ ^
1^ 1^
When ^< 1, lim^ w = 0, and then
1
lim# w = :
When \g\ >1, lim# w is not 0, and hence by 80, 3, & is not conver
gent.
2. The series   JL^ J^ J_ , /g
82 SERIES
is called the general harmonic series of exponent p. When /* = 1,
it becomes j, 1 + ^ + ^ + { + ... (4
the harmonic series. We show now that
The general harmonic series is convergent when /i > 1 and is di
vergent for /*<!.
Let /*>!. Then
l + l^l+l ^J^ . a< l
2* 3<* 2* 2* 2^ 2'" 1 ' *
1, 1 + 1 + .1 <1  1 , 1 + 1 = 1 = ,2
4,* ^ 5/ot ^ tj,x ^ 7/01 ^ 4/a ^ 4/m ^ 4^ ^ 4^ 4^ ^ '
< + + ... + = = ^3 etc.
15" 8^ 8^ 8^ 8* <J/ '
Let 7i < 2". Then
Thus lim H n exists, by I, 109, arid
1
Let ^<1. Then .. .
Thus 3) is divergent for //,< 1, if it is for /i = 1.
But we saw, I, 141, that
lim J n = GO,
hence 7 is divergent.
It is sometimes useful to know that
In fact, by I, 180,
^" =1. (6
log n
lira ^a = lim "~ ""' = lim
log n log n  log (n  1) i og f n
PRELIMINARY DEFINITIONS AND THEOREMS 83
Since * n > log n > l^n  we have
. (7
n l r n
Another useful relation is
For log(l f w) logm = log( 1 + ~ ) < ~.
\ w/ m
(8
Let w = l, 2>n. If we add the resulting inequalities we
get 8).
3. Alternating Series. This important class of series is defined
as follows. Let a l > # 2 > # 3 > =0.
Then A = a j a 2 + 3 4 f (9
whose signs are alternately positive and negative, is such a series.
The alternating series 9) is convergent and
For let p> 3. We have
A n>f = ( l)]a n+1  a n+2 + . (
(l)P.
If /? is even,
^ = (n^l  fln+ 2 ) +  + (n4p
If jo is odd,
P = ( a n+i  n+ 2 ) + +(a n+p
Thus in both cases,
^><*n+ln+ 2 >0. (11
Again, if p is even,
* In I, 461, the symbol " lim " in the first relation should be replaced by lira.
84 SERIES
If p is odd,
P = a n+ 1 (n+2 ^n+s) (#n+pl ~ #+p)
Thus in both cases,
P < a n+l  (a n+2  a n + 8 ) < n+1 . (12
From 11), 12) we have
< n+l ~ n+2 <  ^, p  < n fl ~ ( n+2  "n+a)
Hence passing to the limit p = oo,
moreover, _._ n
^n M "
Example 1. The series
ii + *i+
being alternating, is convergent. The adjoint series is
which being the harmonic series is divergent. Thus 13) is an
example of a convergent series which is not absolutely convergent.
Example 2. The series
V2  1 V2 + 1 V3  1 V3 + 1
is divergent, although its terms are alternately positive and nega
tive, and a n == 0.
For
2 77?  1,
If now A were convergent,
lira A n = lim .A a
by I, 103, 2.
PRELIMINARY DEFINITIONS AND THEOREMS 85
4. Telescopic Series. Such series are
A = (!  a a ) 4 ( 2  8 ) + (03  a 4 ) +
We note that
A n = O 1 tf 2 )+ + (a n a n41 )
= ! a nfl . (14
Thus the terms of any A n cancelling out in pairs, A n reduces to
only two terms and so shuts up like a telescope.
The relation 14) gives us the theorem :
A telescopic series is convergent when and only when lim a n exists.
A = a l f # 2 + denote any series.
Then a n = A n A n ^ , 4,0.
Hence A = (A l  A ) + (^ 2  AJ + (A 3  AJ + 
This shows us that
Any series can be written as a telescopic series.
This fact, as we shall see, is of great value in studying the
general theory of series.
Example 1.
1 2 2 3 3 4
; l n.
Thus A is a telescopic series and
1
n
Example 2. Let a v # 2 , a 8 , > 0. Then
A *^ a n
A __ > n
^A / , ~ ~ '
1 + !> (!+_,)
is telescopic. Thus
and A is convergent and < 1 .
86 SERIES
Examples. 4=X  \   x^ 0, 1, 2,
* i O + rc
1 # f
is telescopic.
1 11
x x + n x
82. Dints Series. Let A = a l + # 2 + fo a divergent positive
term series. Then
is divergent.
For
2
(^mHl "I" '" "H
m ~ m, p m+p
Letting m remain fixed and jt? = oo, we have D m >l, since
mhp ==oo. Hence D is divergent.
Let 4=1 + 1 + 1+ ... Then A n = n.
Hence j) = i + + i+... is divergent.
Let 4=1 +  + J
Then . 
is divergent, and hence, a fortiori,
But A n _^ > log n.
Hence i j
21og2 31og3
is divergent, as Abel first showed.
PRELIMINARY DEFINITIONS AND THEOREMS 87
83. 1. Abel's Series.
An important class of series have the form '
B = a^j + a^ + a B t B + (1
As Abel first showed how the convergence of certain types'of
these series could be established, they may be appropriately called
in his honor. The reasoning depends on the simple identity
(AbeFs identity),
where as usual A n ^ m is the sum of the first m terms of the re
mainder series A n . From this identity we have at once the fol
lowing cases in which the series 1) converges.
2. Let the series A = a^ + a% f and the series 2f n+1 t n \
converge. Let the t n be limited. Then B = a l t 1 + a 2 t^ f converges.
For since A is convergent, there exists an m such that
\A ntp \<e; n>m, jt? = 1, 2, 3 ...
Hence
i^nJ<e{*n+lW2 + *n+2^+3+  +  *n + p \ } 
3. Let the series A = a^  a 2 f converge. Let t v t^ t^' be a
limited monotone sequence. Then B is convergent.
This is a corollary of 2.
4. Let A = a l + # 2 f 5e 3M<?A ^Aa^ \A n \ < Gr, n = 1, 2,
2 1 n+1 ^ n  converge and t n = 0. 2%e^ JS is convergent.
For by hypothesis there exists an w such that
' ^+l*+2 + *n+2*iM.3+ "+IWl<
for any n > m.
5. Let \A n \<& and t^ > 2 > f 3 > =0. TAen 5 i convergent.
This is a special case of 4.
88 SERIES
6. As an application of 5 we see the alternating series
is convergent. For as the A series we may take J. = 1 1 + 1
1+ ... as \A n \<l.
84. Trigonometric Series.
Series of this type are
= # + a l cos x + a 2 cos 2 x + # 8 cos 3 a? + (1
$ = flj sin x + a 2 sin 2 x f # 3 sin 3 x + (2
As we see, they are special cases of Abel's series. Special cases
of the series 1), 2) are
F = , + cos x + cos 2 x f cos 3 x + (3
2 = sin x 4 sin 2 # + sin 3 x + (4
It is easy to find the sums F n , 2 n as follows. We have
. . , 2ral 2m + l
2 sin mx sin x cos  # cos   x.
Letting m = 1, 2, w and adding, we get
2 sin x 2 n = cos \ x cos n #. (5
^
Keeping # fixed and letting w = oo, we see 2 n oscillates between
fixed limits when x^ 0, 2 TT,
Thus 2 is divergent except when x= 0, TT,
Similarly we find when x 3= 2
r  s
*1 v
sin \x
Hence for such values T n oscillates between fixed limits. For
the values x = 2 mir the equation 3) shows that T w = + oo.
From the theorems 4, 5 we have at once now
If 2 1 a n+l a n  converges and a n = 0, and hence in particular if
a i> a 2 '" ^> t'h e series 1) converges for every x, and 2) converges
for x^2 WITT.
If in 3) we replace x by x 4 TT, it goes over into
A = cos x f cos 2 # cos 3 x f (7
PRELIMINARY DEFINITIONS AND THEOREMS 89
Thus AH oscillates between fixed limits if x^ (2 m I)TT,
when n ^ oo . Thus
J/ 2 1 +!  a n converges and a n = 0, and hence in particular if
a l > # 2 >  =0, the series a a l cos # f a 2 cos 2 x a% cos 3 x f
converge* for # = (2 m 1) TT.
85. Power Series.
An extremely important class of series are those of the type
P = a + ^(z a) + a^(x a) 2 f a 3 (>a) 3 + (1
called power series. Since P reduces to # if we set x = #, we see
that every power series converges for at least one point. On the
other hand, there are power series which converge at but one
point, e.g.
a + l](x a ) + 2\(xa)* + Xl(za)* +  (2
For if x*f* a, lim n\ x a \ n = oo, and thus 2) is divergent.
1 . If the power series P converges for z=b, it converges absolutely
within n , .
AGO > X=ai.
Jf jP diverges for x = b, it diverges without J9 A (a).
Let us suppose first that P converges at b. Let # be a point in
Z> A , and set  x a \ =f. Then the adjoint of P becomes for this
u L X
lim n X n = 0,
since series P is convergent for x = b.
Hence ^ n ^ i .
and II is convergent. X
If P diverges at x = 6, it must diverge for all V such that
 J' a \ > X. For if not, P would converge at b by what we have
just proved, and this contradicts the hypothesis.
90 SERIES
2. Thus we conclude that the set of points for which P con
verges form an interval (a p, a + p) about the point a, called
the interval of convergence ; p is called its norm. We say P is
developed about the point a. When a = 0, the series 1) takes on
the simpler form + ^ + 0,2? +
which for many purposes is just as general as 1). We shall
therefore employ it to simplify our equations.
We note that the geometric series is a simple case of a power
series.
86. Cauchy's Theorem on the Interval of Convergence.
The norm p of the interval of convergence of the power series,
P = a Q + a l x + a t2 x* +
is given by i
 = hmVa n ^n #nr
p
We show II diverges if >/>. For let
p
Then by I, 338, l, there exist an infinity of indices i^ t a for
which
Hence
and thus ,
since /3>l. Hence y p n
n
is divergent and therefore H.
We show now that II converges if < p. For let
Then there exist only a finite number of indices for which
Let m be the greatest of these indices. Then
V n </3 n>m.
TESTS OF CONVERGENCE FOR POSITIVE TERM SERIES 91
Hence
and
Thus
and IT is convergent.
Example 1. 2 .
1 4 4 4 4 ...
1! 2! 31
Here vX^L^O by I, 185, 4.
vn!
Hence /? = oo and the series converges absolutely for every x.
Example 2. x a? tf>
Here ^ = ^ = 1 by I, 185, 8.
Vn
Hence p= 1, and the series converges absolutely for  x < 1.
Tests of Convergence for Positive Term Series
87. To determine whether a given positive term series
is convergent or not, we may compare it with certain standard
series whose convergence or divergence is known. Such com
parisons enable us also to establish criteria of convergence of
great usefulness.
We begin by noting the following theorem which sometimes
proves useful.
1. Let A, B be two series which differ only by a finite number of
terms. Then they converge or diverge simultaneously.
This follows at once from 80, 2. Hence if a series A whose
convergence is under investigation has a certain property only
02 SERIES
after the wth term, we may replace A by A m , which has this
property from the start.
2. The fundamental theorem of comparison is the following :
Let 4 = a x 4 </ 2 4 > B = &! 4 J 2 4 i &00 positive term series.
Let r > denote a constant. If a n < rb n , A converges if B does and
A < rB. If a n > rb n , A diverges if B does.
For on the first hypothesis
A n <rB n .
On the second hypothesis
A>rB n .
The theorem follows on passing to the limit.
3. From 2 we have at once :
Let A = a l 4 # 2 + '" ^ == ^i "^ ^2 "^ "" ^ e ^ wo p s ^i ve term series.
Let r, s be positive constants. If
r<~<i* ra= 1, 2,
or if
lim ~
exists and is * 0, A and B converge or diverge simultaneously. If
B converges and == 0, A also converges. If B diverges and ^ =00,
A also diverges.
4. Let A = a x 4 ^ 2 "^" '"* B = b l \ b 2 + i^ positive term series.
If B is convergent and
A converges. If B is divergent and
a n+\ ^ g_nf I
j^, * i
rt n ft n
yl diverges.
For on the first hypothesis
TESTS OF CONVERGENCE FOR POSITIVE TERM SERIES 93
We may, therefore, apply 3. On the second hypothesis, we
have
and we may again apply 3.
Example 1. A = ^ + ^ + ^f ~
is convergent. For
tf!SS fi.n + l < n
and V is convergent. The series JL was considered in 81, 4, Ex. 1.
^w 2
Example 2. A== e~ x cos $ + e~ 2x cos 2 # +
is absolutely convergent for x > 0.
For la <
which is thus < the nth term in the convergent geometric series
Example 3. A = ]P  log ^t
is convergent.
Thus A is comparable with the convergent series ]~V
^< n 2
88. We proceed now to deduce various tests for convergence
and divergence. One of the simplest is the following, obtained
by comparison with the hyperharmonic series.
Let A = ! + a 2 f be a positive term series. It is convergent if
lira a n n* < oo , ^ > 1,
and divergent if
lim na n > 0.
94 SERIES
For on the first hypothesis there exists, by I, 338, a constant
G > such that
a< (? nssl g ...
Thus each term of A is less than the corresponding term of the
^^A 1
convergent series Gr2^
On the second hypothesis there exists a constant c such that
, o
#n> n = 1, z,
and each term of A is greater than the corresponding term of the
divergent series c V  .
** n
Example 1. A = V m >0.
^log m n
Here ^ = ^ = 400, by I, 463.
log m n
Hence A is divergent.
Example 2. A = V  .
^ n log n
Here
.
log n
Thus the theorem does not apply. The series is divergent
by 82.
Example 3.
=2Z n = 21og(l + +
\ n n r
where /A is a constant and  6 n \ < G.
From I, 413, we have, setting r = 1 + a,
Hence nl n = fJL , if /A ^= 0,
TESTS OF CONVERGENCE FOR POSITIVE TERM SERIES 95
and L is divergent. If /* > 0, L is an essentially positive term
series. Hence = + <x>. If /A < 0, L~ oo.
Let /i = 0. Then
which is comparable with the convergent series
T r I
^f n r
Thus L is convergent in this case.
Example 4 The harmonic series
is divergent. For li m = !.
Example 5. <
4 = V : yS arbitrary.
^ ft a loer ft
Ail a
w  = oo , a<l
Here
by I, 463, l. Hence A is divergent for a< 1.
Example 6.
Here I
na n = ^ = 1 by I, 185, Ex. 3.
Example 7.
log ft
Here, if /* > 0,
_
log 71
(l\ n
1 + ] s .
n)
Hence A is divergent.
96 SERIES
89. D'Alemberfs Test. The positive term series A = a l + a% f
converges if there exists a constant r < 1 for which
It diverges if
Tins follows from 87, 4, taking for B the geometric series
1+ r +
Corollary. Let ?*l==Z. //' ?<1, .A converges. If Z>1, i
Example 1. The Exponential Series.
Let us find for what values of x the series
is convergent. Applying D'Alembert's test to its adjoint, we find
x n n ^ [
a n
r n\
Thus ^ converges absolutely for every x.
Let us employ 80, 9 to estimate the remainder E n . Let x >0.
The terms of J? are all > 0. Since
(w+/>)! n\ n h 1 n + 2 n+ p n!\nfV
we have _ _ __
(2
However large x may be, we may take n so large that x<n + 1.
Then the series on the right of 2) is a convergent geometric series.
Let #<0. Then however large \x\ is, JE n is alternating for
some m. Hence by 81, 3 for w>m,
TESTS OF CONVERGENCE FOR POSITIVE TERM SERIES 97
Example 2. The Logarithmic Series.
Let us find for what values of x the series
is convergent. The adjoint gives
a n+l\ = n . I 3,1^1 x ^
a n I n H" 1
Thus i converges absolutely for any #<1, and diverges lor
When x = 1, L becomes
which is simply convergent by 81, 4.
When x = 1, L becomes
which is the divergent harmonic series.
Examples. ^ = ~+ ~ + ~+ 
As A is convergent when /*>! and divergent if /n.<l, we see
that D'Alembert's test gives us no information when I = 1. It is,
however, convergent for this case by 81, 2.
Example 4*
ao . f
vp n.
^r _. _
Here ..
a a nflf^
and D'Alembert's test does not apply.
Example 5.
A = 2n M ^.
Here
98 SERIES
Thus A converges for #<1 and diverges for \x\ > 1. For
 x  = 1 the test does not apply. For x = 1 we know by 81, 2
that A is convergent for JJL < 1, and is divergent for p > 1.
For x = 1, A is divergent for /i > 0, since a n does not = 0. A
is an alternating series for JJL < 0, and is then convergent.
90. Cauchy's Radical Test. Let A = a l + # 2 f be a positive
term series. If there exists a constant r < 1 such that
ya~ n <r n=l, 2, ^
A is convergent. If, on the other hand,
v^>i
A is divergent.
For on the first hypothesis,
a n <r n
so that each term of A is <; the corresponding term in
r {. 7.2 __ r s _j_ ... a convergent geometric series. On the second
hypothesis, this geometric series is divergent and a n > r n .
Corollary. If lim Va n = /, and I < 1, A is convergent. Ifl>\,
A is divergent.
Example 1. The series
*4 log n n
is convergent. For
n/ 1
n
logn
Example 2.
. f\
= 0.
1 .
""
is convergent. For
Example 8. In the elliptic functions we have to consider series
of the type
0(t>) = 1 + 2 Sj*' cos 2 Trnv < q < 1.
TESTS OF CONVERGENCE FOR POSITIVE TERM SERIES 99
This series converges absolutely if
? + ? 4 + 9 9 + 
does. But here
A/^ = V ? 2 :=2 n = 0.
Thus 6(v) converges absolutely for every v.
Example 4 Let < a <b < 1. The series
.A = + 6 2 + <i 8 + 6* + ...
is convergent. For if rt
& n2m
^/a n = 2 V^ = 6.
If rc = 2 m + 1, 2//J+1
V0 n = va m+1 = a.
Thus for all n ni ,
Vn < * < 1
Let us apply D'Alembert's test. Here
Thus the test gives us no information.
91. Cauchy's Integral Test.
Let <f>(x) be a positive monotone decreasing function in the interval
(a, oo ). The series
is convergent or divergent according as
y^QO
I (f> (x) dx
*Ja
is convergent or divergent.
For in the interval (n, n f 1), n>m> a,
100 SERIES
Hence n+l
Letting n = w, TW + 1, w jt?, and adding, we have
, p+l
Passing to the limit jt? = oo, we get
which proves the theorem.
Corollary. When 4> t8 convergent
Example 1. We can establish at once the results of 81, 2. For,
taking *(*) = !,
is convergent or divergent according as /i>l, or /*<!, by I,
036, 636.
We also note that if
~" " "*""""
then
Example 2. The logarithmic series
^ 1
8 = 1,2,
are convergent if JJL > 1; divergent if /* < 1.
We take here ^
and apply I, 637, 638.
TESTS OF CONVERGENCE FOR POSITIVE TERM SERIES 101
92. 1. One way, as already remarked, to determine whether
a given positive term series A = a^ f a 2 f is convergent or
divergent is to compare it with some series whose convergence or
divergence is known. We have found up to the present the
following standard series S:
The geometric series
The general harmonic series
1
The logarithmic series
1 + 1+.. . (2
2" 3" V
(3
S * , (4
^W W/ 'W/^W
We notice that none of these series could be used to determine
by comparison the convergence or divergence of the series follow
ing it.
In fact, let a n , b n denote respectively the nth terms in 1), 2).
Then for#<l, /t>0,
^^oo by I, 464,
a n+l
or using the infinitary notation of I, 461,
t> n > a n .
Thus the terms of 2) converge to infinitely slower than the
terms of 1), so that it is useless to compare 2) with 1) for conver
gence. Let g > 1. Then
a n >b n .
This shows we cannot compare 2) with 1) for divergence.
102 SERIES
Again, if # n , b n denote the nth terms of 2), 3) respectively, we
have, if /i > 1,
^ = ^ = 00 by I, 463,
a n log* n
or 7
*n > <*'
> ..
Thus the convergence or divergence of 3) cannot be found
from 2) by comparison. In the same way we may proceed with
the others.
2. These considerations lead us to introduce the following
notions. Let A = a l f a 2 + , B = b 1 f J 2 f be positive term
series. Instead of considering the behavior of a n /b n , let us gen
eralize and consider the ratios A n : B n for divergent and A n : B n
for convergent series. These ratios obviously afford us a measure
of the rate at which A n and B n approach their limit. If now A^
B are divergent and . ^
^n ~ Hf*
we say A, B diverge equally fast ; if
A diverges slower than jS, and B diverges faster than A. From
I, 180, we have :
Let A, B be divergent and
According as I is 0, =0, oo, A diverges slower, equally fast, or
faster than B.
If A, B are convergent and
we say A, B converge equally fast ; if A converges and
B, <A n ,
TESTS OF CONVERGENCE FOR POSITIVE TERM SERIES 103
B converges faster than A, and A converges slower than B. From
I, 184, we have:
Let A, B be convergent and
According as I is 0, = 0, oo, A converges faster, equally fast, or slower
than B.
Returning now to the set of standard series >S T , we see that each
converges (diverges) slower than any preceding series of the set.
Such a set may therefore appropriately be called a scale of con
vergent (divergent) series. Thus if we have a decreasing positive
term series, whose convergence or divergence is to be ascertained,
we may compare it successively with the scale S, until we arrive
at one which converges or diverges equally fast. In practice such
series may always be found. It is easy, however, to show that there
exist series which converge or diverge slower than any series
in the scale S or indeed any other scale.
Foplet A, B, a,... ( 2
be any scale of positive term convergent or divergent series.
Then, if convergent,
11 > 51 >!>...;
if divergent, A n > B n > C n > ...
Thus in both cases we are led to a sequence of functions of the
type
Thus to show the existence of a series H which converges (di
verges) slower than any series in X, we have only to prove the
theorem :
3. (Du Bois Reymond.^) In the interval (a, oo) let
denote a set of positive increasing functions which =00
Moreover^ let f
J\
104 SERIES
Then there exist positive increasing functions which == oo slower than
Foras/ 1 >/ 2 there exists an a l >a such that /i>/ 2 4l for
x> a r Since / 2 >/ 3 , there exists an a 2 > a 1 such that / 2 >/ 8 + 2
for x>a%. And in general there exists an a n >a n _ l such that
f n >/n+i + n for x > a n . Let now
n
Then g is an increasing unlimited function in (, oo) which
finally remains below any f m (x) + m 1, m arbitrary but fixed.
Thus ff
Hence
< lim *i Q
/ w + W  1
93. From the logarithmic series we can derive a number of
tests, for example, the following :
1. (Bertrams Tests.) Let A = a l + a 2 f 6e a positive term
series.
Let , 1
log
/. x N a n nLn L_ift ^ ^ 7 ^
G.(^)=  V  LJ  = 1, 2, ... Z n=l.
L 8 + \ n
If for some s and m,
(?,O)>A*>1 n>m, (1
is convergent. If, hoivever,
<?.(n)<l, (2
i divergent.
For multiplying 1) by /, +1 n, we get
or t
log  > ^ log ^ra = log Zyn.
ajiltfi fr,_j?i
Hence 
or
TESTS OF CONVERGENCE FOR POSITIVE TERM SERIES 105
Thus A is convergent.
The rest of the theorem follows similarly.
2. For the positive term series .A=a 1 f a 2 f to converge it is
necessary that, for % = oo,
lim a n = 0, lim na n = 0, lim naj^n = 0, lim na^nl^n = 0,
We have already noted the first two. Suppose now that
lim %#{}% I 8 n > 0.
Then by I, 338 there exists an m and a c > 0, such that
na n l^n I 8 n > c , n > wi,
or
Hence A diverges.
Example 1. * __
n a
We saw, 88, Ex. 5, that J. is divergent for < 1. For = 1,
^1 is convergent for /3 > 1 and divergent if y8 ^ 1, according to
91, Ex. 2.
Then if ^8 > 0,
and A is convergent since V is. If /3 < 0, let
ra a n*
But log*' w < n*' by I, 463, 1 ;
and A is convergent since ^\  is.
*i n *
106 SERIES
Example 2.
Here 1
log
Q r=
1
by 81, 6).
Hence A is convergent for /x>0 and divergent for /i<0. No
test for /i = 0.
But for /i = 0, j
lo g r rr i
= 00,
since l^n > l$n. Thus A is divergent for p = 0.
94. A very general criterion is due to Kummer, viz.:
Let A == a l 4 a 2 f  be a positive term series. Let k r k^ be a
set of positive numbers chosen at pleasure. A is convergent, if for
some constant k > 0.
^4 is divergent if
t 2
is divergent and
K n <0 w = l, 2,
For on the first hypothesis
TESTS OF CONVERGENCE FOR POSITIVE TERM SERIES 107
Hence adding,
1 ' r, v l l
and A is convergent by 80, 4.
On the second hypothesis,
a n Ti ~~^n '
or ._!
Hence 4 diverges since R is divergent.
95. 1. From Rummer's test we may deduce D'Alembert's test
at once. For take
Then A = a^ 4 # 2 + '" converges if
K n = ^\
i.e. if
a n  P
Similarly A diverges if  n1 >.!.
2. To derive Uaabe's test we take
k n = n.
Then A converges if
i.e. if
Similarly A diverges if
108 SERIES
96. 1. Let A = a 1 f a 2 f be a positive term series. Let
A converges if there exists an s such that
it diverges if \ / \ ^ t ^
y J X,( w ) < 1 for n > m.
We have already proved the theorem for X (n). Let us show
how to prove it for Xj(V). The other cases follow similarly.
For the Kummer numbers k n we take
k n = n log n.
Then A converges if
k n = n log n  n   O 1 1) log (n f 1) > k > 0.
As
n+l
j
nj
Thus A converges if X x (^) > S > 1 for n>m.
In this way we see that A diverges if Xj(n) < 1, n > m.
2. Cahen's Test. For the positive term series to converge it is
necessary that , , \ ^
limn Uf J!._i)_il = 400.
n=o ( \a n+1 J }
TESTS OF CONVERGENCE FOR POSITIVE TERM SERIES 109
For if this upper limit is not f 00,
for all n. Hence I n
But the right side = 0. Hence Xj(w) < 1 for n > some ?w, and
A is divergent by 1.
Example. We note that Raabe's test does apply to the harmonic
l '* es 1 4. i 4. i 4. (\
Here
Hence 'P n = 0, and
lim P n = 0.
Hence the series 1) is divergent.
97. Gauss' Test. Let A = ctj + ocgf be a positive term series
such that
where s, a l b 1 do not depend on n. Then A is convergent if
a l b > ] , and divergent if a b l < 1.
Using the identity I, 91, 2), we have
i . JL t 7 >
Thus limX (w)=a 1 6j. Hence, if a 1 &j>l, ^L is conver
gent; if ! i 1 <l, it is divergent. If ^ ^ = 1, Raabe's test
does not always apply. To dispose of this case we may apply
the test corresponding to X 1 (n). Or more simply we may use
Cahen's test which depends on \(n). We find at once
lim P n == 2 J 2 b 1 < oo ;
and A is divergent.
110 SERIES
98. Let A = a l + # 2 f be a positive term series such thtt
5a =:! +  + * /x>l, /S n <oc.
Then A is convergent if a > 1 and divergent if a < 1 .
For
and ^L converges if a > 1 and diverges if < 1 . If = 1,
and A is divergent.
EXAMPLES
99. The Binomial /Series. Let us find for what values of x and
LL the series
converges. If ^ is a positive integer, f is a polynomial of degree /t.
For fji = 0, J?= 1. We now exclude these exceptional values of p..
Applying D'Alembert's test to its adjoint we nnd
n f 1
= \x\.
Thus B converges absolutely for \x\ < 1 and diverges for \x\ > 1 .
Letx*=l. Then
B=I+ +^'^"" 1 i ^/ A  1 '/ i  2
M 1.2 123
Here D'Alembert's test applied to its adjoint gives
As this gives us no information unless M< 1, let us apply
Raabe's test. Here
T
, for sufficiently large n
TESTS OF CONYEU<;ENTK FOR POSITIVE TERM SERIES 111
Thus B converges absolutely if M>0, and its adjoint diverges
if /i<0. Thus B does not converge absolutely for /A<0.
But in this case we note that the terras of B are alternately
positive and negative. Also
1 
so that the \a n \ form a decreasing sequence from a certain term.
We investigate now when a n = 0. Now
. ~
In I, 143, let a = /*, = 1 . We thus find that lim a n = only
when /i> 1. Thus jB converges when ^> 1 and diverges
when /&< !
a?=l. Then
i n 2
~~' ^" 12 123
Jf IJL > 0, the terms of B finally have one sign, and
/ V
Hence B converges absolutely.
If /A < 0, let fj. = \. Then B becomes
Here
n
1.2 '
iV
1.23
l\ ^_ l
^ / i
\l
n
Hence J5 diverges in this case. Summing up :
The binomial series converges absolutely for #<1 and diverges
for \x\>\. When x 1 it converges for p, > 1 and diverges for
p < 1 ; it converges absolutely only for /i > 0. When x = 1, it
converges absolutely for p, > awo? diverges for /x < 0.
112 SERIES
100. The Hyper geometric Series
7
2 7 7
o
1.23.7.7 + 1.7 + 2
Let us find for what values of x this series converges. Passing
to the adjoint series, we find
x = x .
(i
Thus F converges absolutely for  x \ < 1 and diverges for  x \ > 1.
Let x = 1. The terms finally have one sign, and
a n +i w 2 f n(l + 7) + 7
'
Applying Gauss', test we find _F converges when and only when
Let x = 1. The terms finally alternate in sign. Let us find
when a n = 0. We have
= a P (a + 1) Q
'
f
Now
m
( t+
\ m
= m(l + *}
y + m = m
Thus
But by I, 91, 1),
1 _
, . 1
mm 2
mm*
m
where o m == 1, r m = 7 2 as m == oo.
PRINGSHEIM'S THEORY 113
Hence
ff71 ,.
Hence
and thus
L = lira log  a n+2 \ = 2J l m
Now for a n to == it is necessary that L n === oo. In 88, Ex. 3,
we saw this takes place when and only when + /3 7 1<0.
Let us find now when  a n+} \<\a n . Now 1) gives
*n+2
n
Thus when a f /3 7 1 < 0, a n+2 \ <  a n+l . Hence in this
case J 7 is an alternating series. We have thus the important
theorem :
The Tiy per geometric series converges absolutely when \ x \ < 1 and
diverges when \x > 1 . When x = 1, F converges only when a f j3
7<0 and then absolutely. When x = 1, f 7 converges only
when tff/3 7~1<0, a9?cZ absolutely if a + /3 7 < 0.
Pringsheinis Theory
101. 1. In the 35th volume of the Mathematische Annalen
(1890) Pringsheim has developed a simple and uniform theory oi
convergence which embraces as special cases all earlier criteria,
and makes clear their interrelations. We wish to give a brief
sketch of this theory here, referring the reader to his papers for
more details.
Let M n denote a positive increasing function of n whose limit
is H QO for n = CXD . Such functions are, for example, p > 0,
nf" , log* 1 /i , Ifn , l^nltfi Z.
114 SKRIKS
An, where A is any positive term divergent series.
B n ~* where B is any positive term convergent series.
It will be convenient to denote in general a convergent positive
term series by the symbol
<7= aB <? 1 + <? 2 + ...
and a divergent positive term series by
Z> = rf 1 + </ 2 + ...
2. The series
Is convergent, and conversely every positive term convergent series
may be brought into this form.
*S
= j. JL^JL
M l M m +\ M\
and is convergent.
Let now conversely C=c l \ c^\ be a given convergent
positive term series. Let
Then ^ .
^i ~"~~  ~~ ~^"~
8. TA series
V = %(M n+l M n ) (2
?' divergent* and conversely every positive term divergent series may
be brought into this form.
For
PRINGSHEIM'S THEORY 115
Let now conversely D = d l f <7 2 f be a given positive term
divergent series. Let M n
lYL n JJ n _ j.
Then d M M
a n irj. n+l ^a n .
102. Having now obtained a general form of all convergent
and divergent series, we now obtain another general form of a
convergent or divergent series, but which converges slower than
1) or diverges slower than 101, 2). Let us consider first con
vergence. Let M' n < M n , then
is convergent, and if M' n is properly chosen, not only is each
term of 1) greater than the corresponding term of 101, 1), but 1)
will converge slower than 101, 1). For example, for M' n let us
take M*, < p < 1. Then denoting the resulting series by
0' = 2^, we have
ji/f\ii _ n i fe)
~"r^7 " ' w~ c
1 r JUn+l
Thus O r converges slower than C. But the preceding also
shows that O 1 and
converge equally fast. In fact 2) states that
Since M n is any positive increasing function of n whose limit
is oo, we may replace M n in 3) by l r M n so that
is convergent and a fortiori
^l r M n+l l r M n r = 12 , ... (4
^ V*M^ ^
is convergent.
116 SERIES
Now by I, 413, for sufficiently large n,
log M n+l  log M n =  log(l  &%=&) > **f*
\ 1VJ n+\ J 1YL n+l
Replacing here M n by log M n , we get
7 M 1 M ^ loRj^tiJ=J2j?^ ^ M n+i^L M n^
c '2 irx n + l *V u n ^ i" 7i/r " ^ TUT ~i 7i/r~ " '
log^n + l M n + l \^M n ^
and in general
Thus the series
V
converges as is seen by comparing with 4). We are thus led to
the theorem :
The series ^M n ^M n y^^ .
^ M n M n+l ' ^ ^i^
an infinite set of convergent series; each series converging
slower than any preceding it.
The last statement follows from I, 463, l, 2.
Corollary 1 (Abel). Let D = d t + J 2 f denote a positive term
divergent series. Then
z convergent.
Follows from 3), setting M n+l = D n .
Corollary 2. If we take ^f n = n we get the series 91, Ex. 2.
Corollary 3. Being given a convergent positive term series
(7 = (?j f c? 2 f we can construct a series which converges slower
than C.
PRINGSHEIM'S THEORY 117
For by 101, 2 W e may bring to the form
Then any of the series 7) converges slower than C.
103. 1. Let us consider now divergent series. Here our
problem is simpler and we have at once the theorem :
Tfie series M iw
diverges slower than
zC (2
That 1) is divergent is seen thus : Consider the product
, M m+ i  M m \ _ M m+l
~~~'
which obviously = oo.
N W P n = (
n ..
2
Hence J9 n = oo and D is divergent.
AS d = 1 =Q
* Jf
we see that 1) converges slower than 2).
2. ^4w^ given positive term series D = d + d^ + can be put in
the form I).
For taking M l >0 at pleasure, we determine M v M z by the
relations jur
118 SERIES
Then M n+l > M n and
Moreover M n = oo. For
> 1 + A, by I, 90, 1.
But J9 B = oo.
3. 2%e *m'e8
an infinite set of divergent series, each series divergent slower
than any preceding it. l Q M n = M n .
For log M nn  log M n  log l +
M n '
This proves the theorem for r = 0. Hence as in 102 we find,
replacing repeatedly M n by log M n ,
1 M _ 7 M ^ M n+l M n ,o
Corollary 2. If we take M n = n, we get the series 91, Ex. 2.
Corollary 2 {Abel). Let D = d l + d 2 f fo a divergent positive
term series. Then ,
'$ divergent.
We take here J!J, = .Z) n .
Corollary 3. Being given a positive term divergent series D, we
can construct a series which diverges slower than D.
For by 101, 3 we may bring D to the form
Then 1) diverges slower than D.
PRINGSHEIM'S THEORY 119
104. In Ex. 3 of I, 454, we have seen that M n +i is not always^
M n . In case it is we have
1. The series
is convergent.
Follows from 102, 3).
2. The series
 M
is convergent if p > 0; it is divergent if /*< 0.
For #"* > i ^Ml  M. I /* > 0.
Thus
3. If M n+l ~ MM we have
I M I M ^
i r+l JXL n +i Lr+i^n
For by 102, 5), 103, 3),
M + M n
i ]u ___j M
Lr+ i Mn+l l r+\ m *
Now since M n+l ~ M n , we have also obviously
. l m M n ~l m M n+l m=l, 2, ...r.
105. Having obtained an unlimited set of series which converge
or diverge more and more slowly, we show now how they may be
employed to furnish tests of ever increasing strength. To ob
tain them we go back to the fundamental theorems of comparison
of 87. In the first place, if J.= a a f a 2 f is a given positive
term series, it converges if
120 SERIES
It diverges if
^>#. (2
d n ^
In the second place, A converges if
^n C n
and diverges if ,
a n d n ~~
The tests 1), 2) involve only a single term of the given series
and the comparison series, while the tests 3), 4) involve two
terms. With Du Bois Reymond such tests we may call respec
tively tests of the first and second kinds. And in general any
relation between p terms
of the given series and JP terms of a comparison series,
0m tfn+ti '" <Vf P ii or d n , d n+1 d w +pi
which serves as a criterion of convergence or divergence may be
called a test of the p th kind.
Let us return now to the tests 1), 2), 8), 4), and suppose we
are testing A for convergence. If for a certain comparison
series
not always <.6r , n > m
it might be due to the fact that c n = too fast. We would then
take another comparison series O f = ^c' n which converges slower
than C. As there always exist series which converge slower than
any given positive term series, the test 1) must decide the con
vergence of A if a proper comparison series is found. To find
such series we employ series which converge slower and slower.
Similar remarks apply to the other tests. We show now how
these considerations lead us most naturally to a set of tests which
contain as special cases those already given.
106. 1. General Criterion of the First Kind. The positive term
series A = a + a% f converges if
n+l
PRINGSHEIM'S THEORY 121
It diverges if M n
*I + *! (2
This follows at once from 105, 1), 2); and 101, 2; 103, 1.
2. To get tests of greater power we have only to replace the
senes
M n
just employed in 1), 2) by the series of 102 and 103, 3 which con
verge (diverge) slower. We thus get from 1 :
The positive term series A converges if
__ 7lf 7L/> __ If 7 JIT ... 7 TIT n+VM
llm "" rlim " a ' <00 
It diverges if M n LM n l r M n
 ~1tf. + " jr. " an>
Sonnet's Test. The positive term series A converges if
lira nl^n l r ^7il\^n a n < oo , //. > 0.
*7
Follows from the preceding setting 7Jf n = n.
3. J 7 ^ positive term series A converges or diverges according as
'"">'
"' "
For in the first case
.5
and in the second case
< 1 , ,.>0, (3
The theorem follows now by 104, 2.
4. The positive term series A converges if
log^i* 1  lo gl
lim _^2 >0 or Mm
122 SERIES
It diverges if
M n+ ,  M n , M n+ ,  M n
P <0 oriS g
hm  TJJ  < u or nm
r = 0, 1, 2, and as before l Q M n = 3f n .
For taking the logarithm of botli sides of 3) we have for con
log M +i  M
As /i is an arbitrarily small but fixed positive number, A con
verges if lim q > 0. Making use of 104, 3 we get the first part
of the theorem. The rest follows similarly.
Remark. If we take M n = n we get Cauchy's radical test 90
and Bertram's tests 93.
/T
^l= log {'I =  log Va n
it is necessary that n/ ^ .
Also if
log     f log _
a n nl^i l r n __ a n nl^n / r _ 1
? r+1 n / r+1 n
log
= 1 
it is necessary that j
107. In 94 we have given Rummer's criterion for the conver
gence of a positive term series. The most remarkable feature
about it is the fact that the constants k r & 2 which enter it are
subject to no conditions whatever except that they shall be positive.
On this account this test, whicli is of the second kind, has stood
entirely apart from all other tests, until Pringsheim discovered its
counterpart as a test of the first kind, viz. :
PRINGSHEIM'S THEORY 123
Pringsheims Criterion. Let p v p% be a set of positive numbers
chosen at pleasure, and let P n = p l f 4 p n > The positive term
series A converges if
log_!L
15m _ ^L > 0. (1
* n
For A converges if
Jim  Ji2  > , by 106, 4. (2
 M n
But M n+l M n = d n is the general term of the divergent series
/)= rfj + c^f
Thus 2) may be written
log
lim _ ^>0. (3
Moreover A converges if
that is, if lim5t>0 f
n
where as usual (7= <?jf <? 3 h is a convergent series.
Hence J. converges if Cn
lim^>0. (4
^n
But now the set of numbers p v p 2 gives rise to a series
P=p l f p^ 4 ... which must be either convergent or divergent.
Thus 3), 4) show that in either case 1) holds.
108. 1. Let us consider now still more briefly criteria of the
second kind. Here the fundamental relations are 3), 4) of 105,
which may be written :
6' nM ^  c n > for convergence; (1
4 .j   t/ n < for divergence. (2
124 SERIES
Or in less general form :
The positive term series A converges if
It diverges if
0. (4
Here as usual C=c l \c^^~ is a convergent, and D=d l
a divergent series.
2. Although we have already given one demonstration of
Rummer's theorem we wish to show here its place in Pringsheim's
general theory, and also to exhibit it under a more general form.
Let us replace c n , c nn in 1) by their values given in 101, 2.
We get
or snce
n +2  n 1 . n __ n+1 ~ n > Q
M n + 1 <! ^n
or by 103, 2 a n , Q
^n+l  a n > v i
^n+1
where D = d 1 +d 2 {' is a^y divergent positive term series.
Since any set of positive numbers &j, & 2 , gives rise to a series
&i 4 ^2 "+" '" which must be either convergent or divergent, we see
from 1) that 5) holds when we replace the eTs by the Fs. We
have therefore:
The positive term series A converges if there exists a set of positive
numbers k v k 2 such that
(6
a n+\
It diverges if
where as usual d l 4 d% f denotes a divergent series.
ARITHMETIC OPERATIONS ON SERIES 125
Since the k's are entirely arbitrary positive numbers, the rela
tion 6) also gives
A converges if
as is seen by writing
if
n ^F
K n
reducing, and then dropping the accent.
3. From Rummer's theorem we may at once deduce a set of
tests of increasing power, viz. :
The positive term series A is convergent or divergent according as
~M 7 TIT ... / /I/ fi 7VT 1 1\T ... / 7l/
LfJ n+\ l \ LJ n+\ t/ r L ' J n+\ "'n+l LY  L n l \ L  L n L r L1J n
is > or is j< 0.
For & a , & 2 ... we have used here the terms of the divergent
series of 103, 3.
Arithmetic Operations on Series
109. 1. Since an infinite series
is not a true sum but the limit of a sum
A = limA n ,
we now inquire in how far the properties of polynomials hold for
the infinite polynomial 1). The associative property is expressed
in the theorem :'
Let A = a^ 4 # 2 4 be convergent. Let b^ = a l 4 4 a m ,
^2 = a m l +i 4 4 , , Then the series B = b l 4 & 2 + '" l8 C(m ~
vergent and A = . Moreover the number of terms which b n em
braces may increase indefinitely with n.
For *^
and limA mn = A by I, 103, 2.
126 SERIES
This theorem relates to grouping the terms of A in parentheses.
The following relate to removing them.
2. Let B = b l + 6 2 f be convergent and let b l = a + + <*< m *
* a = / Wl +i+ ' +, v y 1<D ^ = a i + a 2 + " ** Convergent,
A = B. 2 /f f/i* terww n >0, 4 i* convergent. 8 ^ eacA
in n m n _i <^p a constant, and a n = 0, A is convergent.
On the first hypothesis we have only to apply 1, to show
A = jB. On the second hypothesis
> 0, ra, B n < e, w > w.
Then A.< e , s>m n .
On the third hypothesis we may set
A. = B r +b' r+l
where b' r+1 denotes a part of the aterms in b r+l . Since b r+1 con
tains at most p terms of A, b' r+l = 0.
Hence
.= m r , or = .
Example 1. The series
B = (11) + (11) + (11)+..
is convergent. The series obtained by removing the parentheses
4 = 11 + 11+ .
is divergent.
Example 2.
^ vfi __ )= y __ ^
^ ^
As J^ is comparable with 5],, it is convergent. Hence A is
^rr
convergent by 3.
110. 1. Let us consider now the commutative property.
Here Riemann has established the following remarkable
theorem :
ARITHMETIC OPERATIONS ON SERIES 12?
The terms of a simply convergent series A = a l + a a + can be
arranged to form a series S, for which lim S n is any prescribed
number, or 00.
For let p , ,
/> = fl f f  
be the series formed respectively of the positive and negative
terms of A, the relative order of the terms in A being preserved.
To fix the ideas let / be a positive number ; the demonstration
of the other cases is similar. Since B n == + oo, there exists an m l
such that
, > i. (i
Let m l be the least index for which l)is true. Since (7 n = oo,
there exists an /w 2 such that
An, + O m , < I. (2
Let z 2 be the least index for which 2) is true. Continuing,
we take just enough terms, say m 3 terms of B, so that
Then just enough terms, say m t terms of (7, so that
S mt 4O mt + B mi . mt +O m ^ mt <l,
etc. In this way we form the series
&'=B mi +C m , + m ,, m ,+ 
whose sum is L For
a, i < s > a ;
r 1 Qr
Hence
'2. Let A = aj f a 2 H 6e absolutely convergent. Let the terms
of A be arranged in a different order, giving the series B. Then B
is absolutely convergent and A = B.
For we may take m so large that
A m < e.
128 SERIES
We may now take n so large that A n B n contains no term
whose index is <. m. Thus the terms of A n n taken with
positive sign are a part of A m and hence
A n  B n  < A m < e n > m.
Thus B is convergent and B = A.
The same reasoning shows that B is convergent, hence B is
absolutely convergent.
3. If A = a l f a 2 f "enjoys the commutative property, it is
absolutely convergent.
For if only simply convergent we could arrange its terms so as
to have any desired sum. But this contradicts the hypothesis.
Addition and Subtraction
111. Let A = a l + a% f , B = b l f 6 2 h 6e convergent.
The series
<?= (^^ + (^62) + ...
are convergent and 0= AB.
For obviously O n = A n B n . We have now only to pass to the
limit.
Example. We saw, 81, 3, Ex. 1, that
is a simply convergent series. Grouping its terms by twos and
by fours [109, l] we get
W3 4w2 4wl
Let us now rearrange J., taking two positive terms to one nega
tive. We get
ADDITION AND SUBTRACTION 129
We note now that
_!_V!
+
twS 4w2 4wl
, _J
t n  3 4n2
1 1 1 \
: n  1 "*" r?r^~3 "" 2w/ ^
= by 109, 2.
Thus B =  ^4.
This example, due to Dirichlet, illustrates the noncommutative
property of simply convergent series. We have shown the con
vergence of B by actually determining its sum. As an exercise let
us proceed directly as follows :
The series 1) may be written :
8/i3 ^Y ~
n\ n
Comparing this with
we see that it is convergent by 87, 3. Since 1) is convergent, 5
is also by 109, 2.
112. 1. Multiplication. We have already seen, 80, 7, that we
may multiply a convergent series by any constant. Let us now
consider the multiplication of two series. As customary let
denote the infinite series whose terms are all possible products
a, b K without repetition. Let us take two rectangular axes as in
analytic geometry ; the points whose coordinates are x = *, y = K
are called lattice points. Thus to each term aJt> K of 1), cor
130 SERIES
responds a lattice point t, K and conversely. The reader will find
it a great help here and later to keep this correspondence in mind.
Let A = a l f a a H , B = b l f b z f 6e absolutely convergent.
Then (J = # A is absolutely convergent and A B = (7.
Let w he taken large at pleasure ; we may take n so large that
F n A m B m contains no term both of whose indices are <. m.
Then T n  A m E m < ,B m + ,B m + ... + m B m
+ /8 1 A m +/3 2 A ro +... + ft m \ m
< A m E m + B m A m
< e for m sufficiently large.
Hence
and is absolutely convergent.
To show that (7= A B, we note that
 C n  A m m < F,,  A m B m < e n > n,.
2. We owe the following theorem to Mertens.
If A converges absolutely and B converges (not necessarily abso
lutely*), then
(7= a^j h O^ + a J>i) ^ OA f #2*2 +
is convergent and C = A  B.
We set 0= 6*j 4 <' 2 + c z+ '"
where ^ a l b l
c 2 = a^ f a^
c z = a^g I a 2 i 2 f aa
Adding these equations gives
C n = a^ + a^B n ^ f a 3
ADDITION AND SUBTRACTION 131
But 
B m =BB m m = l,2, ...
Hence
where
The theorem is proved when we show d n = 0. To this end let
us consider the two sets of remainders
B l , B<i , ... B n ^
_ Wj 4 n 2 = n.
DO ]5
"wt+1 ' "n t +2 ' '" ^fii+n,
Let * $ach one in the first set be  <  M^ and each in the second
set < M Then since
Now for each e > there exists an n such that
also a i>, such that
Thus 1) shows that , , ,
I <*;i I <. .
3. When neither J. nor B converges absolutely, the series
may not even converge. The following example due to Cauchy
illustrates this.
^ = JL_J_ + _L__L + ...
Vl V2 V3 V4
5 = J = _L + J_J_ + ... = A .
VI V2 V3 V4
*The symbols  < ,  <  mean numerically <, numerically <.
132 SERIES
The series A being alternating is convergent by 81, 3. Its
adjoint is divergent by 81, 2, since here /* = . Now
ViVi WiV2
WT V3 V2 V2 Vrt VT
=J T 1 1 1
A
By I, 95,
== G' a + <? 3 f <? 4 4
and
_
VI Vw  1 V2 Vre  2 V  1 VI
Vw ( >) < ''
Hence 1_ > 2
m) n n
Hence is divergent since c n does not = 0, as it must if C
were convergent, by 80, 3.
4. In order to have the theorems on multiplication together,
we state here one which we shall prove later.
If all three series A, B, Q are convergent, then = A B.
113. We have seen, 109, l, that we may group the terms of a
convergent series A a l + a% f into a series B = b^ 4 ^ 2 +
each term b n containing but a finite number of terms of A. It is
easy to arrange the terms of A into a finite or even an infinite
number of infinite series, jB', B n ', B' n For example, let
B" = a 2 f a p+2 f a 2p + 2 4
Then every term of vl lies in one of these p series B. To decom
pose A into an infinite number of series we may proceed thus :
In B 1 put all terms a n whose index n is a prime number ; in B n
put all terms whose index n is the product of two primes ; in
TWOWAY SERIES 183
B (m) all terms whose index is the product of m primes. We ask
now what is the relation between the original series A and the
series JS', B ff
If A = a l 4 # 2 4 is absolutely convergent, we may break it up
into a finite or infinite number of series B* , fl , fn , Each of
these series converges absolutely and
That each J9 (m) converges absolutely was shown in 80, 6. Let
us suppose first that there is only a finite number of these series,
say p of them. Then
A n = B' ni + B\ +  + B% M = n, +  + n r
As n = oo, each n v 7i 2 .=oo. Hence passing to the limit
n = QO , the above relation gives
Suppose now there are an infinite number of series B (m \
We take v so large that A B n , n>v, contains no term a n of
index < w, and m so large that
4 ^
\ m
M5.<A,
Twoway Series
114. 1. Up to the present the terms of our infinite series have
extended to infinity only one way. It is, however, convenient
sometimes to consider series which extend both ways. They are
of the type
which may be written
a O+"l + a 2+  1" a ~l + a 2+
or
134 SERIES
Such series we called twoway series. The series is convergent
if
lira 2a B (2
r,s=ao n=~r
is finite. If the limit 2) does not exist, A is divergent. The ex
tension of the other terms employed in oneway series to the
present case are too obvious to need any comment. Sometimes
n = is excluded in 1) ; the fact may be indicated by a dash,
<x
thus 2'a w .
00
2. Let m be an integer ; then while n ranges over
... _ 8 , 2, 1,0,1,2,3...
v = n h m will range over the same set with the difference that v
will be m units ahead or behind n according as w^O. This
shows that
W,= oo H a
Similarly,  _ a
,, ^ a  n
ft= tX> Jl=zGQ
3. Example 1. = nx+an2
This series is fundamental in the elliptic functions.
Example 2.
l , , ^ ^
 T / \  )
x ~ \x f n nj
The sum of this series as we shall see is TT cot TTX.
TWOWAY SERIES 135
115. For a twoway series A to converge, it is necessary and
sufficient that the series formed with the terms with negative indices
and the series C formed with the terms with nonnegative indices be
convergent. If A is convergent, A = B + C.
It is necessary. For A being convergent,
\AB r C.\</2 , AB,C B ,\<e/2
if s, s f > some <r and r > some p. Hence adding,
\C 8 C 8 ,\<e,
which shows C is convergent. Similarly we may show that B is
convergent.
It is sufficient. For B, C being convergent,
j5 r <e/2 , <7<7.<e/2
for r, s > some p. Hence
TVma n=
inus lim 2 n = B + O.
Example 1. The series
x T^ \x + n n
is absolutely convergent if x = 0, 1, ^,
For
a n \ =
1
x { n n
\x I
~H
Hence s ,
^*a n and 2<a n
00
OP ^
are comparable with 5]
Example 2. The series
(#) = l^nx+an' x arbitrary (2
00
is convergent absolutely if a < 0. It diverges if a > 0.
136 SERIES
__
n > 0, Vtf n = e x e an = if a <
= 00 if a>0;
w = w', /i'>0 \/ n = '*"*' = ifa<0
== QO if a > 0.
The case a = is obvious.
Thus the series defines a onevalued function of x when a < 0.
As an exercise in manipulation let us prove two of its properties.
1 (*)(#) is an even function.
For
e(z)=26r w * +a 2 . (3
x>
If we compare this series with 2) we see that the terms corre
sponding to n = m and n m have simply changed places, as the
reader will see if lie actually writes out a few terms of 2), 3).
Of. 114, 2.
2 O + 2 ma) = e~"**+ ma) (x). m = 1, 2, ...
For we can write 2) in the form
r 2 ^ (.r+Swtf) 2
<H)(=e < 2<T 4 "" (4
HPl
18 8 
which with 4) gives 3).
CHAPTER IV
MULTIPLE SERIES
116. Let x = Zj, x m be a point in wway space 9t m . If the
coordinates of x are all integers or zero, x is called a lattice point,
and any set of lattice points a lattice system. If no coordinate of
any point in a lattice system is negative, we call it a nonnegative
lattice system, etc. Let f(x l # w ) be defined over a lattice
system 1 = ^,...^. The set J/(^OS * s called an mtuple
sequence. It is customary to set
/('i v) = ., .s.
Then the sequence is represented by
The terms Um A , lira
as 4 1 t m converges to an ideal point have therefore been denned
and some of their elementary properties given in the discussion
of I, 314328 ; 336338.
Let r = .TJ *m V = Vi '" Mm be two points in 9t m . If
y\ ^ x \ '" Hm ^L x m we shall write more shortly y > x. If x
ranges over a set of points x r > x n > x'" we shall say that x is
monotone decreasing. Similar terms apply as in I, 211.
If now
when y >_ x, we say /is a monotone increasing function. If
<* '" ^ >
we say /is a monotone decreasing function.
Similar terms apply as in I, 211.
138 MULTIPLE SERIES
117. A very important class of multiple sequences is connected
with multiple series as we now show. Let # tl ... tw be defined over
a nonnegative lattice system. The symbol
or X# t ... lm , or A. v ^... Vm
o
denotes the sum of all the a's whose lattice points lie in the rec
tangular cell 0<x l <v l 0<x m <p m .
Let us denote this cell by M v ^.. vm or by R v . The sum 1) may be
effected in a variety of ways. To fix the ideas let m = 3. Then
etc. In the first sum, we sum up the terms in each plane and
then add these results. In the second sum, we sum the terms on
parallel lines and then add the results. In the last sum, we sum
the terms on the parallel lines lying in a given plane and add the
results; we then sum over the different planes.
Returning now to the general case, the symbol
u4 = 2a tl ... lm i v ... * m =0, 1, QO,
or A = 2a tl ... twi
o
is called an wtuple infinite series. For m = 2 we can write it
out more fully thus
In general, we may suppose the terms of any wtuple series dis
played in a similar array, the term a tl ... lm occupying the lattice
point t = (6 1 "t m ). This affords a geometric image of great
service. The terms in the cell R v may be denoted by A v .
If limA,, l ... Vm = limA v (2
GENERAL THEORY 139
is finite, A is convergent and the limit 2) is called the sum of the
series A. When no confusion will arise, we may denote the series
and its sum by the same letter. If the limit 2) is infinite or does
not exist, we say A is divergent.
Thus every mtuple series gives rise to an 7ntuple sequence
\A v ^... vm \. Obviously if all the terms of A are >0 and A is diver
gent, the limit 2) is 4 oo. In this case we say A is infinite.
Let us replace certain terms of A by zeros, the resulting series
may be called the deleted series. If we delete A by replacing all
the terms of the cell R v ^... vm by zero, the resulting series is called
the remainder and is denoted by J.^...^ or by A v . Similarly if
the cell R v contains the cell R^ the terms lying in R v and not in
R^ may be denoted by A^ .
The series obtained from A by replacing each term of A by its
numerical value is called the adjoint series. In a similar manner
most of the terms employed for simple series may be carried over
to 7wtuple series. In the series 2a tl ... lm the indices i all began
with 0. There is no necessity for this; they may each begin with
any integer at pleasure.
118. The Geometric Series. We have seen that
= 1 + a + a 2 +  a \ < 1,
= ! + &+ 6 2 +  6<1.
1 b
Hence
1
(1  a)(l  b) o
for all points a, b within the unit square.
In general we see that
is absolutely convergent for any point x within the unit cube
< K < 1 *= 1, 2, W,
and 1
140 MULTIPLE SERIES
119. 1. It is important to show how any term of A = 2a tl ... lw can
be expressed by means of the ^l, v .. l//t .
Then /),, ..... ,_,, = '*,,,, . IK,,.  4,, ..... ii, i ( 2
I^et D, lft ..... ,_,= A,, ,...,._, A,K, ..... .,, (3
Similarly
A,, a ..... , . = ^,,v .. A v , ..... _,,, (4
Finally D V} = D v ^ D v ^ . p (6
If now we replace the Z>\s by their values in terms of the As,
the relation 7) shows that a v ^... Vm may be expressed linearly in
terms of a number of ^4 Mi ... Mm where each JJL,, = v r or v r 1.
For in = 4 2 we find
!2. From 1 it follows that we may take any sequence \A Ll ... l}n \
to form a multiple series
A ?a
^  'tt tl ..., m .
This fact has theoretic importance in studying the peculiarities
that multiple series present.
120. We have now the following theorems analogous to 80.
1. For A to be convergent it is necexsary and sufficient that
6>0, p, \A^ \< R P <R<R V .
2. If A is convergent, so is A^ and
A^ = A A^ = lim A^ v .
Conversely if A^ is converyent, so is A.
GENERAL THEORY 141
3. .For A to converge it is necessary and sufficient that
lim A v = 0.
V <X>
4. A series whose adjoint converges is convergent.
5. Let A be absolutely convergent. Any deleted series K of A is
absolutely convergent and \ B \ < A.
6. If A = 2# ti ... tm is convergent, so is B = 2^ ti ..., w a
B = A^4, A: a constant.
121. 1. jFor ^4 to converge it is necessary that
A^ 2 ... ,_! , A,,, ... ,. w  2 , A, < <V 2 ... ,, = 0, as v = oo .
For by 120, 1 , .  ,
^ ! A 1 ...A m ~A Ml ... MMI  <6
if X 1 A m , ^ fin > p.
Thus by 119, 1)
[I> Vl , z .... m  l \<e v>p>
Hence passing to the limit p = oo ,
lim D,,,...^,^ e.
r = QO
As e is small at pleasure, this shows that ^...^.^ 0. In this
way we may continue.
2. Although ,. A
5 hm^ i ..., m =
n ... ^=*
when vl converges, we must guard against the error of supposing
that a v = when v = (v l i/ m ) converges to an ideal point, all of
whose coordinates are not oo as they are in the limits employed
iul.
This is made clear by the following example due to Pringsheim.
Then by 119, 8) . _ 1 , 1
a rM ~T* *
1 a r a 8
142 MULTIPLE SERIES
As
lim A r ^ g =
r, 5=00
A is convergent. But
lim I a r J = , lim I a rt \ =
r=oo ' a* *=oo a r
That is when the point (r, ) converges to the ideal point
(oo, ), or to the ideal point (r, oo ), a T9 does not = 0.
3. However, we do have the theorem :
converge. Then for each e > there exists a \ such that tl ... t <
for any i outside the rectangular cell R^.
This follows at once from 120, l, since
122. 1. Letf{x l # m ) be monotone. Then
Xn) = 1 x l < a v x m < a m , a may be ideal. (1
exists, finite or infinite. If f is limited, I is finite. If f is unlim
ited, I = f GO when f is monotone increasing, and I = oo whenf is
monotone decreasing.
For, let/ be limited. Let J. = j < o^ < = a.
Then
is finite by I, 109.
Let now B = /3 r /3 2 , = a be any other sequence.
Let Km/08.) = ' 11^/08.) 
,  5
Then there exists by I, 338 a partial sequence of B> say
(7= 7 X , 7 2 such that
also a partial sequence J9= Sj, S 2 such that
lim/(S n ) = i.
GENERAL THEORY 143
But for each a n there exists a 7 >. a n ;
hence
and therefore Z >. I. (2
Similarly, for each d n there exists an a <n > S n ;
hence /a> </(<o
and therefore 7 < / f3
Thus 2), 3) give ]im/ ~ = L
B
Hence by I, 316, 2 the relation 1) holds.
The rest of the theorem follows along the same lines.
2. As a corollary we have
The positive term series A = 'a i . is convergent if A v ... is
limited.
123. 1. Let A = 2# tl ... l = 2a t , B 26 tl ... lg S5 t 6e two non
negative term series. If they differ only by a finite number of
terms, they converge or diverge simultaneously.
This follows at once from 120, 2.
2. Let A, B be two nonnegative term series. Let r> denote
a constant. If a L < rl\ , A converges if B is convergent and A j< rB.
If a t > rb t , A diverges if B is divergent.
For on the first hypothesis
and on the second A
3. Let A, B be two positive term series. Let r, s be positive
constants. If
or if
limf'
1=00 6 t
exists and is = 0, A and B converge or diverge simultaneously. If
converges and * == 0, A is convergent. If B diverges and ~ == oo,
* ' *n
A is divergent.
144
MULTIPLE SERIES
4. The infinite nonnegative term series
2^,...,. and 2 log (1 f ,,...,.)
Converge or diverge simultaneously.
This follows from 2.
5. Let the power series
converge at the point a = (a r  #,), A#? tV converges absolutely for
all points x within the rectangular cell H whose center is the origin,
and one of whose vertices is a; that is for \ x, \ <  a t  , t= 1, 2,  s,
For since P converges at a,
lim * Wimi ...af ...<= 0.
M=oo
Thus there exists an Tiff such that each term
Hence
<M
Thus each term of P is numerically < than TUf times the cor
responding term in the convergent geometric series
We apply now 2.
We shall call R a rectangular cell of convergence.
124. 1. Associated with any mtuple series .<4 = 2a li ... l are
an infinite number of simple series called associate simple series,
as we now show.
Let R>, , R^ , R x , ...
be an infinite sequence of rectangular cells each lying in the
following. Let
^\ 1 *> ' " ' ' *1
be the terms of A arranged in any order lying in J? AI . Let
GENERAL THEORY 145
he the terms of A arranged in order lying in R^ R^ and so on
indefinitely.
Then s^ = ai + a ^ + ... + ^ + rtfi+i + ...
is an associate simple series of ^L.
2. Conversely associated with any simple series 21 = 2a n are an
infinity of associate mtuple series. In fact we have only to arrange
the terms of 21 over the nonnegative lattice points, and call now
the term a n which lies at the lattice point i l L m the term <7 4 ... lw .
3. Let$\ be an associate series of A = S^ ti ... lw . Zf 21 ^8 convergent,
so is A and ^ __ s ^
For A Vl ... Vm = % n .
Let now v = oo, then n = oo. But ?I W = 31, hence J.^ ... Vm = 31.
4. If the associate series ?l ?'s absolutely convergent, so is A.
Follows from 3.
5 If A = Sa^ ... ^ m i a nonnegative term convergent series, all its
associate series 21 converge.
For, any 2l,, NP lies among the terms of some A^ v . But for X
sulliciently large ^ < X < M < ^
Hence
6. Absolutely convergent series are commutative.
For let 5 be the series resulting from rearranging the given
series A.
Then any associate 93 of B is simply a rearrangement of an
associate series 21 of A. But 21 = 33, hence A = B.
7. A simply convergent mtuple series A can be rearranged,
producing a divergent series.
For let 21 be an associate of A. 21 is not absolutely convergent,
since A is not. We can therefore rearrange 21, producing a series
33 which is divergent. Thus for some 33
lim SB W
does not exist. Let 33' be the series formed of the positive, and
33" the series formed of the negative, terms of 33 taken in order.
146 MULTIPLE SERIES
Then either 33^ = + 00 or JB(J = oo, or both. To fix the ideas
suppose the former. Then we can arrange the terms of 33 to
form a series & such that S n == + oo. Let now S be an associate
series of 0. Then
^v = ^v^t vtn ~ &n
and thus
lim G v = lim 6 n = + oo.
Hence (7 is divergent.
8. If the multiple series A is commutative, it is absolutely con
vergent.
For if simply convergent, we can rearrange A so as to make the
resulting series divergent, which contradicts the hypothesis.
D. In 121, 2 we exhibited a convergent series to show that
a ti mlm does not need to converge to if ^ i m converges to an ideal
point some of whose coordinates are finite. As a counterpart we
have the following :
Let A be absolutely convergent. Then for each e > there exists
a \, such that any finite set of terms B lying without R^ satisfy the
relation \ T>\ ^ s\
\B\<e\ (1
and conversely.
For let SI be an associate simple series of Adj A. Since 21 is
convergent there exists an n, such that
<.
But if X is taken sufficiently large, each term of B lies in 2l w ,
which proves 1).
Suppose now A were simply convergent. Then, as shown in 7,
there exists an associate series ) which is infinite.
Hence, however large n is taken, there exists a p such that
Hence, however large X is taken, there exist terms B= ( $) n ^ p which
do not satisfy 1).
10. We have seen that associated with any mtuple series
GENERAL THEORY 147
extended over a lattice system 9K in 9t m is a simple series in 9? r
We can generalize as follows. Let 2ft = \i\ be associated with a
lattice system 9ft = \j\ in 9t w such that to each L corresponds &j and
conversely.
If i~ j we set a, ,= a.
J 4... im ./!>
Then .A gives rise to an infinity of wtuple series as
We say JS is a conjugate ntuple series.
We have now the following :
Let A be absolutely convergent. Then the series B is absolutely
convergent and A = .
For let A\ B' be associate simple series of A, B. Then A, B 1
are absolutely convergent and hence A f = '. But A = A r , B = B f .
Hence A B, and B is absolutely convergent.
11. Let A = 2a tl ... lm be absolutely convergent. Let B = Sa^...^
be any ptuple series formed of a part or all the terms of A. Then
B is absolutely convergent and
For let A, B 1 be associate simple series of A and B. Then B f
converges absolutely and jB ; < Adj A.
125. 1. Let ASo,...^. (1
in the cell
Then
..
1 ..., m = f /*r 1 .*r m . (2
c//2...,
Let 72 denote that part of 3t m whose points have nonnegative
coordinates. Let
(3
If e/is convergent, A = J. We cannot in general state the con
verse, for A is obtained from A v by a special passage to the limit, viz.
148 MULTIPLE SERIES
by employing a sequence of rectangular cells. If, however,
a v >_ we may, and we have
For the non negative term series 1) to converge it is necessary and
sufficient that the integral 3) converges.
2. Let f(2\ :r w ) > he a monotone decreasing function of
x in R, the aggregate of points all of whose coordinates are non
negative. Let *f x
* ^....m =/Oi '"O.
The series A ^
" '*% ITO
is convergent or divergent with
J= ( fd*\ " dx m .
J n
For let JKj, J? 2 , be a sequence of rectangular cubes each R n
contained in R n +\ .
Let R n ^R s R n *>n.
Then X, /i being taken at pleasure but > some v, there exist an
I, m such that
But the integral on the right can be made small at pleasure if J
is convergent on taking I > m > some n. Hence A is convergent
if t/is. Similarly the other half of the theorem follows.
Iterated Summation of Multiple Series
126. Consider the finite sum
2a lt ... lm ^ = 0, 1, /ij " i m = 0, 1, n m . (1
One way to effect the summation is to keep all the indices but
one fixed, say all but ij, obtaining the sum
Then taking the sum of these sums when only * 2 is allowed to
vary obtaining the sum m m
ITERATED SUMMATION OF MULTIPLE SERIES 149
and so on arriving finally ;it
m n
(2
whose value is that of 1). We call this process iterated summa
tion. We could have taken the indices L I i m in any order
instead of the one just employed; in each case we would have
arrived at the same result, due to the commutative property of
finite sums.
Let us see how this applies to the infinite series,
^. = 2tf 4i ... lw , ^ ^ = 0, l,...oo. (3
The corresponding process of iterated summation would lead us
to a series .^ ^ ^
which is an mtuple iterated series. Now by definition
*,n v m 1 v \
21 = lim 2 lim 2 lim 2 (i ... lw (5
= lim lim lim A v ^.,. Vrn , (6
while A i * /T
A = lim <4 , ...^. (7
Thus A is defined by a general limit while 21 is defined by an
iterated limit. These two limits may be quite different. Again
in 6) we have passed to the limit in a certain order. Changing
this order in 6) would give us another iterated series of the type
4) with a sum which may be quite different. However in a large
class of series the summation may be effected by iteration and this
is one of the most important ways to evaluate 3).
The relation between iterated summation and iterated integra
tion will at once occur to the reader.
127. 1. Before going farther let us note some peculiarities of
iterated summation. For simplicity let us restrict ourselves to
double series. Obviously similar anomalies will occur in mtuple
series.
150 MULTIPLE SERIES
+ a o2 + + a 10 + a n +
be a double series. The m ih row forms a series
,
n=0
and the n ih column, the series
0= (7" = a mn
nQ n~0 w
are the series formed by summing by roivs and columns, respec
tively.
2. A double series may converge although every row and every
column is divergent.
This is illustrated by the series considered in 121, 2. For A
is convergent while 2 M , 2a rj are divergent, since their terms are
not evanescent. ' <s ~
8. A double series A may be divergent although the series R ob
tained by summing A by rows or the series G obtained by summing
by columns is convergent.
Forlet A^Q if r or 8 =
if r, s > 0.
r H s
Obviously by I, 318, lim A T8 does not exist and A = 2# r , is di
vergent.
On the other hand,
R = lim lim A r8 = 0,
(7= lim lim A r8 = 1.
Thus both R and (7 are convergent.
ITERATED SUMMATION OF MULTIPLE SERIES 151
4. In the last example R and converged but their sums were
different. We now show :
A double series may diverge although both R and C converge and
have the same sum.
For let A rtS = if r or s =
= ^ T) ifr.'.X).
Then by I, 319, Km A r8 does not exist and A is divergent. On
the other hand, fl = H m l im 4, = 0,
C=\\m lim.A r , = 0.
=00 r=00
Then R and S both converge and have the same sum.
128. We consider now some of the cases in which iterated sum
mation is permissible.
CD
Let A = S/7, t ... be convergent. Let t' v ^, i r m be any permutation
of the indices i r i%, i m . If all the m \tuple series
00 00 00
2 2 ... S a.,... .
t '=rO t '=() t ' .=30 * "
23 m
are convergent, A= l "\2 tl ... lw .
l i~ l ~
This follows at once from I, 324. For simplicity the theorem
is there stated only for two variables ; but obviously the demon
stration applies to any number of variables.
129. 1. Let f(%i  # m ) be a limited monotone function. Let the
point a = (# x a nl ) be finite or infinite. When f is limited, all the
stuple iterated limits jj m .
exist . When s = m, these limits equal
lim/toaw). (2
#==
limits we suppose z<a.
152 MULTIPLE SERIES
For if /is limited, Hm/ ^ ^^ (3
.Tt a =rti a
exists by 122, 1. Moreover 3) is a monotone function of the re
maining m 1 variables.
Hence similarly Um lim y
^i_ l =flt_ l xi a =ai s
exists and is a monotone function of the remaining m 2 vari
ables, etc. The rest of the theorem follows as in I, 324.
2. As a corollary we have
Let A be a nonnegative term mtuple series. If A or any one of
its mtuple iterated series is convergent, A and all the ml iterated
mtuple series are convergent and have the same sum. If one of these
series is divergent, they all are.
fS. Let a be a nonnegative term mtuple series. Let s < m. All
the s tuple iterated series of A are convergent if A is, and if one of
these iterated series is divergent, so is A.
130. 1. Let A = Sa ti ... tm be absolutely convergent. Then all its
s tuple iterated series s = l, 2m, converge absolutely and its
mtuple iterated series all = A.
For as usual let a ti ... tw = a tl ... tw . Since A = Adj A is con
vergent, all the 8tuple iterated series of A are convergent.
QO oo
Thus $! = S tl ... lm i g convergent since 2 tl . . i w = <r t . Moreover
t
I s l I < <r r Similarly 2 Sa tl ... lw = Ssj is convergent since
1^=0 tt =0 l a
2 2a lt ... lm =s 20j is convergent; etc. Thus every stuple iter
, a =o i t =o l2
ated series of A is absolutely convergent. The rest follows now
by 128.
2. Let A = 2a tt ... lm . If one of the mtuple iterated series B
formed from the adjoint A of A is convergent, A is absolutely con
vergent.
Follows from 129, 2.
3. The following example may serve to guard the reader against
a possible error,
ITERATED SUMMATION OF MULTIPLE SERIES 158
Consider the series
a 2 a*
as ... !=s
and R = e" + e 2(t + ^ f
This is a geometric series and converges absolutely for a < 0.
Thus one of the double iterated series of A is absolutely conver
gent. We cannot, however, infer from this that A is convergent,
for the theorem of 2 requires that one of the iterated series formed
from the adjoint of A should converge. Now both those series
are divergent. The series A is divergent, for \a n \ == oo , as
r, s = oo .
131. 1. Up to the present the series
2., ...... (1
have been extended only over nonnegative lattice points. This
restriction was imposed only for convenience ; we show now how
it may be removed. Consider the signs of the coordinates of a
point x (x v x m ). Since each coordinate can have two signs,
there are 2 m combinations of signs. The set of points x whose
coordinates belong to a given one of these combinations form a
quadrant for m = 2, an octant for m = 3, and a 2 m tant or polyant
in 9? m . The polyant consisting of the points all of whose coordi
nates are > may be called the first or principal polyant.
Let us suppose now that the indices i in 1) run over one or more
polyants. Let JS A be a rectangular cell, the coordinates of each of
its vertices being each numerically < X. Let A^ denote the terms
of A lying in # A . Then I is the limit of A K for X = oo, if for each
e > there exists a X such that
\A,A^\<e X>X . (2
154 MULTIPLE SERIES
If lim A k (3
A=
exists, we say A is convergent, otherwise A is divergent. In a
similar manner the other terms employed in multiple series may
be extended to the present case. The rectangular cell R Xo which
figures in the above definition may without loss of generality be
replaced by the cube
K<A,  km<V
Moreover the condition necessary and sufficient for the exist
ence of the limit 3) is that
 A  Ap  < X, /JL > \ Q .
132. The properties of series lying in the principal polyant
may be readily extended to series lying in several polyants. For
the convenience of the reader we bring the following together,
omitting the proof when it follows along the same lines as before.
1. For A to converge it is necessary and sufficient that
lim A x = 0.
A=
2. A series whose adjoint converges is convergent.
3. Any deleted series of an absolutely convergent series A is
absolutely convergent and
\B\< Adj A.
4. If A = 2a h ... tn is convergent, so is B = TLka^... ln and A = kB.
5. The nonnegative term series A is convergent if A^ is limited,
= oo.
6. If the associate simple series 21 of an m tuple series A converges,
A is convergent. Moreover if 21 is absolutely convergent, so is A.
Finally if A converges absolutely, so does 21.
7. Absolutely convergent series are commutative and conversely.
8. Let ./(#! # m ) >.0 be a monotone decreasing function of the
distance of x from the origin.
Let
ITERATED SUMMATION OF MULTIPLE SERIES 155
Then 4 v
A = ^,.. lm
converges or diverges with
the integration extended over all space containing terms of A.
133. 1. Let B, C, D  denote the series formed of the terms of A
lying in the different poly ants. For A to converge it is sufficient
although not necessary that B, C, converge. When they do,
A = B+ C + D+  (1
For if 7? A , CA denote the terms of B, which lie in a
rectangular cell R^
Passing to the limit we get 1).
That A may converge when B, (7, do not is shown by the
following example. Let all the terms of A= 2a tl .. tm vanish ex
cept those lying next to the coordinate axes. Let these have the
value +1 if i v i 2  i m >0 and let two a's lying on opposite sides
of the coordinate planes have the same numerical value but opposite
signs. Obviously, A^ = 0, hence A is convergent. On the other
hand, every B, is divergent.
2. Thus when B, converge, the study of the given series
A may be referred to series whose terms lie in a single polyant.
But obviously the theory of such series is identical with that of
the series lying in the first polyant.
3. The preceding property enables us at once to extend the
theorems of 129, 130 to series lying in more than one polyant.
The iterated series will now be made up, in general of twoway
simple series.
CHAPTER V
SERIES OF FUNCTIONS
134. 1. Let i = (* r i. 2 ) run over an infinite lattice system ?.
Let the onevalued functions
be defined over a domain 21, finite or infinite. If the jt?tuple series
extended over the lattice system 8 is convergent, it defines a one
valued function F(x l # m ) over 21. We propose to study the
properties of this function with reference to continuity, differen
tiation and integration.
2. Here, as in so many parts of the theory of functions depend
ing on changing the order of an iterated limit, uniform convergence
is fundamental.
We shall therefore take this opportunity to develop some of its
properties in an entirely general mariner so that they will apply
not only to infinite series, but to infinite products, multiple inte
grals, etc.
3. In accordance with the definition of I, 325 we say the series
1) is uniformly convergent in 21 when F^ converges uniformly to its
limit F. Or in other words when for each e>0 there exists a \
such that
 ,
I ff T jUl <
for any x in 31. Here, as in 117, F^ denotes the terms of 1) lying
in the Fectangular cell R^, etc.
As an immediate consequence of this definition we have :
Let 1) converge in 21. For it to converge uniformly in 21 it is
necessary and sufficient that \ F>,  is uniformly evanescent in 21, or in
other words that for each e > 0, there exists a X such that F^ \ > e for
any x, in 21, and /A>\.
156
GENERAL THEORY 157
135. 1. Let
lim /(^ x m < ^ 't n ) = $(x l a; m )
tT
in 21. Here 21, r may be finite or infinite. If there exists an
77 >0 such that /==</> uniformly in V^a), a finite or infinite, we
shall say f converges uniformly at a ; if there exists no rj < 0. we
say / does not converge uniformly at a.
2. Let now a range over 21. Let 93 denote the points of 21 at
which no rj exists or those points, they may lie in 21 or not, in
whose vicinity the minimum of 77 is 0. Let D denote a cubical
division of space of norm d. Let 93^ denote as usual the cells of
D containing points of 93. Let (/> denote the points of 21 not in
93^. Then/=< uniformly in (# however small d is taken, but
then fixed. The converse is obviously true.
3. Iff converges uniformly in 21, and if moreover it converges at a
finite number of other poi tits 33, it converges uniformly in 21 f 33.
For if / = (f> uniformly in 21,
/ 0 < e x in 21, t in F 6o *(T).
Then also at each point b a of 93,
I/ <l<e x = b. inr 4j *(T).
If now S < 8 , 8j, S 2 these relations hold for any x in 21 + 93
and any t in F 5 *(r).
4. Let f(x " x m , j n ) == </> (^ # m ) uniformly in 21
/ Je limited in 21 /or eac/i m PyO"). 7%ew $ is limited in 21.
for any x in 21 and t in V s *(r). Let us therefore fix t. The
relation 1) shows that <J> is limited in 21.
f>. jfjf 2 / ti ... ^(^ XM)  converges uniformly in 21, so dWs 2/ tl ... v
For any remainder of a series is numerically < than the corre
sponding remainder of the adjoint series.
6. Let the stuple series
158 SERIES OF FUNCTIONS
converge uniformly in 21. Then for each e > there exists a X
such that , p I ,+
for any R v > R^ > jR A . When 8=1, these rectangular cells re
duce to intervals, and thus we have in particular
f n (x 1 x m )  < e for any n > n'.
When 8 > 1 we cannot infer from 1) that
\f ll ...t.(ix l x m )\<e , in 21, (2
for any i lying outside the above mentioned cell J? A .
A similar difference between simple and multiple series was
mentioned in 121, 2.
However if f t > in 21, the relation does hold. Cf. 121, 3.
in
136. 1. Let f \x 1 #,, t l n ) fo defined for each x in 21, #wd t
. .Z^ ,. ,, . . w
hm/= 0^ a: m ) ^n 81,
t=r
finite or infinite. The convergence is uniform if for any x in 21
O tin ^*(r
lim ir = 0.
For taking e>0 at pleasure there exists an ?;>0 such that
l/r<6 , *in r,*(T).
But then if 8< ?;,
i/*i<
for any ^ in Fi*(r) and any ^ in 21.
Example.
T . sin x sin y A , or ^A x
hm T^ T2 = = ^ m 21 = (0, oo).
y==![ l + #tan 2 y
Is the convergence uniform ?
Let
* = ;
then w = 0, as y =
GENERAL THEORY 159
Then
sin x cos u
\+x cot 2 u
sin x sin^u
SHI x cosi4sin 2 t6
. A
u = 0.
I x cos 2
Hence the convergence is uniform in 21
2. As a corollary we have
Weierstrass Test. For each point in SI, let \f^... lp \^.M^... lp
The series ^Lf ll ... lp (x l #,) is uniformly convergent in 21 if
is convergent.
Example 1.
'
Here
and F is uniformly convergent in 21 since
y L
^y On
is convergent. ^
Example %.
JF T (o;) = 2a 7l sin X n a?
is uniformly convergent for ( 00, oo) if
2  a n 
is convergent.
137. 1. The power series P = '2a mi ... mp x'T l *'"!* converges
uniformly in any rectangle R lying within its rectangle of con
vergence.
For let b = (J r l p ) be that vertex of R lying in the principal .
polyant. Then P is absolutely convergent at 6, i.e.
is convergent. Let now # be any point of R. Tlion each term in
2<v ...,,? &"'
is < than the corresponding term in 1).
160 SERIES OF FUNCTIONS
an
2. If the power series P a^\ a^x \ a^x 1 \ converges at a
end point of its interval of convergence, it converges uniformly at
this point.
Suppose P converges at the end point x = R > 0. Then
however large n is taken. But for < .> < H
<e by Abel's identity, 83, i.
Thus the convergence is uniform at x~R. In a similar
manner we may treat x = R.
3. Let ,/*(#! :r m ), M = 1, 2 be defined over a set 21. If each
\f n < some constant c n in 21, / is limited in 31. If moreover the
r n are all < some constant f, we say the f n (x) are uniformly
limited in 21. In general if each function in a set of functions
{/! defined over at point set 21 satisfy the relation
/ 1 < a fixed constant 6 Y , x in 21,
we say the jf s are uniformly limited in 21.
The series F= ^gji n is uniformly convergent in 21, if G = c/ l h// 2 f
is uniformly convergent in 21, while 2 1 h n+l h n \ and \ h n \ are
uniformly limited in Jl.
Tliis follows at once from Abel's identity as in 83, 2.
4. The series F=^.o n h n is uniformly convergent in 21, if in 21,
2  h n+l h n  is uniformly convergent, h n is uniformly evanescent,
and the Gr n uniformly limited.
Follows from Abel's identity, 83, l.
5. The series F= ^g n h n i* uniformly convergent in 21 if
Gr = g l f g% f is uniformly convergent in 21 while h v h%  are
uniformly limited in 21 and \h n \ is a monotone sequence for each
point of 21
FOP by 88,1, .
liKXKltAL THEORY l(it
(). The series F~ 7LgJ l n ^ s uniformly convergent in ?l if Gr l = <j r
(jr., <j l 4 # 2 , are uniformly limited in 31 and if h v h^ not only
form a monotone decreasing sequence for x in 91 hut also are uni
formly evanescent.
For by 83, 1, , F , , ff
I ^w, P \ < "'n+l tr '
Example. Let A = a l + a. z 4 be convergent. Let b v {)% =
be a limited monotone sequence. Then
converges uniformly in any interval 21 which does not contain a
point of
A
For obviously the numbers
A.=
form a monotone sequence at each point of ?I. We now apply 5.
7. As an application of these theorems we have, using the re
sults of 84,
The series , ,
tf f a t cosxh 2 cos %x 4
converges uniformly in any complete interval not containing one of
the points 2 mir provided 2  a n ^ a n \ is convergent and a n == 0,
and hence in particular if a l > a 2 > == 0.
8. , ,
^ a x cos x 4 a 2 cos 2 #
converges uniformly in any complete interval not containing one of
the points (2 m I)TT provided 2 a n+1 4 a n \ is convergent and
a n = 0, 6?wJ A^n^ in particular if a l > a, 2 > = 0.
9. The series . , . .
j sin # 4 a 2 sin 2 # 4 3 sin 6 x 4
converges uniformly in any complete interval not containing one of
the points 2 mn provided 2  a n+l a,,  i convergent and a n = 0,
^///(^ hence in particular if a l >^ a 2 >^ ==0.
162 SERIES OF FUNCTIONS
10. Theories ^ ^ x _ ^ g a; + a 3 sin Sx _ ...
converges uniformly in any complete interval not containing one of
the points (2m I)TT provided 2  #n+i+ a w  convergent and
a n = 0, and hence in particular if a^>_a^ _>= 0.
138. 1. Let FV/* f* .. *^
* ~ ~/ii i.V 1 ! ^m7
6e uniformly convergent in 21. ie .A, jB Je ^o constants and
i uniformly convergent in ?I.
For then
But F being uniformly convergent,
I^A,J<6.
2. Let f^Zf^fr...^ />0
converge uniformly in 31.
i uniformly convergent in 21. Moreover if F is limited in 21, so
is L.
or / t > in 21, hence
for any t outside some rectangular cell 72 A .
Thus for such i
4/c < log (1 +/ t ) < Bf, in 21.
139. 1. Preserving the notation of 136, let g^ # 2 , g m be chosen
such that if we set
formly in 21,
lira A = Km \f(g l "(/ m ,t l O
tT 1 = T
GENERAL THEORY 103
For if /= $ uniformly in 31,
e>0, 8>0 /_4> <
for any x in 21 and any t in Fa*(r), S independent of x.
But then  A  < e t in F 6 *(V).
2. As a corollary we have :
Let #}, # 2 ) == # ^^ ^ = ^/ & e uniformly convergent at a.
Then iU0o.
140. Example 1.
r ,. r sin w sin 2 w ., (2for#=0,
hm/ = lim  =<f>(#j=J '
u =o w =o sin 2 !/ h x cos 2 16 1 for x^ 0.
The convergence is not uniform at # = 0. For
/.__ _ _
\ + x cot 2 ^
Hence if we set x = v?
lim/= 1, since w 2 cot 2 u^= 1.
M =
Thus on this assumption
Urn /</ = 1 12 = 1.
Example 2. F =\  x + x(\  x)+ x*(l  a:)f 2^(1  x) h
Here ^ .
J 7 = E(l #) a: n .
o
Hence F is uniformly convergent in any ( r, r), < r < 1, by
136, 2.
We can see this directly. For
Hence ^ is convergent for l<z<l, and then
except at x = 1 where F = 0.
Thus  F n (x)  =  x  n , except at # = 1.
But we can choose m so large that r m <e.
Then  J^ m (a;)  < for any a? in (A r).
164 SERIES OF FUNCTIONS
We show now that F does not converge uniformly at z~
For let 1
a n = 1 
n
and F does not converge uniformly at x = 1, by 139, 2.
Example S. <*> ,2
Here ,
f ^
J n 
and JP is telescopic. Hence
rob
<T ' ^ y"
1 +
0 , * = 0.
Thus __ l
Let us take ^
Then __ i
and JP is not uniformly convergent at x=0. It is, however, in
(00, GO) except at this point. For let us take x at pleasure
such, however, that I x \ > S. Then
n \ i , f
1 f
We now apply 136, 1.
Example 4.
f n
GENERAL THEORY 165
Here , t i \
f n n f 1 I
and F is telescopic. Hence
x (n + l)a
rb '"* = <*
The convergence is not uniform at x = 0.
For set a n = ^. Then
n + 1
It is, however, uniformly convergent in 21 except at 0. For
if  x  > 8,
(n + l)a?
< e for n > some m.
141. Let us suppose that the series .F converges absolutely and
uniformly in 21. Let us rearrange JP, obtaining the series Gr.
Since F is absolutely convergent, so is Gr and F = Gr. We can
not, however, state that Gr is uniformly convergent in 21, as Bocher
has shown.
Example. . __
x
F 2n =* 0.
Hence ^is uniformly convergent in 21 = (0, 1).
Let
X
Then
a?
166 SERIES OF FUNCTIONS
Let j
n
Then
== f 1  ) as n = QO.
e\ ej
Hence O does not converge uniformly at x = 1.
142. 1. Let f =* (f> uniformly in a finite set of aggregates 21^
2J 2 ,  2l p . Then f converges uniformly in their union (2lj, 2l p ).
For by definition
> 0, . > 0, /  (f> I < e a? in ., * in JV(r). (1
Since there are only p aggregates, the minimum 8 of Sj, S p
is > 0. Then 1) holds if we replace S, by 8.
2. The preceding theorem may not be true when the number
of aggregates 2lp 21 2 is infinite. For consider as an example
which converges uniformly in 21 = (0, 1) except at x = 1. Let
Then .F is uniformly convergent in each 21,, but is not in their
union, which is 21
3. Letf^ <f>, g = ^ uniformly in 21.
Then f g == <f> ty uniformly.
If <, \/r remain limited in 21,
fg~<f)'\lr uniformly. (1
Jf moreover  ^  > some positive number in 21,
L ~ 2L uniformly. (2
* '
The demonstration follows along the lines of I, 49, 50, 51.
GENERAL THEORY 167
4. To show that 1), 2) may be false if <, ^r are not limited.
Let 
Then $ = $ =  and the convergence is uniform.
But 9
Let # = . Then A = 2 as t = 0, and fg does not
uniformly.
Again, let 
the rest being as before.
*=,
Then 1
But setting x =
A I =
= QO as t =
and  does not converge uniformly to 2. .
9 *
143. 1. As an extension of I, 317, 2 we have :
uniformly in 31. Let
Let y^rj in F*(r).
lim/C^  rc m , ^ y p ) = ^(^ 2? m ), uniformly.
t=r
The demonstration is entirely analogous to that of I, 292.
*' Let linni^...^,^... iO^^iO ^ t = l,2
t= T
uniformly in 91. i^f the point*
168 SERIES OF FUNCTIONS
form a limited set 23 Let F(u u p ) be continuous in a complete
set containing 23. Then
lira F^ . u^ = F^ ... p )
l=T
un'formly in 21.
For jP, being continuous in the complete set containing 23, is
uniformly continuous. Hence for a given e > there exists a
fixed cr > 0, such that
 F(it)  jF(v)  < e u in F^O) , v in 25.
But as U L = v t uniformly there exists a fixed S > such that
 U L  v l I < e' , 2; in 81 , in F 6 *(r).
Thus if e' is sufficiently small, W = (M I , w p ) lies in V ff (v)
when x is in 21 and t in F" 6 *(r).
144. 1.
uniformly in 31. ]jm ^ _
f^T
uniformly in 2t, ^y > </> f limited.
This is a corollary of 143, 2.
lim/O^ . a: m , ^  O = <#)(^  a: m )
t=T
uniformly in 21. Z/^f </> J^ greater than some positive constant in 31.
uniformly in $,, if $ remains limited in 21.
Also a corollary of 143, 2.
3. Let f == (f> and g === i/r uniformly, as t === r.
Ze (/>, i/r be limited in 21, awc? c/> > so7/ie positive number. Then
fp == 0^ uniformly in 21. (1
For (2
GENERAL THEORY 169
But by 2), log/=Mog< uniformly in 21; and by 142, 3
= i/r log 0, uniformly in 21. Hence 2) gives 1) by 1.
145. 1. The definition of uniform convergence may be given a
slightly different form which is sometimes useful. The function
/Oi  * m , *i  O
is a function of two sets of variables x and , one ranging in an $R m
the other in an 9t n .
Let iis set now w = (^ # m , ^ n ) and consider w as a point in
m fjp way space.
As # ranges over 21 and over PVOr), let w range over 3j fi .
Then
uniformly in 21 when and only when
e>0, S>0 /0< w in SBa, 8 fixed.
By means of this second delinition \vc obtain at once the follow
ing theorem:
2. Instead of the variables x x m , t l n Z^t tf introduce the
variables ^ y m , Wj w n ^ ^^ * ' ranges over 33s,
ranges over (5s, f/ae correspondence between 33s, CSs being uniform.
Thenf^= $ uniformly in 21 wAe/i '/i^ 0/4^ ^/i^/i
e > 0, 8 > \f <f>\<e , ^ m (3,
3. Example. \
where /a A A
; X
Then </>(o;) = lim/(^, ri) = , in 21 = (0, oo).
71=00
Let us investigate whether the convergence is uniform at the
point x in 2l
First let x > 0. If < a < # < 6, wo have
\f*\<~
170 SERIES OF FUNCTIONS
As the term on the right = as n^= <x> , we see/=0 uniformly
in (a, ft).
When, however, a = 0, or b = ao , this reasoning does not hold.
In this case we set
which gives ^ i og i//i . t
nr/P
As the point (x, ri) ranges over delined by
the point (, w) ranges over a iield X defined by
t > 1 , w > 1,
and the correspondence between and Z i* uniform. Here
The relation 2) shows that when x > 0, t ^ co as 7^ ^ oo ; also
when x 0, = 1 for any n. Thus the convergence at x = is
uniform when ,
The convergence is not uniform at x = when 3) is not satisfied.
For take j
^ = ;, ' w= 1, 2, ...
^A/a
For these values of x ^L A
/ 0 =e a ^
which does not = as n == QO .
146. 1. (Moore, Osyood.) Let
uniformly in 21. i# a be a limiting point of 21
/or each t in Fi*(r).
4> = Km 0(^ . jJ , ^ = lim
^=a / T
exist and are equal. Here a* r are finite or infinite.
GENERAL THEORY 171
We first show <I> exists. To this end we show that
>0 , S>0 ,
n
Now since f(x, f) converges uniformly, there exists an 77 >0
such that for any x', x n in 31
$(x') =/(*', t) + e' t in r,*(r) (2
*(*'0 =/(*", *) + ''. e',e"<*. (8
On the other hand, since / = ty there exists a 8>0 such that
/(*', = *(>+ e"' (4
/(a;", = ^(0 + *' v I '" M *' T  < (5
for any #', x" in FV*(a) ; t fixed.
From 2), 3), 4), 5) we have at once 1). Having established
the existence of <$>, we show now that <J> = Mf. For since f con
verges uniformly to <, we have
/(^0<K*0< > *m a , *iiiF,*(T). (6
o
Since /= i^, we havo
zin Fy*(a) , t fixed in F/(T). (7
3
Since c^> == <t>,
1 ^(a;)  <I> <^ x in IV*O). (8
3
Thus 7), 8) hold simultaneously for 8 < 8', 8".
Hence
or lim
f=T
2. Thus under the conditions of 1)
lim lim f = lim lim f ;
a=a <==T ^=T ai=a
in other words, we may interchange the order of passing to the
limit.
172 SERIES OF FUNCTIONS
3. The theorem in 1 obviously holds when we replace the un
restricted limits, by limits which are subjected to some condition ;
e.g. the variables are to approach their limits along some curve.
4. As a corollary we have :
Let F = S/aO"! x m ) be uniformly convergent in SI, of which x = a
is a limiting point. Let Iimf 8 = / a , and set L = 21 8 . Then
ipft
Urn F = L ; a finite or infinite,
o>=a
or in other words
Urn 2/ a = 2 limf a .
Example 1.
converges uniformly in 21 = (0, oo) as we saw 136, 2, Ex. 1. Here
and i = 2? n =5J = 1.
Hence lim F(x) 1.
<r==oo
Also J21im/ n = 0;
jr = ()
hence jK lim F(x)= 0.
ar =
Example 2.
converges uniformly in any interval finite or infinite, excluding
x = 0, where .F is not defined. For
+
Hence lim F(x) = e.
GENERAL THEORY 173
Example 3.
1
for x =
1 + a?
= fora:=0.
Here lira jP(z) = 1,
o?=0
while Slim /.(a;)= 20 = 0.
35 =
Thus here lim 2/.(*)*21i,n /.(*),
<P = !T =
But F does not converge uniformly at x = 0. On the other
hand, it does converge uniformly at # = oc .
Now lim.FO) = , lim/ n O)=0,
# = ;/, vo 'J
alld liui S
as the theorem requires.
Example 4. rr/ \ _ V f nx * ( n +
  
wliich converges about x = but not uniformly.
However, r v^/\ vv c s \ r\
Inn 2/ n O?) = 2 Inn f n (x) = 0.
#=0 ^^0
Thus the uniform convergence is not a necessary condition.
147. 1. Let lim f^ l x m , ^ f n ) = <(>i ^m) uniformly at
t=T
x~a. Let f(x, t) be continuous at x=afor each t in I r 8 *(r}.
Then </> is continuous at a.
This is a corollary of the MooreOsgood theorem.
For by 146, 1
lira lim /(a f A, ) = lim lim /(a + A, t).
A=0 <=T ^=T /i=o
Hence
im , , + A) = lim /(a, *) = <^(a
174 SERIES OF FUNCTIONS
A direct proof may be given as follows :
f(x, t) = <KaO + e'  e'  < e, x in V^a)
<K*0<K*'0=/<y< /(*", *)+*'
But /O", /(*',  < e , if  z'  x"  < .
2. Z0 ^= 2^ t ..., p (a: 1 # m ) ie uniformly convergent at x=a.
Let each f Sl ... Sp be continuous at a. Then l?(z l # m ) is continuous
at x = a.
Follows at once from 1).
3. In Ex. 3 of 140 we saw that
is discontinuous at x = and does not converge uniformly there.
In Ex. 4 of 140 we saw that
^
(1 + ^ 2 )(1 + (w
does not converge uniformly at x = and yet is continuous there.
We have thus the result : The condition of uniform convergence in
1, is sufficient but not necessary.
Finally, let us note that
is a series which is not uniformly convergent at x = 0, although
F(x) is continuous at this point.
4. Let each term of F ^f fv .^ p (x l x m ) be continuous at x = a
^ itself is discontinuous at a. Then F is not uniformly
convergent.
For if it were, F would be continuous at a, by 2.
Remark. This theorem sometimes enables us to see at once
that a given series is not uniformly convergent. Thus 140,
Exs. 2, 3.
GENERAL THEORY 175
5. The power series P = 2a v ..., m #f i a? is continuous at any
inner point of its rectangular cell of convergence.
For we saw P converges uniformly at this point.
6. The power series P = a f a^ 4 a 2 a: 2 4 is a continuous
function of x in its interval of convergence.
For we saw P converges uniformly in this interval. In par
ticular we note that if P converges at an end point x = e of its
interval of convergence, P is continuous at e.
This fact enables us to prove the theorem on multiplication of
two series which we stated 112, 4, viz. :
148. Let
converge. Ttien AB = C.
For consider the auxiliary series
F(x) = a + r*
Since J., J?, (7 converge, ^, 6r, IT converge for 2:= 1, and hence
absolutely for  a:  < 1. But for all  #  < 1,
Thus L lim HT(aO = i lira F(x) i lim G (a?),
^=1 jr=l Jr=^l
or (7= A A
149. 1. We have seen that we cannot say that /= <f> uniformly
although /and < are continuous. There is, however, an impor
tant case noted by Dini.
Let f(x l ... x m , t n ) be a function of two sets of variables
such that x ranges over 21, and t over a set having r as limiting
point, r finite or ideal. Let
Then we can set
176 SERIES OF FUNCTIONS
Suppose now  ty(x, t')\ <\ *fy(x, )  for any t r in the rectangu
lar cell one of whose vertices is t and whose center is r. We say
then that the convergence of f to <f> is steady or monotone at x.
If for each x in 21, there exists a rectangular cell such that the
above inequality holds, we say the convergence is monotone or
steady in 31.
The modification in this definition for the case that r is an ideal
point is obvious. See I, 314, 315.
2. We may now state Dims theorem.
Let /(#! x m , t l w ) = ^(^i " r m) steadily in the limited com
plete field 31 as t = r; r finite or ideal. Let f and </> be continuous
functions of x in 31. Then f converges uniformly to <p in 31.
For let x be a given point in 31, and
We may take t' so near T that  ^(x, t')\<~
Let x 1 be a point in V^(x). Then
/(*',
As /is continuous in
Similarly,
Tlms \^(x',t')\<e x' in
Hence
, .^ , . ,. / .
, t) I < e for any x in
and for any t in the rectangular cell determined by t 1 .
As corollaries we have :
3. Let Gr = S / tt ...i,(^i Zm) I converge in the limited complete
domain 31. Let Gr and each f t be continuous in 31 Then Gr and
a fortiori F= 2/l t ... ta converge uniformly in 31, furthermore f^... l8 =
uniformly in 31.
4. Let #=2 \f^... l8 (x l x^  converge in the limited complete
domain 31, having a as limiting point. Let Gr and each f, be con
GENERAL THEORY 177
tinuous at a. Then Gf and a fortiori F = 2/ tl ... t , converge uniformly
at a.
5. Let Gr = 2 / tl ...i a (^i 3" m )  converge in the limited complete
domain 21, having a as limiting point. Let lim Gr and each lim/ t
exit. Moreover, let lim # = 2 lim/i
Then Q is uniformly convergent at a.
For if in 4 the function had values assigned them at x = a dif
ferent from their limits, we could redefine them so that they are
continuous at a.
150. 1 . Let lim f(x l x m , t l t n ) = < (x l # m ) uniformly in
t^T
the limited field 21. I>6tf </> i^ limited in 91.
Km r/=
For let y=<j>^r.
Since /= <j> uniformly 1^1 <
for any t in some V*(r) and for any x in 21.
Thus f r
J f~J <f> <
Remark. Instead of supposing <J> to be limited we may suppose
that/(^r, t) is limited in 21 for each t near r.
2. As corollary we have
Let lim/(o; 1 x m , ^ ^ n ) = ^>(^! ^ /n ) uniformly in the limited
field 91. Let f be limited and integrable in W for each t in
Then is integrable in 91 and
lim f/= f(/>= film/.
t~T *S% J% ^21 =T
3. From 1, 2, we have at once:
Let F='Sf ll ... lt (x l "'X m ) be uniformly convergent in the limited
field 91. Let eachf ti ... ts be limited and integrable in 2L Then F is
integrable and p _, /
X I / tl ...i..
JK
178 SERIES OF FUNCTIONS
If thef tl ... Lt are not integrdble^ we have
I yar **
Example. _
does not converge uniformly at x = 0. Of. 140, Ex. 3.
Here ^ =1 l
and p n for #=^0,
~~ JO fora: = 0.
Hence C l TU 1
I Fax J ,
^o
r^ =i Tr
dx
Thus we can integrate J 7 term wise although F does not converge
uniformly in (0, 1).
151. That uniform convergence of the series
with integrable terms, in the interval 31 = (a < 6) is a sufficient
condition for the validity of the relation
X6 rb /6
^d/:= I ^^4 fdx
J,, * l Ja ' '*
is well illustrated grai)lii(;ally, as Osgood has shown,*
Since 1) converges uniformly in 91 by hypothesis, we have
F n (x^ = F(x}F n (x) (2
and
iP n <<6 n>m (3
for any x in 21.
* Bulletin Amer. Math. Soc. (2), vol. 3, p. 69.
GENERAL THEORY
179
In the figure, the graph of F(x*) is drawn heavy. On either
Bide of it are drawn the curves F e, F+e giving the shaded
band which we call the eband.
From 2), 3) we see that the graph of
each F n , n>m lies in the eband. The
figure thus shows at once that
/'
Ja
Fdx
and
F n dx
can differ at most by the area of the
eband, i.e. by at most
152. 1. Let us consider a case where the convergence is not
uniform, as
Here
nx
~^'
If we plot the curves y = F n (x)^ we observe that they flatten
out more and more as n = oo, and approach the 2;axis except
near the origin, where
they have peaks which
increase indefinitely in
height. The curves
F n (x), n>m, and m suf
ficiently large, lie within
an eband about their
limit F(x) in any inter
val which does not in
clude the origin.
If the area of the
region under the peaks
could be made small at
pleasure for m sufficiently large, we could obviously integrate
termwise. But this area is here
180
SERIES OF FUNCTIONS
r 71 j i c a d r * ~b if IT vi i
Jo ^2 Jo &^ aL"^*^ 1 ^
as n === QO .
Thus we cannot integrate the ^ series termwise.
2. As another example in which the convergence is riot uniform
let us consider
Here
The convergence of J 7 is uniform in 31 = (0, 1) except at x = 0.
The peaks of the curves F n (x) all have the height e* 1 .
Obviously the area of the
region under the peaks can be
made small at pleasure if m is
taken sufficiently large. Thus
in this case we can obviously
integrate termwise, although
the convergence is not uniform
in 21.
We may verify this analytically. For
C x n 7 C x nx 7 1 1 f nx . A
I F n dx =1 dx =  ^ = as n = oo .
/o */o e nx n ne nx
3. Finally let us consider
+
Here
n*x
The convergence is not uniform at x = 0.
The peaks of F n (x) are at the points x = w~ 2 , at which points
GENERAL THEORY 181
Their height thus increases indefinitely with n. But at the
same time they become so slender that the area under them == 0.
In fact
Jf >.(*)& J^ id log
2n\_ Jo 2 n
We can therefore integrate term wise in (0 < a).
153. 1. Let Urn Gr(x, ^ t n ) = #(V) in 21 = (a, a f 8), T
or infinite. Let each Gr'^x, t) be continuous in 21 ; also let Gr r x (x,
converge to a limit uniformly in 21 as t = r.
Km (?i<>, =^(^) m 21, (1
tT
and g 1 (x) is continuous.
For by 150, 2,
lim ra f x dx= f'lim ^^.
/= T *^a ^a /= T
By 1, 538, r ,
I G' z dx= G(x, t)G(a, t).
c/a
Also by hypothesis, Hm { & ^ () _ & f) ; = g ^ _ g ^
t=T
Hence ~ x
g(x)g(a)=\ lim &' x (x, t)dx. (2
/a t=r
But by 147, 1, the integrand is continuous in 21.
Hence by I, 537, the derivative of the right side of 2) is this in
tegrand. Differentiating 2), we get 1).
2. Let F(x) = 2/ tl ... la (V) converge in 21 = (a, a
fl(x) be continuous, also let
uniformly convergent in 21.
J ? '(a;)
This is a corollary of 1.
182 SERIES OF FUNCTIONS
3. The more general case that the terms / t ,... t , are functions of
several variables x r x m follows readily from 2.
154. Example.
Here F n * xa
c x
a function whose uniform convergence was studied, 145, a. We saw
F(x)=z$ foranyz>0.
Hence f'Cx)
Let Q, X ,
Then ^ , , _ JT,,^
I*>0,
hence ^ ; (^)= 2/JOc), (2
and we may differentiate the series termwise.
If z=0, and = 1, X>0; (? n (0)= n A = oo as n = oo.
In this case 2) does not hold, and we cannot differentiate the
series termwise.
For a;=0, and >1, (? n (0)=0, and now 2) holds; we may
therefore differentiate the series termwise. But if we look at the
uniform convergence of the series 1), we see this takes place only
when
ft
155. l
x
converge in SI = (a, 6). Jor 0t;ery x in 21 Z# //(X) < <7p constant.
Let Q = 2^ 4 converge. Then F(x) has a derivative in 21 and
or i^e Tway differentiate the given series termwise.
GENERAL THEORY 183
For simplicity let us take s = 1. Let the series on the right of
1) be denoted by $0*0 ^ or eac ^ x * n ^ we suow that
< e,  Arc  < 8.
e>0, S>0, D
AJP
A A*
where  n lies in F(z).
Thus
But (3 1 being convergent, 6r m < e/3 if 77^ is taken sufficiently large.
Hence
On the other hand, since ^ ==/n(^) and since there are only rw
L\X
terms in D m , we may take S so small that
A,<e/3.
Thus J>< forAa;<S.
2. Example 1. Let
This series converges uniformly in 91 = (0 < 6), since
Also
Hence n
As 2# n converges, we may differentiate 1) termwise. In
general we have
OP X
valid in 21.
184 SERIES OF FUNCTIONS
3. Example 2. The ? functions.
These are defined by
flj (z) = 2 2 (  1 ) 0< n+J)i sin (2 tt + 1) TTZ
o
= 2 ql sin TTZ 2 9$ sin 3 THE h
iV 2 (a?) = 2 ! ? <+i> f cos (2 7i + 1) irx
o
= 2 * cos THE J 2 ^2 cos 3 THE f
= 1 + 2 2g n2 cos 2
= 1 f 2 <? cos 2 TTX + 2 5 4 cos 4 TTX+
V (a;) = 1 + 2 2 (  lj) n j n2 cos 2
= 1 25' cos 2 THE f 2 j 4 cos 4 TTX
Let us take , , < ^
Then these series converge uniformly at every point x. For
let us consider as an example v r The series
is convergent since the ratio of two successive terms is
and this == 0. Now each term in >v t is numerically
and hence < the corresponding term in T.
Thus #j (x) is a continuous function of x for every x by 147, 2.
The same is true of the other v's. These functions were discovered
by Abel, and were used by him to express the elliptic functions.
Let us consider now their derivatives.
Making use of 155, 1 it is easy to show that we may differentiate
these series termwise. Then
f>{ O) = 2 TT! (  l) n (2 n + 1) (f+W cos (2 n f 1) TTX
o
= 27r(}i cos TTX 3 9* cos STTX f ).
GENERAL THEORY 185
^ (V) =  2 TrI (2 n + 1) ?< B+ S sin (2 w + 1) trx
= 2?r(^ sin TTX f 3j* sin 3 THE 4 ).
00
V 3 ' (.r) = 4 7r^,nq n2 sin 2 mrx
i
= 4 TT (g sin 2 7r# f 2 j 4 sin 4 TTX + ) .
oo
iS' a; = 4 TT] 1 ) n n n * sin
i
= f 4 TT (^ sin 2 THE 2 q* sin 4 TTO; h ).
To show the uniform convergence of these series, let us con
sider the first and compare it with
The ratio of two successive terms of this series is
2n+1
_
2 ra + 1  j , w8 2 n f 1
which = 0. Thus S is convergent. The rest follows now as
before.
156, 1. Let
uniformly for <  A  < T;, T finite or infinite.
Let G' r (a, exist
for each t near T. Then <f (a) exists and
This is a corollary of 146, 1. Here
G(a + h f) &(<*,
A
takes the place of f(x, f).
2. From 1 WQ have as corollary :
186 SERIES OF FUNCTIONS
converge for each x in 31 which has x = a as a proper limiting point.
Letf((a) exist for each t = (i> v n ). Let
I
converge uniformly with rexpect to h. Then
CHAPTER VI
POWER SERIES
157. On account of their importance in analysis we shall
devote a separate chapter to power series.
We have seen that without loss of generality we may employ
the series , ,+
a 4 a^x + a^ + ... (1
instead of the formally more general one
# 4 a,i(x a) f 2 (^ <* ) 2 4
We have seen that if 1) converges for r = c it converges abso
lutely and uniformly in (7,7) where < 7 <  c \. Finally,
we saw that if c is an end point of its interval of convergence, it
is unilaterally continuous at this point. The series 1) is, of course,
a continuous function of x at any point within its interval of
convergence.
158. 1. Let P(x) = # 4 a^x 4 a^x 2 4 converge in the interval
9( = ( , a} which may not be complete. The series
P n = 1 2 . ... na n + 2 3 . ... O 4 1X +1 * 4 
obtained by differentiating each term of P n times is absolutely and
uniformly convergent in S3 = ( & /3), /3< a,
For since P converges absolutely for a; = /8,
Let now x lie within 93. Then the adjoint series of P^x) is
^ + 2^ +
Now its ?7i th term
187
188 POWER SERIES
But the series whose general term is the last term of the pre
ceding inequality is convergent.
2. Let P = a 4 ^x f a z x 2 h
converge in the interval 21. Then
Q= I Pdx = I a*dx\ I a^xdx f
*^a ^ a ^a
where a, # 7/0 iw 21. Moreover Q considered as a function of x con
verges uniformly in 21.
For by 137, P is uniformly convergent in (a, x). We may
therefore integrate termwise by 150, 3. To show that Q is uni
formly convergent in 21 we observe that P being uniformly con
vergent in 21 we may set
P = P + P
* * m i * m
w ieie
Then
where
on taking <r sufficiently small.
> / o , o small at pleasure.
<<72l<
159. 1. Let us show how the theorems in 2 may be used to
obtain the developments of some of the elementary functions in
power series.
The Logarithmic Series. We have
1 x
for any x in 21 = ( 1*, 1*). Thus
1 x
Hence
lo !)=
This gives also
z) = z + ... ; a: in
GENERAL THEORY 189
The series 1) is also valid for x = 1. For the series is conver
gent for x = 1, and log (1 f x) is continuous at x = 1. We now
apply 147, 6.
For x = 1, we get
2. 2%0 Development of arcsin x. We have by the Binomial
Series i i i Q i o c

V 24 24.
for x in 21 = ( 1*, 1*). Thus
C* dx , 1 , 13 * , ,o
i  = arcsin x = x H  or* f ^  arJ (:i
/o y i _ ^ 28 245
It is also valid for x = 1 . For the series on the right is conver
gent for x = 1. We can thus reason as in 1.
For x = 1 we get
7T , 1 13 1.35
2 23 245 2467
3. 7/40 Arctan fteries. We have
for x in 31 = (  1*, 1*). Thus
C x dx C' r C x
\  ' = arctan x = I dx \ x 2 dx +
x 3 , a; 5 xo
= ,_ + _... (3
valid in 21. The series 3) is valid for x = 1 for the same reason as
in 2.
For x = 1 we get ^ 11 1
4 ~S + 5~7 + '"
4. The Development of e x . We have seen that
*<* ) = 1+ ii + ii + iT + 
converges for any #. Differentiating, we get
190 POWER SERIES
Hence E'(x) = E(x) (a)
for any x. Let us consider now the function
Wehave
e >2x e*
by (a). Thus by I, 400,/(V) is a constant. For # = 0,/(X) = 1.
Hence r ^ ~3
" 1+ fi + ^ + li + 
valid for any x.
5. Development of cos #, sin x.
The series .
converges for every x. Hence, differentiating,
c> = *+*' *L+.
+ ^
Hence adding, C+ C" = 0. (b)
Let us consider now the function
/()= sin x+ C' cos #.
Then Q , gin ^ _ ^^ gin ^ + g,,, cog f
= ((7+ <7")co8a;
= by (b).
Thus /(a;) is a constant. But O 1, C" = 0, fora; = 0, hence
/O)=o,
or (7 sin x f (7' cos a; = 0. (c)
In a similar manner we may show that
or #(X)= C^cosa; O 1 sin 2= 1. (d)
GENERAL THEORY 191
If we multiply (c) by sin # and (d) by cos x and add, we get
(7= cos x. Similarly we get 0' = sin x. Thus finally
, z 2 x 4
oo.*. !_ + _...
valid for any x.
160. 1 . Let P = a m x m f a m+l x m+l f , a m = 0, converge in
some interval 21 about the origin. Then there exists an interval
SB < 21 in which P does not vanish except at x = 0.
Obviously Q converges in 31. It is thus continuous at x = 0.
Since Q = at x = it does not vanish in some interval 33 about
a?*0by I, 351.
In analogy to polynomials, we say P has a zero or root of order
m at the origin.
2. Let P = a 4 a^x f a^ f vanish at the points b v ?> 2 , ... = 0.
2% m a^ Ae coefficients a n = 0. 2%^ points b n are supposed to be
different from each other and from 0.
For by hypothesis P(bn) = 0. But P being continuous at x = 0,
Hence P(0)=0,
and thus A
a = 0.
Hence Px^.
Thus P l vanishes also at the points b n . We can therefore
reason on P l as on P and thus a l = 0. In this way we may
continue.
3. If P = o +
192 POWER SERIES
be equal for the points of an infinite sequence B whose limit is x = 0,
tlien a n = b n , n = 0, 1, 2
For P Q vanishes at the points B.
Hence _ , __ OT 0 1 2...
a n o n u , n u, i, A
4. Obviously if the two series P, $ are equal for all x in a
little interval about the origin, the coefficients of like powers are
equal; that is ^ = ^ ^ n = 0,1,2...
161. 1. Let y = as + a l x
converge in an interval 21. As x ranges over 21, let y range over
an interval 33. Let
converge in 33 Then 2 may be considered as a function of x de
fined in 21. We seek to develop z in a power series in x.
To this end let us raise 1) to the 2, 3, 4 ... powers ; we get
series 2 , , 2 ,
y = a 2o + a 2i x + V +
which converge absolutely within 21.
We note that a mn is a polynomial.
in # , # n with coefficients which are positive integers.
If we put 3) in 2), we get a double series
+ V21' r + *2^22^ ;2 + ' ' ' ( 4
+ ?> 3^31^ + VV 2 + ' ' '
If we sum by rows, we get a series whose sum is evidently 2,
since each row of D is a term of z. Summing by columns we get
a series we denote by
= CQ + C^X f CyX 2 f  (5
GENERAL THEORY 193
c l = Vi + V 2 i + /> 3 a 31 f  (6
We may now state the following theorem, which is a solution of
our problem.
Let the adjoint yseries,
converge for = f to the value rj = r? . Let the adjoint z series
converge for 7; = ?; . Then the z series 2) can be developed into a
power series in x, viz. the series f>), which is valid for \ x \ <  () .
For in the first place, the series 8) converges for 77 <_?? . We
show now that the positive term series
f
converges for < f . We observe that ) differs from Adj D,
at most by its first term. To show the convergence of ) we
have, raising 7) to successive powers,
We note that ^l mn is the same function F m%n of , j, n as
mn is of a , a n , i.^.
^m,n= ^nCfloi "* n)
As the coefficients of F m% n are positive integers,
<*m,n = m f  < 4, n (9
194 POWER SERIES
Putting these values of rj, rj 2 , ?; 3 in 8), we get
A = (/3 4 /^io) ~t~ $i w i? + $\u< 2 ' 4
4
Summing by rows we get a convergent series whose sum is
or 8). But this series converges for ( < since then ?/ < ?/ ,
and 8) converges by hypothesis for 77 = rj Q . Now by 9) each
term of <J) is < than the corresponding term in A. Hence )
converges for f < f .
2. As a corollary of 1 we have :
Let
y = /i 4 ^ f a 2 2: 2 f ...
converge in 21,
converge for all cc < ;y < 4 oo. 7%0w 2 <?aw 6e developed in a
power series in x,
z = <? 4 c^ 4 ^ 4 ... == C Y
/'or aZZ a; within 91.
3. j/0 ^Ae series
y = a m a; m 4 m + i^ m+1 4 , m>_\
converge for some x > 0. If the series
z = fy) + i>\y + *2# 2 +
converges for some y > 0, i (%m i^ developed in a power series
z= r? 4 ^^ 4 r 2 .r' 2 f ...
convergent for some s > 0.
For we may take =  r j > so small that
i? = mf w lm +1 f wlfl f 
has a value which falls within Ihr interval of convergence of
4. Another corollary of 1 is the following :
Let
y= a 4 ! 4 ^ 2 4
GENERAL THEORY 195
converge in 21 = ( A, A). Then y can be developed in a power
series about any point c of 21,
y= e? + c l (x c') f <? 2 0> <
which is valid in an interval 93 whose center is c and lying within 21.
162. 1. As an application of the theorem 161, 1 let us take
" i+ ff + H + fi+~
__ X_ __ X 3 X 5 _
y ~n sT + 57
As the reader already knows,
z = e v , y = sin x,
hence z considered as a function of x is
z = 6 8ln *.
We have
z = lf z + Qx* ^ 3 f Oz 4 f
+ Ja^+  i^ 4 4
+ i* 3 + o  T v^
+ ^+
Summing by columns, we get
^ _ ^,8in x _ 1 i /v. i 1 , r 2 1 ^,4 1 ^.5 1 ^6 . . .
Ze ltXtigX gX ^X 23"$ 3T
2. As a second application let us consider the power series
z =
convergent in the interval 21 = ( 72, 72). Let a; be a point in 31
Let us take 77 > so small that y = x + h lies within 21 for all
 h  <. rj.
Then . , , ,,
s = f ^ (x 4 A)
h a 3 (^+3 2;% f 3 zA 2 + A 8 )
190 POWER SERIES
This may be regarded as a double series. By 161, 1 it may be
summed by columns. Hence
P(x 4 h) = a 4 ax 4 a^x 2 4 a^x 3 4
f A(aj 4 2 a 2 :r 4 3 a^ 4 )
7.2
**+...) (2
21 a !
on using 158, l.
This, as the reader will recognize, is Taylor's development of
the series 1) about the point x. We thus have the theorem :
A power series 1) may be developed in Taylor '# series 3) about
any point x within its interval of convergence. It ix valid for all h
such that x+ h lies within the interval of convergence ofV).
163. 1. The addition, subtraction, and multiplication of power
series may be effected at once by the principles of 111, 112. We
have if P /^.L/T^J.
X I*Q  tl<jC ~
W/ "~~ f) I Q y, J
converge in a common interval 31 :
P  Q = 'v\) + ("A + ^(A)* f
These arc valid within 31, and the first two in 31.
2. Let us now consider the division of P by M. Since
^=P 1 '
R R
the problem of dividing P by R is reduced to that of finding the
reciprocal of a power series.
I** P == a 4 a^ 4 a^ 4 , *=
converge absolutely in R =( 7?, 7Z). Z/^f
$ = a^f a 2 a^4
/>^ numerically < \ a  w 33 = ( ^ /4 ) r < R.
GENERAL THEORY 197
TJieit \/P can be developed in a power series
valid in 33. The fir 8t coefficient <? = .
1
"
a l "i!
for all # in S3* We have now only to apply 1<I1, 1.
8. Suppose n...
To reduce this case to the former, we remark that
P = :rQ
where n .
C = + ^m ^ +
Then 1^1 j_
P .r ' Q'
But l/^ has been treated in 2.
164. 1. Although the reasoning in 161 affords us a method of
determining the coefficients in the development of the quotient of
two power series, there is a more expeditious method applicable
also to many other problems, called the method of undetermined
coefficients. It rests on the hypothesis that/(#) can be developed
in a power series in a certain interval about some point, let us say
the origin. Having assured ourselves on this head, we set
f(x) = a + ap + atf? +
where the a's are undetermined coefficients. We seek enough
relations between the a's to determine as many of them as we
need. The spirit of the method will be readily grasped by the
aid of the following examples.
Let us first prove the following theorem, which will sometimes
shorten our labor.
198 POWER SERIES
2 ' V f(x) = 0,0 + 0,^ + 0,^+ ...; R<x<R, (1
is an even function, the righthand Me can contain only even powers
of x; iff(x) is odd, only odd powers occur on the right.
For if /is even, f( ^ ==/( _ x) . (2
But
/( x) = a  ajtf + <* 2 z 2   (3
Subtracting 3) from 1), we have by 2)
= 2 (a^ + atf? f a$? f )
for all # near the origin. Hence by 160, 2
i = 3 ==a 5 == ' ==:0 
The second part of the theorem is similarly proved.
165. Example 1. /(*) = tan *.
Since sina .
tan ic =  ,
cos x
and ^ & a*
we have
Since cos x > in any interval SB = ( ^ + 8, ^ & J , 3 > 0, it
follows that \Q\<1 in S3.
Thus by 163, 2, tana: can be developed in a power series about
the origin valid in S3. We thus set
tan x = a<iX f ^x 3 + agx 6 f (2
GENERAL THEORY 199
since tana; is an odd function. From 1), 2) we have, clearing
fractions,
Comparing coefficients on each side of this equation gives
a 1 = l.
a, 1 1
a 2! 3!
9 2 ! 4 !
Thus
17
>
62
C ' ^ ' 1 6 ' 315 7 "" 2 nl?^ " V
v^^W in ( , ).
2 ' 2
Example 2. */ ^ 1
^ j (x) = cosec ^ = T
sin a;
1
3! 5!
Since ^ , sin 2:
^
we see that i m ^ 1
when x is in 33=(~7r + S, TT  8), 8>0. Thus xf(x) =
can be developed in a power series in 33. As /(#) ^ s an
function, xf(x) is even, hence its development contains only even
powers of x. Thus we have
f
200 POWER SERIES
Hence
Comparing like coefficients gives
00=1.
k
Thus I 1^1 ^ " 3J . 1
I '}' I *>l> 1 '
"~" I . *' I .  . **' I , .,*
166. Let J p (rc)=/i(A . )+/2 (,. )+ ...
where / , N , ,, , ,
f n (x) = a nQ f a nl x + a n2 x~ + M = 1, 2
Let the adjoint series
*0 + "nl + 2  2 + *
converge for  = R and have c/> n ii.s sums for this value of .
Let cj> = 1 + ^ . h ...
converge. Then jP converges uniformly in ?[ = ( /^ /^) and ^
may be developed as a power series, valid in 31, by summing by
columns the double series
GENERAL THEORY 201
F converges uniformly in 31. For as \x\ <,
We now apply 136, 2 as ^<f> n is convergent for = It.
To prove the latter part of the theorem we observe that
(< 10"f n ft 4 12 A >2 4 '
4 , 20 4 <% ft 4 22^ 2 + ' * '
4 .......
is convergent, since summing it by rows it has <l> as sum. Tlius
tlie double series 1) converges absolutely for #<, by 1*28, 2.
Thus the series 1) may be summed by columns by 130, l and has
JF(jc) as sum, since 1) has .Fas sum on summing by rows.
167. Example.
This series we have seen converges in 21 = (0, J), 6 positive and
arbitrarily large.
Since it is impossible to develop the/ n (V) in a power series about
the origin which will have a common interval of convergence, let
us develop Fin a power scries about o? >0.
We have
1 1 1
1 f a n x 1 f a n x Q , a n (x XQ
"
t+a%l l+a% (l+%) 2
where j __ ( l) K <y n<
202 POWER SERIES
Thus F give rise to the double series
JD = A^ + A'^x  a? ) f A'^(x  z ) 2
"J" "10 ~^~ ^llv 2 ' "" ^o) "^" ^12 v^ ~ *^o)
where * f __ ( l) n j
"WK """" i **n K.
The adjoint series to/ n (V) is, setting f = \x a? ,
This is convergent if
. a * <1 or if $<x y
1 f a n a;
that is, if Q ^
For any # such that # < a; < 2 ^ , g = x x Q .
Then for such an x
A 1 1
(f) =
' n  ! 1 i /<.! X
and the corresponding series
is evidently convergent, since <f> n < 
ft I
We may thus sum D by columns ; we get
F(x) = lB K (xx Q Y
K=0
where
The relation 1) is valid for < x < 2 z .
GENERAL THEORY 203
168. Inversion of a Power Series.
Let the series , , 7 . . , . 2 , ^1
v = b 4 o^ 4 V +  (1
have b 1 ^= 0, and let it converge for t= t Q . If we set
, , f  J o
= 2rt , w =:__fi,
Mo
it goes over into a series of the form
u = x a%x* agfi (2
which converges for x = 1. Without loss of generality we may
suppose that the original series 1) has the form 2) and converges
for x= 1. We shall therefore take the given series to be 2). By
I, 437, 2 the equation 2) defines uniquely a function x of u which
is continuous about the point u = 0, and takes on the value x= 0,
for u = 0.
We show that this function x may be developed in a power
series in w, valid in some interval about u = 0.
To this end let us set
x = u f c 2 u 2 4 c s ifi f (3
and try to determine the coefficient c, so that 3) satisfies 2)
formally. Raising 3) to successive powers, we get
x 2 = w 2 + 2 <? 2 tt 3 f (V + 2 ^X + ( 2 6 4 + 2 <V3> 6 +  '
x 3 = n 3 f 3 V 4 + (3 <? 2 2 f 3 <? 3 > 5 + . (4
z 4 = M 4 + 4 <? 2 w 5 4
Putting these in 2) it becomes
u = w + (6' 2 a 2 )w 2 f (<? 3 2 a 2 tf 2 a 3 )w 3
+ (<? 6  2 2 (c 4 + 6> 2 c 3 )  3 a 3 (e' 2 2 + <? 8 )  4 a 4 <? a  a^u* (5
f ...............
Equating coefficients of like powers of u on both sides of this
equation gives
2 a 2
<? 3 = 2 a 2 c 2 f a 3
<?4 = ^(^ 3 2 + 2 * 8 )+ 8 a 3 ^ 2 f 4 (6
(? 6 = 2 a 2 ( 4 + (? a (? 8 ) 4 3 3 (c 2 2 4 c 8 ) 4 4 <* 4 2 4 5 .
204 POWER SERIES
This method enables us thus to determine the coefficient c in
8) such that this series when put in 2) formally satisfies this
relation. We shall call the series 3) where the coefficients c have
the values given in 6), the inverse series belonging to 2).
Suppose now the inverse series 3) converges for some w ^=0 ;
can we say it satisfies 2) for values of u near the origin ? The
answer is, Yes. For by 101, 3, we may sum by columns the
double series which results by replacing in the right side of 2)
a, # 2 , 3?,
by their values in 3), 4). Hut when we do this, the right side of
2) goes over into the right side of 5), all of whose coefficients
by 0) except the first.
We have therefore only to show that the inverse series con
verges for some u = 0. To show this we make use of the fact that
2) converges for A = 1. Then a n = 0, and hence
 tt n  < some a n = 2, 3, (7
On the other hand, the relations ft) show that
^n=/n(" 2 ' r/ 3' '"<*) ( 8
is a polynomial with integral positive coefficients. In 8) let us
replace a 2 , a 3 * by a, getting
7n=/n(i , ) 0*
Obviously a n \ < y n . (10
Let us now replace all the a\s in 2) by a ; we get the geometric
SOL 1CS if O A / 4 4
u = x w (U, a.t, ... (11
(12
1  x
The inverse series belonging to 11 ) is
x = u + 7 2 2 + 7s w3
where obviously the 7's are the functions 9).
We show now that 11) is convergent about u = 0. For let us
solve 12) ; we get
GENERAL THEORY 205
Let us set 1 2(2 a f l)w f ^ 2 = 1 v. For u near u = 0,
v \ <l. Then by the Binomial Theorem
Vl  v = 1 4 d^v + dtfp H 
Replacing v by its value in w, this becomes a power series in u
which holds for u near the origin, by 161, 3. Thus 14) shows that
x can be developed in a power series about the origin. Thus 13)
converges about u = 0. But then by 10) the inverse series 3)
converges in some interval about u = 0.
We may, therefore, state the theorem:
Let u=b + V + V + V + .&!* 0, (15
con re rye about the point .r = 0. Then this relation defines r as a
function of u which admits the development
r = (M />) P
about the point u b. The coefficients a may be obtained from 15)
/>// the method of undetermined coefficient*.
Example. We saw that
u=iog (1 + ^) = ,.+'^ +  . (1
If we set
,u = / 4 //a* 2 f <V ;>{ 4 rt 4 .* 4 4 ("2
we liavc _ , __  _ i
If we invert 2), wo got
x u 4 <? 2 ?/ 2 4 c,^ f
where e's arc given by (>) in 1<>8. Thus
.. e? 2 = .
120'
206 POWER SERIES
Thus we get
But from 1) we have
2
l + z = e = l + +
which agrees with 8).
Taylor s Development
169. 1. We have seen, I, 409, that if f(x) together with its
first n derivatives are continuous in 31 = (a < 6), then
where ^ ,7^1 n^/i^i
a <.a + h <_b , 0<a<l.
Consider the infinite power series in h.
*'=/<) + ^/'() + f'/'W + " (2
We call it the Taylor's series belonging to f(x). The first n
terms of 1) and 2) are the same. Let us set
(3
n \
We observe that R n is a function of n, A, a and an unknown
variable 6 lying between and 1.
Wehave /( + A)r. + ^.
From this we conclude at once :
If 1, /O) and ite derivatives of every order are continuous in
21 = (a, 6), <md 2
lira fi n = lira ^/<">(a+0A) = , n = oo, (4
n \
a<a + h<b < < 1.
TAYLOR'S DEVELOPMENT 207
Then , r 2
/(a 4 A) /() + Af(a) + y/"(<0 +  (5
The above theorem is called Taylor's theorem; and the equa
tion 5) is the development of f(x) in the interval 21 by Taylor's
series.
Another form of 5) f
/ O)
When the point is the origin, that is, when a = 0, 5) or 6)
gives ^2
/CO =/(0)W'(0) + !/'(()) + . (7
This is called Maclaurins development and the right side of 7)
Maclaurirfs series. It is of course only a special case of Taylor's
development.
2. Let us note the content of Taylor's Theorem. It says :
If 1 f(x) can be developed in this form in the interval
51 = (a < 6) ;
2 if f(x) and all its derivatives are known at the point
x = a ;
then the value of / and all its derivatives are known at every
point x within 31.
The remarkable feature about this result is that the 2 condi
tion requires a knowledge of the values of f(x) in an interval
(a, a f S) as small as we please. Since the values that a func
tion of a real variable takes on in a part of its interval as (a < <?),
have no effect on the values that/(#) may have in the rest of the
interval (c < 6), the condition 1 must impose a condition on f(x)
which obtains throughout the whole interval 31.
170. Let f(x) be developable in a power series about the point a,
viz. let
/ (n) () n i ,9
a n = J y w = 0, 1, (2
n !
i.e. the above series is Taylor's series.
208 1'OWKIl SK1UKS
For differentiating 1) n times, we get
f^(x)= t i\a n + 7 llla n ^
Setting here x= a, we get 2).
The above theorem says that if f(x) can be developed in a
power series about x = a, this series can be no other than Taylor's
series.
171. 1. Tjet f (n) (x) exist and be numerically less than some con
stant M for all a < x < ft, and, for every n. Tien f(x) can be
developed in Taylors series for all x in (, ft).
For then I 7? I M '*"
n !
But obviously v h n A
7 Inn = 0.
>,*, n !
2. 'I'he application of the preceding theorem gives at once:
(2
which are valid for ererif r.
Since * = ^^", tf>0,
we have
, . IOP a , 9 lop <2 </
' 
, .
(4
valid for all x and a > 0.
172. 1. To develop (1 f x)^ and log (1 + #) we need another
expression of the remainder R n due to Cauchy. We shall con
duct our work so as to lead to a very general form for R n .
From 169, 1 we have
TAYLOR'S DEVELOPMENT 209
We introduce the auxiliary function defined over (a, b).
Then
and
M 1 !
Hence r>
/ \ , >
</() (2
If we differentiate 1), we (hid the terms cancel in pairs, leaving
^(^C*^ V^CO (8
We apply now Cauchy's tlieorein, I, 448, introducing another
arbitrary auxiliary function (7(^) which satisfies the conditions
of that theorem.
Then
= / (l a<e<x
" '
Using 2) and 3), we get, since .r = + //,,
where < < 1 .
2. Ifwenet
we have a function which satisfies our conditions. Then 4) becomes
n n = i .f<*\a + eh\ (5
/# I I fJL
a formula due to ISehlo milch and Roche.
For /A = 1, this becomes
n
which is ( y au<Jii/8 formula.
210 POWER SERIES
For /i = ?i, we get from 5)
n I
which is Lagrange's formula already obtained.
173. 1. We consider noiv the development of
(1 f x)* x^>l , p arbitrary.
The corresponding Taylor's series is
We considered this series in 99, where we saw that :
T converges for  x \ < 1 and diverges for  x \ > 1.
When x = 1, T converges only when /JL > 1 ; when x = 1,
T converges only when p 5; 0.
We wish to know when
The cases when I 7 diverges are to be thrown out at once. Con
sider in succession the cases that T converges. We have to
investigate when lim R n = 0.
Case 1. 0<aj<l. It is convenient to use here Cauchy's
form of the remainder. This gives
1 2 n
settinef ! , ^
S M'^1 ' MM + 1
* '
NowinTT n ,
hence lim W n = 0.
TAYLOR'S DEVELOPMENT 211
In u " \i + ex\<i + \x\,
which is finite. Hence U n is < some constant M.
To show that lim S n = 0, we make use of the fact that the series
T converges for the values of x under consideration. Thus for
every /*
since the limit of the n th term of a convergent series is 0. In
this formula replace ^ by p 1, then
1 2 n 1
Hence lim
Thus li mj R
Hence 1) is valid for  x \ < 1.
Case 2. x = 1, /x> 1. We employ here Lagrange's form of
the remainder, which gives
J.
Consider W n . Since increases without limit, p n becomes
and remains negative. As 9 >
lim W n = 0.
For U n , we use I, 143. This shows at once that
lim U n = 0.
Hence
and 1) is valid in this case, i.e. for x = 1, p > 1.
212 POWER SERIES
Case 8. x = 1, JJL 5 0. We use here for /i > the Schlomilch
Roche form of the remainder 172, 5). We set a = 0, h = 1 and get
n
Applying I, 143, we see that lim R n = 0.
Hence 1) is valid here if /t >0.
When /i = equation 1) is evidently true, since both sides
reduce to 1.
Summing up, we have the theorem :
The development of (1 f x)* in Taylor s series is valid when
 x < 1 for all JJL. When x == 4 1 it is necessary that JJL > 1 ;
when x~ \ it is necessary that
2. We note the following formulas obtained from 1), setting
= 1 and 1.
174. 1. We develop now log (1 h #) The corresponding
Taylor's series is
We saw, 89, Ex. 2, that I 7 converges when and only when
 x  < 1 or x 1.
< # < 1 . We use Lagrange's remainder, which gives here
Thus 1
!"'<
Hence lim R n = 0.
TAYLOR'S DEVELOPMENT 213
Let 1 < x < 0. We use here Cauchy's remainder, which
gives, setting x = , < f < 1,
if a. =
n 1
W n =
a 
Evidently j. /S =
Also .
7^7 ^ 1
^n ^ ^ Z
Finally 1
liin TT n = since   < 1.
1 #
We can thus sum up in the theorem :
Taylor's development of log (1 f x) is valid when and only when
\ x  < 1 or x = 1. That is, for such values of x
2. We note the following special case :
I  i + i  i +  = log 2.
The series on the left we have already met with.
175. We add for completeness the development of the follow
ing functions for which it can be shown that Km R n = 0.
536
which is valid for ( 1, 1).
arctan x = x  ^ + ^  ~+ (2
o 5 7
which is valid for ( 1*, 1 ).
 1 a* 1 3 a* L3.5 a* , ,,
log(a:+ VI + * 2 ) = * ~ 2 3" + 274 5 ~ ^TTe Y + '" (
which is valid for ( 1*, 1*).
214 POWER SERIES
176. We wish now to call attention to various false notions
which are prevalent regarding the development of a function in
Taylor's series.
Criticism 1. It is commonly supposed, if the Taylor's series T
belonging to a function /(#) is convergent, that then
/<X> = T.
That this is not always true we proceed to illustrate by various
examples.
Example 1. For f(x) take Cauchy's function, I, 335,
For x*Q #00= ""** ; for x = G Y ( = 0.
1 derivative. For x = 0, C'(x) = \ C(x).
6
For x = 0, C> (0) = lim
/<=o
h
2 derivative. x^Q, C f '(x) = C(x) \ 4  4
I X 2J
# = 0, (7"(0) = lim ^A~~" \ L = ii m jl ^ * 8 o.
3 derivative, x ^ 0, (7"' (a?) = (7 (a?) j ~  ^ + 
a:=0, (7"'(0)= lim^^=0.
A
J^i general we have :
On
h terms of lower degree
x 3 "
x = 0, (7 ( ^(0) = 0.
Thus the corresponding Taylor's series is
T= <7(0) +
TAYLOR'S DEVELOPMENT 215
That is, T is convergent for every x, but vanishes identically.
It is thus obvious that C (x) cannot be developed about the origin
in Taylor's series.
Example 2. Because the Taylor's series about the origin be
longing to C(x) vanishes identically, the reader may be inclined
to regard this example with suspicion, yet without reason.
Let us consider therefore the following function,
/O) = 0(x) + e* = G(x) + g(x).
Then /(*) and its derivatives of every order are continuous.
Since /(n)(a;) = O(H)( .^ + ff(n) ^
n = 1, 2 ...
and <7 (n; (0)=0
we have yr.j(0)=l.
Hence Taylor's development for f(x) about the origin is
y = 1 + T! + S + 3 l + 
This series is convergent, but it does not converge to the right
value since T _ x
L e
177. 1. Example 3. The two preceding examples leave noth
ing to be desired from the standpoint of rigor and simplicity.
They involve, however, a function, namely, C(x), which is not
defined in the usual way; it is therefore interesting to have ex
amples of functions defined in one of the ordinary everyday
ways, e.g. as infinite series. Such examples have been given by
Pringsheim.
The infinite series
defines, as we saw, 155, 2, a function in the interval ?l = (0, 6),
b >0 but otherwise arbitrary, which has derivatives in SI of every
order, viz. :
'. (2
216 POWER SERIES
The Taylor's series about the origin for F(x) is
^0) = J ^ (A) (0) ; X! = 1 for X= 0,
A=0^
and by 2)
X! v y 3 w!
Hence
(3
As A >0 and Km A = ^ ^A+I<^A this series is an alternate series
for any x in 21. Hence T converges in 21.
2. Readers familiar with the elements of the theory of func
tions of a complex variable will know without any further reason
ing that our Taylor's series T given in 3) cannot equal the given
function F in any interval 21, however small b is taken. In fact,
F(x) is an analytic function for which the origin is an essentially
singular point, since F has the poles   n= 1, 2, 3 , whose
limiting point is 0.
3. To show by elementary means that F(x) cannot be devel
oped about the origin in a Taylor's series is not so simple. We
prove now, however, with Pringsheim :
If we take a ^( e ^f=.68 *, T(x) does not equal F(x)
\e \J
throughout any interval 21 = (0, 6), however small 6>0 is taken.
We show 1 that if F(x) = T(x) throughout 21, this relation is
true in 33 = (0, 26*).
In fact let 0<# <6.
By 161, 4 we can develop T about # , getting a relation
ro^icu **) (i
valid for all x sufficiently near # . On the other hand, we saw in
167 that
F(x)^E K {xx,Y (2
o
is also valid for Q<x<*2xQ. But by hypothesis, the two power
series 1) and 2) are equal for points near X Q . Hence they are
TAYLOR'S DEVELOPMENT 217
equal for 0<x<2x Q . As we can take # as near b as we choose,
By repeating the operation often enough, we can show that F =
T in any interval (0, 5) where B > is arbitrarily large.
To prove our theorem we have now only to show F 3=. T for
some one x>0.
Since
F(X ^JA ___ i_Wl t ___ a 1 V,
^ Vl+z l + aav/ \2!l + <Ar 8!l + aW
we have i
f :r 1 f
On the other hand
Hence
To find a value of x for which Gr>_ take # = #"*. For this
value of #
Observe that G considered as a function of a is an increasing
function. For //j_i\2 i
=( i) , = 
\6iy e
Hence JP> 3T for >'*.
178. Criticism 2. It is commonly thought if /(a?) and its
derivatives of every order are continuous in an interval 21, that
then the corresponding Taylor's series is convergent in 21.
That this is not always so is shown by the following example,
due to Pringsheim.
It is easy to see that
converges for every x>_Q, and has derivatives of every order for
these values of z, viz. :
218 POWER SERIES
Taylor's series about the origin is
T = l i (  1) A (X + <r
Tlie series 3T is divergent for x > 0, as is easily seen.
179. Criticism 3. It is commonly thought if f(x) and all its
derivatives vanish for a certain value of x, say for x = a, that
then /(a;) vanishes identically. One reasons thus:
The development of/(V) about x= a is
Asf and all its derivatives vanish at a, this gives
f(x) = +  (x  a) + (x  a) 2 f
= whatever x is.
There are two tacit assumptions which invalidate this conclusion.
First, one assumes because f and all its derivatives exist and
are finite at x = a, that therefore f(x) can be developed in
Taylor's series. An example to the contrary is Cauchy's function
C(x). We have seen that C(x) and all its derivatives are at
x 0, yet 0(x) is not identically 0; in fact vanishes only once,
viz. at x = 0.
Secondly, suppose f(x) were developable in Taylor's series in a
certain interval 21 = (a h, a 4 h). Then / is indeed through
out 21, but we cannot infer that it is therefore outside 21. In
fact, from Dirichlet's definition of a function, the values that/ has
in 21 nowise interferes with our giving / any other values we
please outside of 21.
180. 1. Criticism 4 Suppose f(x) can be developed in Taylor's
series at a, so that
for St=O<i).
TAYLOR'S DEVELOPMENT 219
Since Taylor's series T is a power series, it converges not only
in 21, but also within 93 = (2 a #, a). It is commonly supposed
that f(x) = T also in 93. A moment's reflection shows such an
assumption is unjustified without further conditions on f(x).
2. Example. We construct a function by the method considered
in I, 333, viz.
n= 1 + (1 4 #) n
Then /(z) = cos z, in 21 = (0, 1)
= lf sin x, within 93 = (0, 1).
We have therefore as a development in Taylor's series valid
/w=1 _ +f i_ii + ... =3 ,
It is obviously not valid within 93, although T 7 con verges in 93.
3. We have given in 1) an arithmetical expression for jf(#).
Our example would have been just as conclusive if we had said :
Let f(p) == cos x in 21,
and = 1 f sin x within 33
181. 1. Criticism 5. The following error is sometimes made.
Suppose Taylor's development
valid in 21 = (a < i) .
It may happen that 7 is convergent in a larger interval
One must not therefore suppose that 1) is also valid in 93.
2. Example.
and = 6* + sin (x  6) in 93 = ( J, J9) .
Then Taylor's development
/ (a0 .l + iL + + + ... (1
is valid for 21. The series T converging for every x converges in
93 but 1) is not valid for 93.
220 POWER SERIES
182. Let f(x) have finite derivatives of every order in
31 = (<). In order that f(x) can be developed in the Taylor's
series 2
valid in the interval 21 we saw that it is necessary and sufficient
that
Hut R n is not only a function of the independent variable A, but
of the unknown variable 6 which lies within the interval (0, 1)
and is a function of n and h.
Pringsheim has shown how the above condition may be replaced
by the following one in which 6 is an independent variable.
For the relation 1) to be valid for all h such that 0<^A< H, it is
necessary and sufficient that Cauchifsform of the remainder
n 
4
the h and being independent variables^ converge uniformly to zero
for the rectangle D whose points (A, 0) satisfy
1 It is sufficient. For then there exists for each e > an m
such that
I Rn(k 0) I < n^m
for every point (A, 0) of D.
Let us fix h ; then  R n < no matter how 6 varies with n.
2 It is necessary. Let A be an arbitrary but fixed number in
21 = (0, #*).
We have only to show that, from the existence of 1), for A<C A ,
it follows that
**(*,*)<>
uniformly in the rectangle D, defined by
TAYLOR'S DEVELOPMENT 221
The demonstration depends upon the fact that /2 n (A 0) is h
times the w th term / n (, ) of the development of /'(#) about the
point a + a. In fact let A = a + A. Then by 158
/'(a + A) =/
whose n th term is
.,
7i "" JL .
Let = 6h, then
as stated.
The image A , of D is the half of a square of side A ft , below the
diagonal.
To show that R n converges uniformly to in Z> we have only
to show that * s T\ /\ IA x
/ n (, *)= uniformly in A . (2
To this end we have from 1) for all t in 21
f'(a+ t)=f'(a) + tf(a) + f r "(a)+ (3
Its adjoint
=!/'() +/"()! +  (4
also converges in 21.
By 161, 4 we can develop 4) about t = , which gives
,
w 1!
But obviously <?(, A) is continuous in A , and evidently all its
terms are also continuous there. Therefore by 149, 3,
~ (a) = uniformly in A . (5
,
n 1 !
But if we show that
it follows from 5) that 2) is true. Our theorem is then
established.
222 POWER SERIES
To prove 6) we have from 1)
/(>( + )=/ (n) (<0 + */<*+ (a) + ^/ n + 2) (<0+  (7
and from 4)
The comparison of 7), 8) proves 6).
Circular and Hyperbolic Functions
183. 1. We have defined the circular functions as the length
of certain lines; from this definition their elementary properties
may be deduced as is shown in trigonometry.
From this geometric definition we have obtained an arithmeti
cal expression for these functions. In particular
cos* ! + +... (2
valid for every x.
As an interesting and instructive exercise in the use of series
we propose now to develop some of the properties of these func
tions purely from their definition as infinite series. Let us call
these series respectively S and O.
QI rt /Y* 1
Let us also define tan x =  , sec x =  , etc.
cos x cos x
2. To begin, we observe that both S and converge absolutely
for every #, as we have seen. They therefore define continuous
onevalued functions for every x. Let us designate them by the
usual symbols s{ux ^
We could just as well denote them by any other symbols, as
3. Since S =0 , (7=1 for^
we have sin()==0 cos0==L
CIRCULAR AND HYPERBOLIC FUNCTIONS 223
4. Since S involves only odd powers of #, and only even
powers,
sin x is an odd, cos x is an even function.
5. Since S and are power series which converge for every #,
they have derivatives of every order. In particular
dC __#,^_^ 5 ,# 7 __ __ __ o
Tx" 1 8T~5! 7! '" """ '
Hence dsinx dcoxx . ^
 = cos x ,  = sin x. Co
dx ax
6. To get the addition theorem, let an index as a?, y attached to
, indicate the variable which occurs in the series. Then
_ ,. ,
7! 5!2! 3!4! 6 1
2 I I 3 ! 2 ! 4 !
Adding,
_ , +
7! 5!2! 314!
1! 3! 5!
= 'S'lHT
Thus for every a:, y
sin (a; + ^) = sin x cos y + cos x sin y.
In the same way we find the addition formula for cos a;.
224 POWER SERIES
7. We can get now the important relation
sin 2 a? 4 cos 2 x = 1 (4
directly from the addition theorem. Let us, however, find it by
aid of the series. We have
/!_ J^JL_ _!__!_ 1\
' \(T! + 4 ! 2 ! + 2 ! T! "*" 6!/
,
! 6 ! 2 ! 4 ! 4 ! 6 1 2 ! 8 ]
Hence
Now by I, 96,
Thus
== sin 2 x + cos a a; = 1.
8. In 2 we saw sin 2;, cosa; were continuous for x\ 4) shows
that they are limited and indeed that they lie between 1.
For the left side of 4) is the sum of two positive numbers and
thus neither can be greater than the right side.
9. Let us study the graph of sin a:, cos a;, which we shall call 2
and F, respectively.
(t si 11 (r
Since sin x = 0,  = cos x = 1, for x = 0, 2 cuts the a>axis at
dx
under tin angle of 45 degrees.
CIRCULAR AND HYPERBOLIC FUNCTIONS 225
Similarly we see y = 1 for x = 0. F crosses the #axis there
and is parallel to the a>axis.
and each parenthesis is positive for < x 2 < 6,
sinz>0 for < :r<V<3= 2.449
we see _
eosa;>0 for 0<a:<V'2 = 1.414 ...
Since \ x * + x * *Yi
L ~ 2! + 4i ~ T! V 7 
cosx< for x = 2.
Since D x cos a; = sin x and sin ^ > for < x < Vfi, we see
cos x is a decreasing function for these values of x. As it is con
tinuous and > for x = V2, but < for x = 2, cos # vanishes once
and only once in (V2, 2).
This root, uniquely determined, of cos x we denote by As a
first approximation, we have
V2<f<2.
From 4) we have sin 2  = 1 . As we saw sin x > for x< Vti,
we have
sin= + l.
Thus sin x increases constantly from to 1 while cos x decreases
from 1 to in the interval (0, )= L\> We thus know how sin x,
cos# behave in I r
From the addition theorem
sin ( " } x } sin ~ cos x 4 oos sin # = cos x.
cos ( ~ + x ) = cos ^ cos x sin sii\# = sin x.
\2 y 2 2
226 POWER SERIES
Knowing how sin x, cos# march in 1^ these formulae tell us
how they inarch in j^ = f ^, TrV
From the addition theorem,
sin (TT + x) sin #, cos (TT + #) = cos x.
Knowing how sin #, cos x march in (0, TT), these formulae inform
us about their march in (0, 2 ?r).
The addition theorem now gives
sin (x + 2 TT) = sin #, cos (# + 2 TT) = cos x.
Thus the functions sin x, cos # are periodic and have 2?r as period.
The graph of sin x cos x for negative # is obtained now by
recalling that sin x is odd and cos x is even.
10. As a first approximation of ?r we found
V2 < J < 2.
By the aid of the development given 159, 3
, 3? , O? 6 Z 7 , rx
arctg a? = x   g  +  y 4 5)
we can compute TT as accurately as we please.
In fact, from the addition theorem we deduce readily
S in=lz , cosf=4=
4 V2 4 V2
Hence tan  = 1.
This in 5) gives Leibnitz' s formula,
The convergence of this series is extremely slow. In fact by
81, 3 we see that the error committed in stopping the summation
at the n th term is not greater than   . How much less the
error is, is not stated. Thus to be sure of making an error less
than  it would bv' accessary to take ^(10 m 4 2) terms.
CIRCULAR AND HYPERBOLIC FUNCTIONS 227
11. To get a more rapid means of computation, we make use
of the addition theorem.
To start with, let
a = arctgi .
Then5)gives 1 11 11 11
a rapidly converging series.
f .
^
The error J5/ T a committed in breaking off the summation at the
h term is 1 .,
2/il 5 2 " 1
By virtue of the formula for duplicating the argument
t, 9 _ ^ ^ an a
T^tan2a'
wehave tan'2 =1 V
Similarly * < ___ j 2 o
Let
/8 = 4. (7
The addition theorem gives
tan 8 = ^ =
Then 6) gives . = _i__lj_ , 1_J
ft 289 32S9 8 52S9 6 '" C
also a very rapidly converging series.
We find for the error , ^
2nl 2392"''
The formula 7) in connection with 6) and 8) gives  . The
error on breaking off the summation with the th term is
+
228 POWER SERIES
184. The Hyperbolic Functions. Closely related with the cir
cular functions are the hyperbolic functions. These are defined
by the equations
gZ gX
sinh x = (1
^"* (2
sinh a: e x e~ x
, =
cosh x e x + <'~ x
cosecha? =
cosh a; sinh;
Since 9 ?
we have , 6
. + ... (3
x ~ +  (4
J. I
valid for every #. From these equations we see at once :
sinh ( x) = sinh x ; cosh ( x) = cosh a:.
sinh = 0. cosh 0=1.
(5
(6
Let us now look at the graph of these functions. Since sinh x,
cosh x are continuous functions, their graph is a continuous curve.
For x > 0, sinh a; > since each term in 3) is > 0. The relation
4) shows that cosh x is positive for every x.
If x 1 > x > 0, sinh x 1 > sinh a?, since each term in 3) is greater
for x r than for a;. The same may be seen from 5).
a . ! , ,
sum a;= 1 +
dx
/*2 /v4
^T + 4l + '
. . = cosh x.
d , x ,
 cosh a: = r H
aa; 1 !
X a ,3?,
h 3! + 5! +
= sinh x.
THE HYPEKGEOMETRIC FUNCTION 229
Evidently from 3), 4)
lim sinh x = + oo , lim cosh x = 4 oo .
X=+co Xss+ao
At x = 0, cosh x has a minimum, and sinh x cuts the #axis
at 45.
For x > 0, cosh x > sinh x since
e x + e~ x >e x e~ x .
The two curves approach each other asymptotically as a?== 4oo .
For the difference of their ordinates is e~ x which = as x = f oo .
The addition theorem is easily obtained from that of e x . In fact
. , , e x e~ x
sinh x cosh y =  
= (e* +v + e x ~ v e~ x+v e~ x ~ v
Similarly ^ x+y ___ x _ y e _ x+y _
Hence
sinh x cosh y + cosh x sinh y = (e x+v ' e~ (x + y) } = sinh (x +
Similarly we find
cosh (x + #) = cosh ar cosh # + sinh a; sinh y.
In the same way we may show that
cosh 2 x sinh 2 x = 1.
Hyper geometric Function
185. This function, although known to Wallis, Euler, and the
earlier mathematicians, was first studied in detail by Gauss. It
may be defined by the following power series in x:
1 7 1 2 7.741
2 /S 8+ l
, :
1...7.747 +
The numbers a, y8, 7 are called parameters. We observe that
a, /3 enter symmetrically, also when a = 1, y8 = 7 it reduces to
the geometric series. Finally let us note that 7 cannot be zero or
a negative integer, for then all the denominators after a certain
term = 0.
:0 POWER SERIES
Tlie convergence of the series F was discussed in 100. The
main result obtained there is that F converges absolutely for all
 x  < 1, whatever values the parameters have, excepting of course
7 a negative integer or zero.
186. For special values of the parameters, F reduces to ele
mentary functions in the following cases :
1. If a or is a negative integer r n< F is a polynomial of
degree n.
2. F(l, 1, 2; *)==* log (l+:r). (1
For vn , v> ^ , jc x l x 3
J< (1, 1, 4 #) = 1   +   + ..
Also 2
The relation 1) is now obvious.
Similarly we have
l, 1, 2; 3) = log O
, , .
2 x 1 .r
3. !(  , & /8 ; x) = 1  ^ + "j^ 1 ^ 
= (!).
4. zF(%, ., , s: 2 ) = arcsin a;.
5. ^Xl' !' !'  .c 2 ) = a re tan a;.
6. limJ'fa, 1, 1, *")= e.*. (2
=+ V /
a + 1 + 2 1 .3.8/gV
THE HYPERGEOMETRIC FUNCTION 231
Let < G < /3. Then
is convergent since its argument is numerically < 1. Comparing
3), 4) we see each term of 3) is numerically < the corresponding
term of 4) for any  jr \ < Q and any a > $. Thus the series 3)
considered as a function of is uniformly convergent in the
interval (/3 f oo ) by 136, 2; and hereby x may have any value
in (~ (?, (7). Applying now 146, 4 to 3) and letting = +00,
we see 3) goes over into 2).
7. lim xF( a, u< '.* ; ~ } = sin x. (5
a^+oo \ 2 4 V V
For
Let x = (r > and = Cr. Tlien
, , g  g   
is convergent by 185. We may now reason as in 6.
8. Similarly we may show :
/ 1 2 \
lim F[ a, ,  ;  ~ ) = cos x.
a= + oo \ 2 4 V
O 72
, , "\ T^]= si nh ^.
2 4 a 2
i 2
lim J 7 } , , ,  ) = cosh ^.
24
187. Contiguous Functions. Consider two F functions
JF (a,&7; a?) , ^(', ^, y ; S).
If differs from f by unity, these two functions are said to be
contiguous. The same holds for 0, and also for 7. Thus to
F(a/3yx) correspond 6 contiguous functions,
F(al, 1, 71; af).
232 POWER SERIES
Between F and two of its contiguous functions exists a linear
relation. As the number of such pairs of contiguous functions is
65_ 15
r^ 15 '
there are 15 such linear relations. Let us find one of them.
Q n== <*+ '"* t2 y'' > rc +n ."' + f.
Then the coefficient of x n in F(a/3yx*) is
in jP( + 1, ft, 7, x) it is
in F(a, ft, 7 1, x) it is
In'
.
Thus the coefficient of x n in
/3, 7, a) + ^( + 1, A 7, x)
is 0. This being true for each n, we have
(7   1) J^O, ft, 7, a:) + ^ ( + 1, A %
Again, the coefficient of x n in ^(, /9 1, 7, #) is (/5
in ^(a + 1, /3, 7, #) it is 71(7 + n 1) ^ n .
Hence using the above coefficients, we get
( 7 _ a ^ ft)F(a, ft, 7, a:) + (1  x) F (a + 1, ft, 7, a?)
+ ( y 8 7 )^(, ^31, 7, 2:) = 0. (2
From these two we get others by elimination or by permuting
the first two parameters, which last does not alter the value of
the function F(a/3yx).
Thus permuting a, ft in 1) gives
(7   l)P(a, /3, 7, *) + W* /8 + 1, 7, x)
+ (1  i)P(*> ft, 7  1, ) = 0. (3
THE HYPEKGEOMETRIC FUNCTION 233
Eliminating F(ct, /3, 7 1, x) from 1), 3) gives
08  ) J* (, A 7. *) + ^( + 1, A % a;)
F(,/8+l,7,aO = 0. (4
Permuting a, y9 in 2) gives
( 7  a  /3)^(, A % a?) + 0(1  x)F(a, + 1, 7* x)
+ ( a _ 7 )jp( a _ 1, & 7 , 3) = 0. (5
From 3), 5) let us eliminate F (a, ft f 1^ 7i #)i getting
( _ 1  ( 7 _ _ 1)^)^(0, A 7, *) + (7  )^(  1, A 7, *)
+ (1  7)(1  x)F(a, ft 7 !.)= 0 (6
In 1) let us replace a by a 1 and 7 by 7 4 1 ; we get
(7  a + 1)^(~ 1, A 7 + 1, ar) f (a l)JF(a, ^ 7 + 1, a)
7 J F(al,/9,7,a?)=0. (a)
In 6) let us replace 7 by 7 + 1 ; we get
(^A7^) = 0. (b)
Subtracting (b) from (a), eliminates JP( 1, /3, 7 f 1, #) and
gives
7(1  x}F(afax)  7^0  1, 0, % a?)
4 (7  /8>J T (* A 7 + 1,) = 0. (7
From 6), 7) we can eliminate ^(oc 1, /8, 7, ^), getting
f (7  )(7  /3)^( /8, 7 + 1, )
f 7(1 ~ 7)(1  *)y(, /9, 7  1, a;) = 0. (8
In this manner we may proceed, getting the remaining seven.
188. Conjugate Functions. From the relations between con
tiguous functions we see that a linear relation exists between any
three functions
F(& /3, 7, *0 ^O', ', 7', x) F(a", /3", 7", x)
whose corresponding parameters differ only by integers. Such
functions are called conjugate.
234 POWER SERIES
For let j0, q, r be any three integers. Consider the functions
F(aftyx), F(a + l, ft, /,x)>~F(a+p, ft, 7,2?),
F(a + p, /3 + 1, 7, ), F<* + p, ft + 2, 7, )  J F (a+ j>, ft + q, 7, *),
,aO>^( a +^^
We have p f q h r 4 1 functions, and any 3 consecutive ones
are contiguous. There arc thus p + q h r 1 linear relations
between them. We can thus by elimination get a linear relation
between any three of these functions.
189. Derivatives. We have
2 M 7 7
L 2 w + 1 7 7 h 1 7 h /*
 2X + 1) T72TTr M v r7+i" '"
Hence
F" (, /8, 7, z) = ^ 1" ( + 1, /3 + 1, 7 + 1, *)
= " ' " + ~j+'f +  F ^ + 2, /3 + 2, 7 + 2, *)
and so on for the higher derivatives. We see they are conjugate
functions.
190. Differential Equation for F. Since F, F 1 ', F n are conju
gate functions, a linear relation exists between them. It is found
to be
x(x  1)*"' 4 {(a + ft + V)x  7} F' 4 aftF= 0. (1
To prove the relation let us find the coefficient of x n on the left
side of 1). We set
p == CC '^L +1 ' + 'Hl __ y8 + l ' ^Mijl
n " 1 2 ... n 7 7 hi" 7hV 1
THE IIYPERGKOMETR1C FUNCTION 285
The coefficient of x* in x*F u is
in xF" it is
n(>+
" "'
in ( + /8 + 1)2^' it is
M( +
in 7^' it is
in rtS/ 1 it is
Adding all these gives the coefficient of x n in the left side of 1).
We find it is 0.
191. Expression of F^ajB^x) as an Integral.
We show that for  x \ < 1,
3, 7)JF(/87tf)= iO M)v^i(la^) dw (1
/o
where B{p>> y) is the Beta function of I, 692,
/o
For by the Binomial Theorem
xi x i , , tt 4 1 > o , f 1 + 2
(1 xu)~ a =I + xul  T^ x u H  . o
1 1 2! 1 ^ o
for  xu  < 1. Hence
,7= fVH! M)y* 1 (l^)~ a dtt
A)
236 POWER SERIES
Now from I, 692, 10)
Hence
7 7 +
etc. Putting these values in 2) we get 1).
192. Value of F (a, /3, 7, x) for z = 1.
We saw that the F series converges absolutely for x = 1 if
4 ft 7 < 0. The value of F when x = 1 is particularly in
teresting. As it is now a function of a, /3, 7 only, w,e may denote
it by .F(, /?, 7). The relation between this function and the T
function may be established, as Gauss showed, by means of 187, 8)
V17 *
7 ^ 7 _ !+( + + 1
+ (7  ) (7 
Assuming that a + _ 7 < , (2
we see that the first and second terms are convergent for x = 1 ;
but we cannot say this in general for the third, as it is necessary
for this that a f /3 (7 1) < 0. We cap, however, show that
& 7  1>V?) = 0, (3
supposing 2) to hold. For if  x \ < 1,
J^(, & y 1, a:) = + a^x + a^x* + (4
Now by 100, this series also converges f or x = 1. Thus
lim a n = 0. / (5
n=oo
From 4) we have
(1  x)F(a, & y  1, a) = a + Oi 00)* + 0*2  i)a? +
Let the series on the right be denoted by Q(x}. As
# n+1 (1) = a n , we see Gr (1) is a convergent series, by 5), whose
sum is 0. But then by 147, 6, G(x) is continuous at x = 1.
Hence L lim (}(x)=a (1) = 0,
THE HYPERGEOMETRIC FUNCTION 237
and this Establishes 3). Thus passing to the limit x = 1 in 1)
gives
 *<, A 7) =
Replacing 7 by 7 + 1, this gives
etc. Thus in general
^<X ft, 7) =
Gauss se^e now
,, , , _
^ X) ~
Hence the above relation becomes
! n*
w, 7  a  w, 7  / 
N W
W=oo
For the series
n) ' (6
lira ^(a, ^ 7 + ) = 1. (7
l + ... (8
7 7 7
converges absolutely when 2) holds. Hence
1 . 6? 1 2 (?.+ 1
is convergent. Now each term in 8) is numerically < the corre
sponding term in 9) for any 7 > Gr. Hence 8) converges uni
formly about the point 7 = f oo. We may therefore apply 146, 4.
As each term of 8) has the limit as 7 = + oo, the relation 7)
is established.
238 POWER SERIES
We shall show in the next chapter that
lim n (w, x)
n=Go
exists for all x different from a negative integer. Gauss denotes
it by II (x) ; as we shall see,
T (x) = II (x  1) , for x > 0.
Letting n == 20, 6) gives
j. ( , A 7) = n(7  1)11(7  0D.
< 1/3 ' 7; n (7  1)11(7 1)
We must of course suppose that
7, 7  , 7  ft, 7   /3,
are not negative integers or zero, as otherwise the corresponding
n or F function are not defined.
Bessel Functions
193. 1. The infinite series
converges for every #. For the ratio of two successive terms of
the adjoint series is 1^12
which ==s as * == <x> for any given x.
The series 1) thus define functions of x which are everywhere
continuous. They are called Bessel functions of order
n = 0, 1, 2..
In particular we have
2 2 2 2 4 2 2 2 4 2 6 2
1" '>*" / 7*" T 1
U^ U./ . Ux C/
*^ i /^ 'J
2 2 2 4 2 2  4 2 6 ~ 2 2 4 2 . 6 2 8 '"
Since 1) is a power series, we may differentiate it termwise and
+"i C4
' (
BKSSKL FUNCTIONS 239
2. The following linear relation exists between three consecutive
Bessel functions :
(5
/ *""*  jf. 1). ^ +n ~ l (
" ' l 2 '(n  1 ) ! ,=i ) 2 + * !(!+)!
* ,2*tnl
(T
I leuce
i y/^_ l^
2 1 n 1
JC
\\. We show next that
2 ^(a;) = /_,(*) J^Cz) w>0. (8
For subtracting 7) from <>) ^ives
From 8) we get, on replacing e/ n _ f . 1 by its value as given by 5) :
n>0. (9
X
From 5) we also get
J n ^(x^ n>0. (10
4. The Bessel function J n satish'es the following linear homo
geneous differential equation of the 2 order :
n = Q. (11
240 POWER SERIES
This may be shown by direct differentiation of 1) or more sim
ply thus : Differentiating 9) gives
Ttf _ n T n 77 i Tf
J n=^ J n J " + J "l'
Equation 10) gives
Tf n ~ 1 T T
J n\ =  J n\ "*
X
Replacing here J n _^ by its value as given by 9), we get
Putting this in 12) gives 11).
5. e*^ ='uJ n (x) (13
for any x, and for u 3= 0.
For
<T 2" = e* e 2
"9 **" ~*
(91/92 9 ! o/2 I '
A (A> Zj *j : 66 J
Now for any a; and for any u 3* 0, the series in the braces are
absolutely convergent. Their product may therefore be written
in the form _&, /W _J N
2 V2/2I2I "V
+ 2
BESSEL FUNCTIONS 241
194. 1. Expression of ' J n (x) as an Integral.
* f"co 8 (
n + l\Jo
For
2n + l\
V 2 )
Hence ^ __ JN.
cos (a; cos <) = j ^5 , ^ cos2 *
o ( 8 )
and thus
^ _
cos (a: cos <) sin 2n </> = 2 ^ y ^ cos 2 * sin 2n <.
o
As this series converges uniformly in (0, TT) for any value of x,
we may integrate termwise, getting
jfcos O cos <) sin 2n </)J<^ = J? ^^ &* fcos 2 * <^> sin 2n
(2)! 2 ' 2
2+i\ b j 692
' J '
We shall show in 225, 6, that
1.3.5...
~ 2
Thus the last series above
2
Thus
CHAPTER VII
INFINITE PRODUCTS
195. 1. Let S# 4 ...tJ be an infinite sequence of numbers, the
indices 1 = ^ i a ) ranging over a lattice system in 8 way
space. The symbol p = ^ = ^ (1
fi C
is called an infinite product. The numbers a t are its factors. Let
/^ denote the product of all the factors in the rectangular cell
R ^ If lim P^ (2
/UL 00
is Unite or definitely infinite, we call it the value of P. It is
customary to represent a product and its value by the same letter
when no ambiguity will arise.
When the limit 2) is finite and = or when one of the factors
= 0, we say P is convergent, otherwise P is divergent.
We shall denote by P^ the product obtained by setting all the
factors a, 1, whose indices i lie in the cell JB M . We call this the
coproduct of P^.
The products most often occurring in practice are of the type
go
P = a l . a 2  3  ". = IIa n . (3
The factor P M is here replaced by
and the coproduct P^ by
* m =: ^TO+1 ' ^m+2 * ^nif " '"
Another type is +30
P=na n . (4
=:
The products 3), 4) are simple, the product 1) is 8tuple. The
products 3), 4) may be called oneway and twoway simple products
when necessary to distinguish them.
242
GENERAL THEORY 243
2 ' Let P = l A* f 4. 
Obviously the product P = 0, as
w
Hence P=0, although no factor is zero. Such products are
sailed 2m> products. Now we saw in 1, 77 that the product of a
finite number of factors cannot vanish unless one of its factors
vanishes. For this reason zero products hold an exceptional posi
tion and will not be considered in this work. We therefore have
classed them among the divergent products. In the following
theorems relative to convergence, we shall suppose, for simplicity,
that there are no zero factors.
196. 1. For P = Ha^.. lg to converge it is necessary that each JPV
is convergent. If one of these P^ converges, P is convergent and
The proof is obvious.
2. If the simple product P = a l a 2 3 is convergent, its fac
tors finally remain positive.
For, when P is convergent,  P n \ > some positive number, for
n > some m. If now the factors after a m were not all positive, P n
and P v could have opposite signs v > n, however large n is taken.
Thus P n has no limit.
197. 1. To investigate the convergence or divergence of an
infinite product P = n ti ... t , when a t > 0, it is often convenient to
consider the series
called the associate logarithmic series. Its importance in this con
nection is due to the following theorem :
The infinite product P with positive factors and the infinite series
L converge or diverge simultaneously. When convergent, P = e L ,
L = log P.
For logP M = ^, (1
P, = e'<*. (2
244 INFINITE PRODUCTS
If P is convergent, jP M converges to a finite limit = 0. Hence
Lp is convergent by 1). If L^ is convergent, P^ converges to a
finite limit =0 by 2).
2. Example 1.
, 2,..
is convergent for every x.
For, however large  x \ is taken and then fixed, we can take m
so large that
n>m.
n
Instead of P we may therefore consider P m .
^
But by I, 413
log(l + ?}=* + M n ^, \M n \<M.
\ nj n n 2
Hence L n =li t M n x> \
w+i n*
which is convergent.
The product P occurs in the expression of sin x as an infinite
product.
Let us now consider the product
>=1, 2,
ni
The associate logarithmic series It is a twoway simple series.
We may break it into two parts Z/, 7/", the first extended over
positive w, the second over negative n. We may now reason on
these as we did on the series 3), and conclude that Q converges
for every x.
'6. Example 2.
*
n
is convergent for any x different from
0, 1, 2, 3,
GENERAL THEORY 245
For let p be taken so large that  x \ < p. We show that the
coproduct x , v x
(1+ I
a p = n^ nj
P+I l+ x
n
converges for this x. The corresponding logarithmic series is
n
As each of the series on the right converges, so does L. Hence
Gi converges for this value of x.
198. 1. When the associate logarithmic series
i=21oga tl ... lt , a t >
is convergent,
Hm log = ^ by
iti=
and therefore v ^
lira ^...^ = 1.
t=w
For this reason it is often convenient to write the factors
a tl ... lf of an infinite product JP in the form 1 f 6 tl ... v When P is
written in the form
we shall say it is written in its normal form. The series
we shall call the associate normal series of P.
2. The infinite product
and its associate normal series
converge or diverge simultaneously.
246 INFINITE PRODUCTS
For P and T v , /1 ,
// = 2 log (1 + a t )
converge or diverge simultaneously by 197. But A arid I/ con
verge or diverge simultaneously by 123, 4.
3. If the simple product P == a 1  a 2  a 3 is convergent, <e w ==l.
For by 196, 2 the factors a n finally become > 0, say for n > m.
Hence by 197, l the series
I log a n a n >Q
tim
is convergent. Hence log a n == 0. /. a n == 1.
199. Let R^ < 7? A2 < X == oo ^^ a sequence of rectangular
cells. TJien if P u convergent^
For P is a telescopic series and
200. 1. Let P^
We call ^=
the adjoint of P, and write
<P = Ad j P.
2. P converges, if its adjoint is convergent. We show that
e > 0, \ \P fA P v \<e p,v>\.
Since ^JJ is convergent,
is also convergent by 199. Hence
< $  ^ < X < fJL < V.
But P v P fJL is an integral rational function of the a's with
positive coefficients. Hence
IP P I < <B ^ (\
* v L  +V K** V^
GENERAL THEORY 247
8. When the adjoint of P converges, we say P is absolutely
convergent.
The reader will note that absolute convergence of infinite
products is defined quite differently from that of infinite
series. At first sight one would incline to define the adjoint of
P =!!...,.
tobe <P=n  ...,,!.
With this definition the fundamental theorem 2 would be false.
For let P=n ( l);
its adjoint would be, by this definition,
Now $ n = 1. '$ is convergent. On the other hand,
P n =( l) w and this has no limit, as n =^ oo. Hence P is
divergent.
4. Jn order that P = 11(1 f a tj ... tg ) converge absolutely r , z z
necessary and sufficient that y
converges absolutely.
Follows at once from 198, 2.
Example. ^ , 
i V w 2 /
converges absolutely for every #.
For o i
Vi? 2 ^ /2Vl
^n 2 ^ ^n 2
is convergent.
201. 1. Making use of the reasoning similar to that employed
in 124, we see that with each multiple product
P=iK... u
are associated an infinite number of simple products
<^lTa tt ,
and conversely.
248 INFINITE PRODUCTS
We have now the following theorems :
2. If an associate simple product Q is convergent, so is P, and
P=Q.
For since Q is convergent, we may assume that all the a's are
> by 196, 2. Then
= gSlogaiji, by 124, 3,
= P by 197, l.
3. If the associate simple product Q is absolutely convergent, so
is P.
For let PII(l + *,...,>
<?=n(l+a n ).
Since Q is absolutely convergent,
is convergent. Hence 11(1 4 a tl ... t ) is convergent by 2.
4. 7/e _P=H(1 H^i ) ^ absolutely convergent. Then each
associate simple product Q= 11(1 f # n ) ^ absolutely convergent and
P=Q.
For since P is absolutely convergent,
2<V,
converges by 200, 4. But then by 124, 5
2 n
is convergent. Hence Q is absolutely convergent.
5. If P Ha tl ... la is absolutely convergent, the factors a tli ... t9 >0
if they lie outside of some rectangular cell 7? M .
For since P converges absolutely, any one of its simple associ
ate products Q=Tla n converges. But then a n >0 for n>m, by
198, 3. Thus a v .. t > if t, lies outside of some R^
6. From 5 it follows that in demonstrations regarding abso
lutely convergent products, we may take all the factors > 0,
without loss of generality.
GENERAL THEORY 249
For P = P^P^
and all the factors of P^ are > 0, if /* is sufficiently large. This
we shall feel at liberty to do, without further remark.
7. A=II(l + a lt ..... ) a t >0
and Z = 2 log(l + ,...)
converge or diverge simultaneously.
For if A is convergent,
2<v.e,
is convergent by 200, 4. But then L is convergent by 123, 4.
The converse follows similarly.
202. 1. As in 124, 10 we may form from a given mtuple
product 4IIa,,....,
as infinite number of conjugate wtuple products
where a t = 5 7  if i and/ are corresponding lattice points in the two
systems.
We have now :
2. If A is absolutely convergent, so is B, and A = B.
For by 201, 6, without loss of generality, we may take all the
factors > 0.
Then
= B.
an absolutely convergent mtuple product.
be any p tuple product formed of a part of or all the factors of A.
Then B is absolutely convergent.
250 INFINITE PRODUCTS
For S log a, is convergent.
Hence 2 log fy is.
Arithmetical Operations
203. Absolutely convergent products are commutative, and con
versely.
For let A TT
4 = IIa tl ... lBI
be absolutely convergent. Then its associate simple product
21= lla n
is absolutely convergent and A = 21, by 201, 4 . Let us now re
arrange the factors of A, getting the product . To it corre
sponds a simple associate series 93 and B = 93. But 21 = 93 since
21 is absolutely convergent. Hence A = B.
Conversely, let A be commutative. Then all the factors # 4 ... ljw
finally become > 0. For if not, let
jRj < /? 2 < =00 (1
be a sequence of rectangular cells such that any point of 9i m lies
in some cell. We may arrange the factors a t such that the partial
products corresponding to 1),
1 ' ^2 ' 3 "*
have opposite signs alternately. Then A is not convergent, which
is a contradiction. We may therefore assume all the a's > 0.
Then A 2 log
A = e l
remains unaltered however the factors on the left are rearranged.
Hence v ,
21og<v.. lw
is commutative and therefore absolutely convergent by 124, 8.
Hence the associate simple series
is absolutely convergent by 124, 5. Hence
2n
is convergent and therefore A is absolutely convergent.
ARITHMETICAL OPERATIONS 251
204. 1. Let
4. I,
be absolutely convergent. Then the stuple iterated product
is absolutely convergent and A~ where ij i! 9 is a permutation of
*r '2 ' *
For by 202, 3 all the products of the type
Ua t t IIa t t
t t ..l, I, ...I,
i. !* l
are absolutely convergent, and by I, 324
n = nn.
^i i *i i
Similarly the products of the type
n
l i l a l 
are absolutely convergent and hence
n= n n n.
In this way we continue till we reach A and B.
2. We may obviously generalize 1 as follows :
Let A = Ua t .
4'" 1 *
be absolutely convergent. Let us establish a 1 to 1 correspondence
between the lattice system ? over which i = ( L I ... ,) ranges, and the
lattice system 2ft oi^r which
.9 = O 11,7 12 '".721^22 " JrlJrt "'Jrp)
ranges. Then the ptuple iterated product
JS=II II . ... Ha, 7 f
1 2 r J 1?l> ^
z absolutely convergent, and
A = ^.
252 INFINITE PRODUCTS
3. An important special case of 2 is the following:
Let A = Ua n , 71=1,2,..
converge absolutely. Let us throw the a n into the rectangular array
a n , 12 .
converge absolutely, and
4 = J^JBa J? r .
4. 2%0 convergent infinite product
P = (1 + ^X1
is associative.
For let
^
< wi< ... =00.
We have to show that
0=(l+4i)(
is convergent and P = $.
This, however, is obvious. For
But when w = <x> so does z/.
Hence v ^ ,.
lim Q n = hmP n .
Remark. We note that m m+1 m m may = QO with n.
ARITHMETICAL OPERATIONS 253
205. Let A = Ua^..^ , = Ub ti ... Lt
be convergent. Then
(7=na t .6 t , D = U^
o,
are convergent and
C=A.B , D = ^.
>
Moreover if A, B are absolutely convergent, so are (?, D.
Let us prove the theorem regarding (7 ; the rest follows simi
larly. We have A n
J Of. = Ap Bp.
Now by hypothesis A^ == A^ B^ == B as ft = oo.
Hence ^ = ^.A
To show that is absolutely convergent when A, B are, let us
write a, = 1 f a t , 6 t = 1 f b t and set  a t  = t ,  b t  = y8 t .
Since A, B converge absolutely,
2 log (! + .) , 2 log (1 4 A)
are convergent. Hence
S {log (1 + <*) 4 log (1 + A) I = 2 lo ^ C 1 H O C 1 + A)
is absolutely convergent. Hence C is absolutely convergent
by 201, 7.
206. Example. The following infinite products occur in the
theory of elliptic functions :
They are absolutely convergent for all  q\ < 1.
For the series 2  q* n \ , 2 1 9 2 "" 1 1
are convergent. We apply now 200, 4.
As an exercise let us prove the important relation
P *><} I.
254 INFINITE PRODUCTS
For by 206, /> = IT ( 1 + ? 2n ) ( 1 + J 2 "' 1 ) ( 1 
Now all integers of the type i'n, are of the type 4n'2 or 4w.
Hence by 204, 3,
n (1  r/ 2 ") = 1 1 ( 1  ? 4 ") II (1  ?*" 2 ),
x
P = II ^
= 1
Uniform Convergence
207. Jft A0 limited or unlimited domain 31, ^
i = 21og/ ti ... ta (^....r m ) , / t >0
he uniformly convergent and limited. Then
is uniformly convergent in 21.
For ^ /
F A = e^A
Now L* = L uniformly. Hence by 144, l, F is uniformly con
vergent.
208. If the adjoint of
z' uniformly convergent in 21 (finite or infinite^ F is uniformly
convergent.
For if the adjoint product,
is uniformly convergent, \ve have
!$$  <
for any .> in ?(.
UNIFORM CONVERGENCE 255
But as already noticed in 200, 2, 1)
JFV/M<l***r.
Hence F is uniformly convergent.
209. The product
is uniformly convergent in the limited or unlimited domain $[, if
* = 2<k 1 ..M.OvO , <k=l/J
is limited and uniformly convergent in ?l.
For by 138, 2 the series
is uniformly convergent and limited in 21 Then by 207, the
adjoint of F is uniformly convergent, and hence by 208, F is.
210. Let ^ \ n ^ / \
F(x v  x m ) = n./ ti ... ta (#! ;r w )
be uniformly convergent at x = a. If each fi is continuous at a, F
is also continuous at a.
This is a corollary of 147, 1.
211. 1. Let G = S  / tl ... i t (2i ^ w ) I converge in the limited
complete domain 31 having a as a limiting point. Let Q and each
f t be continuous at a. Then
is continuous at a.
For by 149, 4, G is uniformly convergent. Then by 209, F is
uniformly convergent, and therefore by 210, F is continuous.
2. Let Or =2 / tl ... t8 (^i # m ) converge in the limited complete
domain ?l, having x = a as limiting point. Let
lim/ t = a t , lim Q == Sa t .
256 INFINITE PRODUCTS
For by 149, 6, # is uniformly convergent at x = a. It is also
limited near x = a. Thus by 209,
is uniformly convergent at a. To establish 1) we need now only
to apply 146, l.
212. 1. Let J r =n/ ll ... i /ar) , / t >0 (1
converge in 2l=(, afS). Then
log F=L = Slog/.. (2
If we can differentiate this series termwise in 21 we have
Thus to each infinite product 1) of this kind corresponds an infi
nite series 3). Conditions for termwise differentiation of the series
2) are given in 153, 155, 156. Other conditions will be given in
Chapter XVI.
2. JSxample. Let us consider the infinite product
6(x) =2q*Q sin Trail (1  2 q 2tl cos 2 TTX + q* H ) (1
which occurs in the elliptic functions.
Let us set
1  u n =l  2 2n cos 2 TTX + q* n .
Then  u n \ < ;2  q  2 +  q  4n .
Thus if  q \ < 1, the product 1) is absolutely convergent for any x.
It is uniformly convergent for any x and for  q \ < r< 1.
If it is permissible to differentiate termwise the series obtained
by taking the logarithm of both sides of 1), we get
. (2
4n
^<l2q 2n cos 2
If we denote the terms under the 2 sign in 2) by v n we have
THE CIRCULAR FUNCTIONS
257
Now the series 2a w converges if  q \ < 1. For setting 6 W = 
the series 2J n is convergent in this case. Moreover,
.
b n
Thus we may differentiate term wise.
Tfie Circular Functions
213. 1. Sin # and cos a; as Infinite Products.
From the addition theorem
sin (mx + x) = sin (m f 1)# = sin TWO? cos x 4 cos w# sin x
m = 1, 2, 3 we see that for an odd w
sin nx = a sin n a: + a l sin 11 " 1 x + f # n i g i n #
where the coefficients a are integers. If we set t ?= sin #, we get
sin nx = jP n (0 = a/ f a^^ 1 + + a n i^ (1
Now JF n being a polynomial of degree w, it has n roots. They are
A . 7T . . 2 7T
0, sm , sm ,
n n

2 n
corresponding to the values of x which make sin nx = 0. Thus
F.(l)  V ('  si n j) (( + sin J) ...
Dividing through by
_
n
ir>2:
sn
1 7T
and denoting the new constant factor by , 1), 2) give
sin nx = a sin
1
2
sn
1
in2:
sn
1
INFINITE PRODUCTS
To find a we observe that this equation gives
si Ji nx
sin x
1
sin 2 a;"
. 2 7T
sin 2
n
Letting x= we now get a = n. Thus putting this value of
x
in ,T), and replacing x by , we have finally
sin x = n sin  P(.r,
72
where
/*7T
1 2 " 1
 i, ^, ... o
We note now that as n == oo,
Similarly
sin
. x n .
n sin = x = x.
n x
n
snr 
n
It seems likely therefore that if we pass to the limit n = oo in
l\ we shall get n/ , ^ r
71 & sin^ = r/ > (rr) (5
ivhere
r Fhe correctness of 5) is easily shown.
Let us set
L(x, n) = log P(#, w) = 2 log
L(x) = log P(x) = 2 log l 
sin 2 
THE CIRCULAR FUNCTIONS
We observe that
lim P (x, n) = lim ** n > = e**> = P (x)
259
provided
lim L (#, n) = L (x) .
We have thus only to prove 7). Let us denote the sum of the
first m terms in 6) by L m (x, n*) and the sum of the remaining
l>yZ m (*,w). Then
Since for
we have
7T
2'
 < sin x < x,
2*
 ,
n 4 n^ x*
""
\. (8
and hence for an m l so large that < 1, we have,
log 1
snr
"
. o
Sill 2
log(l
a*
But the series
is convergent. Hence for a sufficiently large m
r > m.
Now giving m this fixed value, obviously for all n > some v the
first term on the right of 8) is < e/3, and thus 7) holds.
260 INFINITE PRODUCTS
2. In algebra we learn that every polynomial
a + a v x + a^x* + + a n x n
can be written as a product
0*O  i)O2>> (sO,
where 04, 2 are its roots. Now
x x* , r 6 /Q
81 n*= +.,.. (9
is the limit of a polynomial, viz. the first n terms of 9). It is
natural to ask, Can we not express sin x as the limit of a product
which vanishes at the zeros of sin x ? That this can be done we
have just shown in 1.
3. If we set x = 7r/2 in 5), it gives,
Hence ^ 2r2r = 2 2 4 4 (S (i ... n()
2 U (2rl)(2r+D 1 3 3 5 5~. 7 ' ^
a formula due to
4. From 5) we can get another expression for sin #, viz. :
siux^xU(l~^\e^ r= 1, 2, ... (11
For the right side is convergent by 197, 2. If now we group
the factors in pairs, we have
This shows that the products in 5) and 11) are equal.
5. From 5) or 11) we have
sin x = lim P n (x) = lim x ri' ?J!T
n=oo sn 87T
where the dash indicates that s = is excluded.
THE CIRCULAR FUNCTIONS 261
214. We now show that
To this end we use the relation
sin 2 x = 2 sin x cos x.
Hence
cos = =
from which 1) is immediate.
From 1) we have, as in 213, 4,
^ n = 0, 1, 2, ... (2
215. From the expression of sin ar, cos # as infinite products,
their periodicity is readily shown. Thus from 213, 12)
sin x = lim P n (x).
.
'_ _. _ I as n = Q0>
P n (^;) rwir
Hence lim p ^ (rr + ^ = _ Hm p^^^
sin (x + TT) = sin #.
Hence . ^ , \
sin(a: f 2 TT) = sin #
and thus sin x admits the period 2 TT.
216. 1. Infinite Series for tan #, cosec a?, etc.
If O<#<TT, all the factors in the product 213, 5) are positive.
Thus
262 INFINITE PRODUCTS
Similarly 214, 1) gives
. (2
To get formula) having a wider range we have only to square
the products 213, 5) and 214, 1). We then get
log sin 2 x = log a* + 2 log (l  ^Y, (3
valid for any x such that sin x = ; and
/ 4# 2 \ 2
logeo s ^log^l (28 _ i)%2 j, (4
valid for any x such that cos#^0.
If we differentiate 3), 4) we get
cot * =
x
tan:c=2 T  (6
lid as in 3), 4).
Remark. The relations 5), 6) exhibit cot #, tana; as a series of
rational functions whose poles are precisely the poles of the given
functions. They are analogous to the representation in algebra
of a fraction as the sum of partial fractions.
2. To get developments of sec a?, cosec a?, we observe that
cosec x = tan \ x + cot x.
Hence
COSeC Z=2 2/ TZ^ TTo 12+~
2 a?
_ ,
valid for
THE CIRCULAR FUNCTIONS
3. To get sec #, we observe that
cosec f ~ x J = sec x.
\** 1
263
Now
cosec x =  
x
X 87T f X )
1 1
Hence
2"~ y TT
Let us regroup the terms of 8, forming the scries
I 1 1 1 f 1 11
+  o +o +
7T 7T
2 w  1
  ,
we see that 7 is convergent and = 8. Thus
valid for all x such that cos x = 0.
217. As an exercise let us show the periodicity of cot x from
216, 5). We have
n \
cot x = lim F n (x) = lim V
5= n
' w ^+( + i>.
Letting n = oo we see that
lim 1^(2; + TT) = lim .F n (a;
cot (x f TT) = cot a?.
nir
and hence
264 INFINITE PRODUCTS
218. Development of log siii x, tan z, etc., in power series.
From 216, 1)
If we give to ^L?its limiting value 1 as x = 0, the relation 1)
holds for  x \ < TT.
Now for x < TT
Thus
, sin x x* , 1 a; 4 , 1 .r
_l () o  =3 + 4 + 01
./' 7T 2 2 7T 4 > 7T 6
r l 1 J A 1 ^
i x i l _~ i l _J L ...
^ 32 ^2^2 3 4 7T 4 8 3 6 7T 6
4
provided we sum this double series by rows. But since the series
is a positive term series, we may sum by columns, by 129, 2.
Doing this we get
sn x
whoro
1.1,1,1
? relation 2) i valid for \x\ < TT.
In a similar manner we iind
7T
valid for \ x \ <  Here
THE CIRCULAR FUNCTIONS 265
The terms of Q n are a part of ff n . Obviously
Tliese coefficients put in 3) give
log w*x=(&l)ff
'TT"  ' "TT* " " ' ~7T 6
valid for  x \ <  If we differentiate 4) and 2), we get
7T 2 7T 4 7T 6
valid for \ x \ <  ;
cot *= i  2 # 2 ^ _ 2 # 4 ^  2 JT,^  ... (6
^ 7T 7T* 7T
valid for < \x\ < TT.
Comparing 5) with the development of tan x given 165, 3)
gives
111 _2 1 <) _2 O '2
77 L JL J 7T I ^ 7T 15 ^i 7T
2 ~~ 72 ~" 92 "* ^2 """ * " "" 7r "" H ' *> ! "~ J ' 9 !
JL w > 'J vi rf . w
zr = l , 1 ,1, = Z^ = ?i^ = R.?i^ /?
4 I 4 2 4 3 4 1>0 30 ' 4 ! 3 4! ^
// = j_tj_ = jl_l. ?!^! = **
6 is^^^.s 6 !>4f) 42 ' ti! 5 ' 6!
rr _ 1 , 1 , 1 , 7T 8 1 2' 7T 8 V 7T R
"8  a ' ,.0 ' .>u I"
1 s 2 8 H 8 J t50 30 8 ! 7 8 !
Let us set I72n _ t 2n
TT" ^ """ D /"Q
_/Z 2n :== ~ **' > n 1* V
Then 5) gives
valid for a;< The coefficients B v B% are called
nouillian numbers. From 7) we see
266 INFINITE PRODUCTS
From 6), 8) we get
cotan*^!^^*'"' (10
valid for <  x \ < ?r.
219. Recursion formula for the Bernouillian Numbers.
If we set f() == ^ an Xt >
we have by Taylor's development
where ^2ni) (0 ) _ 2(2 2  I)g 2n _ 2 2 "(2 2  1)
(2nl)!~ 7T 2 " ~ (2)I 2 " 1
Now by I, 408,
From 1), 2) we get
i(2 2  1) D /2n  1\ 2 ln  3 (^" z  1) R
O2nl ~ I 2 J  I7~j  "2n3
n\ i _... =( _ 1) .., (3
We have already found jB r 5 3 , J^ 5 , ^ 7 ; it is now easy to find
successively :
Thus to calculate i? 9 , we have from 3)
98 27(2 8  1) _1_
12 4 '30 1.2.3.4 3 42
2 9 (2 10 ~~ 1) _ 98 27(2 8  1) _1_ 9 8> 7 6 2 5 ( 2 6 ~
'
^ 9.8.7 23(2* 1) .1, 9 . 2r2 2 _ n . 1  l
1.23 2 30 + C 1; 6"
Thus
* = 512 Ao23 51  9 + 168  2016 4 9792}
^ 5 . 7936 = ^
512 1023 66'
THE B AND T FUNCTIONS 267
The B and F Functions
220. In Volume I we defined the B and F functions by means
of integrals: ... x ,
BO, v) = I ^ (1
v ' Jo (l + a: )+* v
Xoo
e*^ 1 ^ (2
which converge only when u, v > 0. Under this condition we saw

We propose to show that F(M) can be developed in the infinite
product / 1 \ u
i( 1 + )
# = ln^ _ n L. (4
J
n
This product converges, as we saw, 197, 3, for any u^Q, 1,
2, From 201, 7 and 207 it is obvious that Gr converges abso
lutely and uniformly at any point u different from these singular
points. Thus the expression 4) has a wider domain of definition
than that of 2). Since Gr = F, as we said, for w>0, we shall ex
tend the definition of the F function in accordance with 4), for
negative u.
It frequently happens that a function f(x) can be represented
by different analytic expressions whose domains of convergence
are different. For example, we saw 218, 9), that tan x can be de
veloped in a power series
valid f or  x \ < . On the other hand,
x_ __ x a _
1~!~~3T 5l "" sin a;
tan x =  5  z  ~ 
x* x* cosx
268 INFINITE PRODUCTS
and ^
tan. = 2S * by 216, 6)
are analytic expressions valid for every x for which the function
tan x is defined.
221. 1. Before showing that Q and F have the same values for
u > 0, let us develop some of the properties of the product # given
in 220, 4). In the first place, we have, by 210:
The function G(u) is continuous, except at the points u = 0, 1,
2,..
Since the factors of 4) are all positive for u > 0, we see that
Q(u) is positive for u > 0.
2. In the vicinity of the point x = m, m = 0, 1,
G(u) = H(u ^
x f m
where H(u) is continuous near this point, and does not vanish at
this point.
For
'! + 
m
where //is tho infinite product Gr with one factor left out. As we
may reason on /fas we did on Q, we see H converges at the point
x = m. Hence H^ at this point. But II also converges uni
formly about this point; hence /Tis continuous about it.
r ... M ^
= lim  ^     n u . (\
n=ao W (u + 1)(W h 2) ..(M + W 1)
To prove this relation, let us denote the product under the limit
sign by P n . We have
THE B AND T FUNCTIONS 269
Also
Thus P n = G n . But Q n = 6r, hence P B , is convergent and (} =
lira P n .
223. Huler's Constant. This is defined by the convergent series
It is easy to see at once that
by 218, 7). By calculation it is found that
C' = .577215
224. Another expression of Gr is
n)
where is the Eulerian constant.
For when a > 0, a u = e u Ioga .
Hence
Now
and
270 INFINITE PRODUCTS
are convergent. Hence
from which 1 ) follows at once, using 223.
225. Further Properties of Q.
1. #O+l)=M#(tt). (1
Let us use the product
employed in 222. Then
AO + i) = MMPn(M)  (2
M fw
= M as /i == QO
u h n
we get 1) from 2) at once on passing to the limit.
2. G(u+ n)=u(u+ 1) (u + n 1)<7(V). (3
This follows from 1) by repeated applications.
3. Gr(n) = 1 2 n 1 = (n 1) ! (4
where n is a positive integer.
sin TTU
For QCL u^ss uGKu} bv 1
eG " , , by 224, 1).
Hence i .C'W.C'M
. e rt
wy
i i
THE B AND T FUNCTIONS 271
We now use 213, 5).
Let us note that by virtue of 1, 2 the value of Q is known for
all u > 0, when it is known in the interval (0, 1). By virtue of
5) Q is known for u < when its value is known for u > 0.
Moreover the relation 5) shows the value of G is known in (, 1)
when its value is known in (0, J).
As a result of this we see & is known when its values in the
interval (0, ) are known ; or indeed in any interval of length \.
Gauss has given a table of log Gr(u) for 1<^<1.5 calculated
to 20 decimal places. A fourplace table is given in " A Short
Table of Integrals " by B. 0. Peirce, for 1 < u < 2.
5. <?() = VT^. (6
For in 5) set u = ^. Then
Hence
We must take the plus sign here, since Gr > when u > 0, by 221.
where n is a positive integer.
, etc.
Thas 2+l\_2nl 2. 8 3 1
mas
226. Expressions for log (?(w), and if Derivatives.
From 224, 1) we have for w > 0,
i(0 = log ff O) =  O^
Differentiating, we get
Iw u + n}
That this step is permissible follows from 155, 1.
272 INFINITE PRODUCTS
We may write 2)
' =  (7+ V f i    1  (3
rf (n u+ n  1 J
That the relations 2), 3) hold for any u^ 0,  1,  2 follows
by reasoning similar to that employed in 216. In general we have
^ , ,>1. (4
In particular,
Z/(l) = _ O. (5
227. Development of log 6r(w) i/i # Power Series. If Taylor's
development is valid about the point ?/ = 1, we have
log #< = < = i(l) + =li L'd)
or using 226, 5), and setting u = 1 4 a;,
log
We show now this relation is valid for Jj r < x < 1, by proving
that
converges to 0, as s == QO .
For, if 0<x<l, then
Also if 
i i
x=o.
The relation 1) is really valid for 1 <#<_ 1, but for our pur
pose it suffices to know that it holds in 31 = ( , 1). Legendre
THE B AND T FUNCTIONS 27:3
has shown how the series 1) may be made to converge more
rapidly. We have for any x in 81
log (1 4 x) = x  I ( 1)"
2 n
This on adding and subtracting from 1) gives
log 0(1 + *) = log(l +*)+ (1 # )*+(
2
Changing here x into x gives
log &(i  *) =  log (i  *)  (i  c> +
Subtracting this from the foregoing gives
log 0(1 +x) log 0(1  x)
From 225, 4
log 0(1 + X) + log 0(1  2?) = log ~
sin TTX
This with the preceding relation gives
log 0(1 + *)
a^t^T^^*^*!^ 1 ^ (2
valid in 31.
This series converges rapidly for 0<#<, and enables us to
compute 0(u) in the interval l<w<. The other values of 6r
may be readily obtained as already observed.
228. 1. We show now with Prinf/sheim* that 0(^) =F(wK for
u>0.
We have for <><_!,
l\u 4 w)=
Math, Anuttlen, vol. .'Jl, p. 466.
274 INFINITE PRODUCTS
Now for any x in the interval (0, w),
x u <n u , x u >xn u ~ l
since u > and u 1 < 0.
Also for any # in the interval (n, oo )
x u <xn u ~ l , x u >n u .
Hence
u  l je x x n dx+n u e' x x n ~ l dx<T(u 4 ft)
/* xao
< rc u I ^^^^^f /i 1 *" 1 I e~ x x n dx.
*/0 /n
Thus
n u
< I e~ x x n ~ l dx\ I e~ x x n dx  I e~ x x n dx.
*so n*/o n^o
Let us call these integrals A, J5, respectively.
We see at once that
= n 1== j ,
n n
Also, integrating by parts,
[e~ x x n ~\ n . 1 C"  x n7 n n , n
\  _j_ _ I e * x n dx =  h O.
L n J n* 70 ?ie n
Thus
Hence
where
Now
THE B AND T FUNCTIONS 275
But
"n>l + 77+ + 7 . 1N n , ; r , for any w
(n + 1) (w + m)
Let us take
or <.
w m
Then
ra m
Since m may be taken large at pleasure,
lim z/ n = QO
an*d hence ,. A
hm j n = 0.
Thus , ,
But from T(u + 1) = wF(^) we have
?i u ^(w 1) ! w n u (n 1) !
also, as n = QO . Thus the relation 1) holds for 1 < w< 2, and in
fact for any w
As
we have
v J ( + !)... ( + !)
Hence using 1), ^ (!)! . T(.
V. y X^ . 1 \ X . IN x'
Letting ^= oo , we get T(u)= Gr(u) for any w>0, making use
of 1) and 222, 1).
2. Having extended the definition of T(u) to negative values
of w, we may now take the relation
as a definition of the B function. This definition will be in
accordance with 220, 1) for w, v > 0, and will define B for negative
w, v when the right side of 2) has a value.
CHAPTER VIII
AGGREGATES
Equivalence
229. 1. Up to the present the aggregates we have dealt with
have been point aggregates. We now consider aggregates in
general. Any collection of welldetermined objects, distinguish
able one from another, and thought of as a whole, may be called
an aggregate or set.
Thus the class of prime numbers, the class of integrable func
tions, the inhabitants of the United States, are aggregates.
Some of the definitions given for point aggregates apply obvi
ously to aggregates in general, and we shall therefore not repeat
them here, as it is only necessary to replace the term point by
object or element.
As in point sets, 31 = shall mean that 31 embraces no elements.
Let 91, 33 be two aggregates such that each element a of 91 is
associated with some one element b of 33, and conversely. We say
that 21 is equivalent to 33 and write
cyr ci(\
l ~ 3O
We also say 31 and 33 are in one to one correspondence or are in
uniform correspondence. To indicate that a is associated with b
in this correspondence we write
a ~ b.
2. If 21 ~ 33 and 33 ~ 6, then 91 2.
For let a^b, b ~ c. Then we can set 91, ( in uniform corre
spondence by setting a ~ c.
3. Let 91 = 33+S + >+
A = S + C+ D 4
// 33  B, 6  C, , then 91^1.
276
EQUIVALENCE 277
For we can associate the elements of 21 with those of A by
keeping precisely the correspondence which exists between the
elements of S3 and J9, of ( and (7, etc.
Example 1. 21 = 1, 2, 3, ...
If we set a n ~ ft, 91 and S3 will stand in 1, 1 correspondence.
ExampleS. 21 = 1, 2, 3, 4,
S3 = 2, 4, 6, 8, ...
If we set n of 91 in correspondence with 2 n of S3, 91 and S3 will
be in uniform correspondence.
We note that S3 is a part of 91 ; we have thus this result : An
infinite aggregate may be put in uniform correspondence with a
partial aggregate of itself.
This is obviously impossible if 9t is finite.
Example 3. 21 = 1, 2, 3, 4, ...
S3 = 10 1 , 10 2 , 10 3 , 10 4 , ...
If we set n ~ 10 n , we establish a uniform correspondence be
tween 9t and S3. We note again that 91 ~ S3 although 91 > S3.
Example 4 Let & = j(, where, using the triadic system,
=& 8  . = 0.2
denote the Cantor set of I, 272. Let us associate with f the point
#=.2:^3 ... (1
where x n = when n = 0, and = 1 when n = 2 and read 1) in
the dyadic system.
Then \x\ is the interval (0, 1). Thus we have established a
uniform correspondence between and the points of a unit interval.
In passing let us note that if < and x, x 1 are the correspond
ing points in {#}, then x <x f .
This example also shows that we can set in uniform correspond
ence a discrete aggregate with the unit interval.
We have only to prove that is discrete. To this end consider
the set of intervals marked heavy in the figure of I, 272. Ob
278 AGGREGATES
viously we can select enough of these deleted intervals so that
their lower content is as near 1 as we choose. Thus
Cont (7=1.
As Cont C < 1, C is metric and its content is 1. Hence is
discrete.
230. 1. Let 91 = a f J., 33 = /? 4 J?, where a, b are elements
0/91, 33 respectively. If^K^ 33, then A ~ B and conversely.
For, since 21 ~ 33, each element a of 21 is associated with some
one element b of 33, and the same holds for 33. If it so happens
that a ~ /?, the uniform correspondence of A, B is obvious. If
on the contrary a. ~ b' and /3 ~ a', the uniform correspondence be
tween A, B can be established by setting a 1 ~~ b r and having the
other elements in A, B correspond as in 3l~ 33.
2. We state as obvious the theorems:
No part 33 of a finite set 91 can be ~ 31.
No finite part 33 of an infinite set 31 can be ~ 31.
Cardinal Numbers
231. 1. We attach now to each aggregate 31 an attribute
called its cardinal number, which is defined as follows :
1 Equivalent aggregates have the same cardinal number.
2 If 91 is ~ to a part of 33, but 33 is not ~ 31 or to any part
of 31, the cardinal number of 91 is less than that of 33, or the
cardinal number of 33 is greater than that of 31. The cardinal
number of 91 may be denoted by the corresponding small letter
a or by Card 31.
The cardinal number of an aggregate is sometimes called its
power or potency.
If 91 is a finite set, let it consist of n objects or elements.
Then its cardinal number shall be n. The cardinal number of
a finite set is said to be finite, otherwise transfinite. It follows
from the preceding definition that all transfinite cardinal num
bers are greater than any finite cardinal number.
CARDINAL NUMBERS 279
2. It is a property of any two finite cardinal numbers a, b that
either t
o = b , or a > b , or a < b. (1
This property has not yet been established for transfmite car
dinal numbers. There is in fact a fourth alternative relative to
31, 33, besides the three involved in 1). For until the contrary
lias been shown, there is the possibility that :
No part of 91 is ~ 53, and no part of 53 is ~ 21.
The reader should thus guard against expressly or tacitly
assuming that one of the three relations 1) must hold for any
two cardinal numbers.
3. We note here another difference. If 21, 53 are finite with
out common element,
Card (21 + 53) > Card 21. (2
Let now 21 denote the positive even and 53 the positive odd
numbers. Obviously
Card (21 + 53) = Card 21 = Card 53
and the relation 2) does not hold for these transfinite numbers.
4. We have, however, the following :
Let 21 > 53, then
Card 21 > Card S3.
For obviously 53 is ~ to a part of 21, viz. 53 itself.
5. This may be generalized as follows :
Let
If Card 53 < Card B , Card < Card <7, etc.,
then Card 21 < Card A.
For from Card 53 < Card B follows that we can associate in 1,
1 correspondence the elements of 53 with a part or whole of B.
The same is true for , tf; , Z>;
Thus we can associate the elements of 21 with a part or the
whole of A.
280 AGGREGATES
Enumerable Sets
232. 1. An aggregate which is equivalent to the system of
positive integers $ or to a part of Q is enumerable.
Thus all finite aggregates are enumerable. The cardinal num
ber attached to an infinite enumerable set is K , aleph zero.
At times we shall also denote this cardinal by e, so that
2. Every infinite aggregate 91 contains an infinite enumerable set 33.
For let a 1 be an element of 2( and
Then 21 x is infinite ; let a 2 be one of its elements and
Then 21 2 is infinite, etc.
Then ^ saa
is a part of 91 and forms an infinite enumerable set.
3. From this follows that
K is the least transfinite cardinal number.
233. The rational numbers are enumerable.
For any rational number may be written
(\
n
where, as usual, m is relatively prime to n.
The equation
admits but a finite number of solutions for each value of
p = 2, 3, 4, ...
Each solution m, n of 2), these numbers being relatively prime,
gives a rational number 1). Thus we get, e.g.
p = 2 , 1.
/> = 8 , 2, J.
jt> = 4 , 3, J.
j = 5 , 4, J ,  .
ENUMERABLE SETS 281
Let us now arrange these solutions in a sequence, putting those
corresponding to p = q before those corresponding to p=sq + 1.
We
r i * r 2 * r a " v, rf
which is obviously enumerable.
234. Let the indices t x , * 2 , * p range over enumerable sets. Then
is enumerable.
For the equation __
where the z/s are positive integers, admits but a finite number
of solutions for each n = p, _p + 1, p + 2, p43 Thus the
elements of ^ _ ,, ,
may be arranged in a sequence
by giving to n successively the values p, p f 1, and putting the
elements b Vi ... Vp corresponding to n = q + 1 after those correspond
ing to n = q.
Thus the set 48 is enumerable. Consider now 31. Since each
index i m ranges over an enumerable set, each value of i m as i' m is
associated with some positive integer as m f and conversely. We
may now establish a 1, 1 correspondence between 21 and $Q by
setting
J >nX";~ ai ; i ;"v
Hence 21 is enumerable.
235. 1. An enumerable set of enumerable aggregates form an
enumerable aggregate.
For let 21, 33, 6 be the original aggregates. Since they form
an enumerable set, they can be arranged in the order
2lj , 2( 2 ? 2ls i * (1
But each 2l m is enumerable ; therefore its elements can be
arranged in the order
282 AGGREGATES
Thus the aelements in 1) form a set
\a mn \ m, n,= 1, 2,
which is enumerable by 234.
2. The real algebraic numbers form an enumerable set.
For each algebraic number is a root of a uniquely determined
irreducible equation of the form
the a's being rational numbers. Thus the totality of real algebraic
numbers may be represented by
\Pn, a,a 3  a n i
where the index n runs over the positive integers and a^*** a n range
over the rational numbers.
3. Let 31, 33 be two enumerable sets. Then
Card 91= Card = K .
Card (H + ) = Ko.
And in general if Slj , 91 2 are an enumerable set of enumerable
aggregates, Card (Slj, 2I 2 , ) = K .
This follows from 1.
236. Every isolated aggregate 21, limited or not, forms an enumer
able set.
For let us divide $R m into cubes of side 1. Obviously these form
an enumerable set^Cp CIj""' About each point a of 21 in any C n
as center we describe a cube of side <r, so small that it contains no
other point of 21. This is possible since 21 is isolated. There are but
a finite number of these cubes in C n of side <r = , z>= 1, 2, 3,
v
for each v. Hence, by 235, l, 21 is enumerable.
237. 1. Every aggregate of the first species 21, limited or not, is
enumerable.
For let 31 be of order n. Then
ENUMERABLE SETS 283
where 2l t denotes the isolated points of 21 and 2lp the proper limit
ing points of 21'
Similarly,
Thus,
qr _. or i or/ i or" i ... i 9f(n)
<t  <lt I ^J0,t I v*p,l I I vlp *
But 2l (n) is finite and ?(<?> < 2J (7l) .
Thus 21 being the sum of n f 1 enumerable sets, is enumerable.
2. If W is enumerable, so is 31.
For as in 1,
and {,<'.
238. 1. Every infinite aggregate 21 contains a part 33
3321.
For let (S = (a 19 a 2 , a 3 ) be an infinite enumerable set in 21,
so that
21 = <g + g.
Let g == a x + ^
To establish a uniform correspondence between J5?, (5 let us
associate a n in ( with a n+1 in E. Thus <&~ E.
We now set
Obviously 21 ^ 33 since E~ @, and the elements of % are common
to 21 and SB.
2. Z^2I~ 33 are infinite, each contains a part 2l x , 33i such that
For by 1, 21 contains a part 2l x such that 2l~2l r Similarly,
33 contains a part 33 a such that 33 ~ 33 r As 2l~S3, we have the
theorem.
284 AGGREGATES
239. 1. A theorem of great importance in determining
whether two aggregates are equivalent is the following. It is
the converse of 238, 2.
Let* l <%, !<. If K^ and ^ ~ ,
then 21 ^ 33.
In the correspondence 2lj ~ 33, let 21 2 be the elements of $1
associated with SS l . Then
31 2 ~ !  21
and hence 91 91 f 1
But as 2lj > 2(3 , we would infer from 1) that also
21 2l r (2
As 2lj ~ 53 by hypothesis, the truth of the theorem follows at
once from 2).
To establish 2) we proceed thus. In the correspondence 1), let
21 3 be that part of 2I 2 which ~ 21 1 in 21. In tfc correspondence
! ~ 21 3 , let 21 4 be that part of 21 3 which ~ 21 2 in 2l t .
Continuing in this way, we get the indefinite sequence
21 > Slj > 21 2 > 3 > 
such that Qr or fty
Letnow a^ + t^ , 3^ =
Then a = ^ + ^ + ^ + ^ + ^ + _ (g
and similarly 9r^a.^u.^j.pra.^a.
^X = A) 4 & 2 4 i> 3 f V24 4" V2, 6 H '
We note that we can also write
a 1 = s> + <E 8 + e a + e 6 + < 4 +  (4
Now from the manner in which the sets 21 3 , 21 4 were obtained,
it follows that
61 ~ 6, , C 8 ^C5 ( s
Thus the sets in 4) correspond uniformly to the sets directly
above them in 3), and this establishes 1).
ENUMERABLE SETS 285
2. In connection with the foregoing proof, which is due to
Bernstein, the reader must guard against the following error. It
does not in general follow from
21^ + Sj , 2t 2 =2*3 + 3 , 2(~21 2 , ^Sls
that K n
(>! ~ 3
which is the first relation in 5).
Example. Let 21 = (1, 2, 3, 4, ).
2l x = (2, 3, 4, 5 .) , 21 2 = (3, 4, 5, 6 )
2( 3 =(5, 6, 7,8 )
Then , e i= l 6 8 =(3, 4).
Now 21, 21 j, 21 2 , 21 3 are all enumerable sets ; hence
a a, , a! a,.
But obviously Sj is not equivalent to S 3 , since a set containing
only one element cannot be put in 1 to 1 correspondence with a
set consisting of two elements.
240. 1.
For by hypothesis a part of 48, viz. @, is ~2l. But a part of 21
is ~$, viz. 35 itself. We apply now 239.
2. Let a be any cardinal number. If
a < ("arc! 33 < oc,
then a = CardS.
For let Card 21 = . Then from
a < Card $
it follows that 21 ~ a part or the whole of 3d ; while from
Card $ <
it follows that 53 is ^ u part or the whole of 21.
3. Any part 38 of an enumerable set 21 is enumerable.
For if 59 is finite, it is enumerable. If infinite,
Card > HO
On the other hand
Card < Card = *.
286 AGGREGATES
4. Two infinite enumerable sets are equivalent.
For both are equivalent to $S ^ ne se t of positive integers.
241. 1. Let @ be any enumerable set in 91 ; set 21 = g + 93.
33 i* infinite, 9193.
For S3 being infinite, contains an infinite enumerable set
Let = g + Then
+ g~g. Hence ~.
2. We may state 1 thus :
Card (21 <)= Card 9t
provided 91 @ is infinite.
3. From 1 follows at once the theorem :
31 ie <my infinite set and ( an enumerable set. Then
Card (31 + <5) = Card 31.
Transformations
242. 1. Let 7 be a transformation of space such that to each
point x corresponds a single point X T , and conversely.
Moreover, let #, y be any two points of space. After the trans
formation they go over into X T ^ y T . If
we call To, displacement.
If the displacement is defined by
^ = 0?! + ^ , x f m ^x m + a m
it is called a translation.
If the displacement is such that all the points of a line in space
remain unchanged by T, it is called a rotation whose axis is the
fixed line.
THE CARDINAL C 287
If 9? denotes the original space, and 9? r the transformed space
after displacement, we have, obviously,
!l\ = tx \ ' " V = tX m , t > 0. (1
Then when a; ranges over the mway space , y ranges over an
mw f dy space 9). If we set x ~ y as defined by 1),
Als Dist (0, y) = t Dist (0, x).
We call 1) a transformation of similitude. If t > 1, a figure in
space is dilated ; if t <1, it is contracted.
3. Let Q be any point in space. About it as center, let us de
scribe a sphere S of radius R. Let P be any other point. On the
join of P, Q let us take a point P' such that
Dist (P', (?) =
Dist (P, (?)
Then P 7 is called the inverse of P with respect to S. This trans
formation of space is called inversion. Q is the center of inversion.
Obviously points without S go over into points within, and con
versely. As P = oo , P r = Q.
The correspondence between the old and new spaces is uniform,
except there is no point corresponding to Q.
The Cardinal c
243. 1. All or any part of space S may be put in uniform cor
respondence with a point set lying in a given cube 0.
For let @ t denote the points within and on a unit sphere S about
the origin, while @ e denotes the other points of space. By an in
version we can transform @ e into a figure @y lying in S. By a
transformation of similitude we can contract @ M (>, as much as we
choose, getting <g[, @J. We may now displace these figures so
as to bring them within G in such a way as to have no points in
common, the contraction being made sufficiently great. The
288 AGGREGATES
correspondence between @ and the resulting aggregate is obviously
uniform since all the transformations employed are.
As a result of this and 240, 1 we see that the aggregate of all
real numbers is ^ to those lying in the interval (0, 1); for example,
the aggregate of all points of 5R m is ~ to the points in a unit cube,
or a unit sphere, etc.
244. 1. The points lying in the unit interval 31 = (0*, 1*) are
not enumerable.
For if they were, they could be arranged in a sequence
Let us express the as as decimals in the normal form. Then
a n = a nl a n2 n3
Consider the decimal
b = b^b^
also written in the normal form, where
J l^ a M ' ^2^^2,2 ' ^3^^3,3 "
Then b lies in 31 and is yet different from any number in 1).
2. We have (0*, 1*) ~ (0, 1) , by 241, 3,
(a, 4) , by 243,
where a, b are finite or infinite.
Thus the cardinal number of any interval, finite or infinite,
with or without its end points is the same.
We denote it by c and call it the cardinal number of the recti
linear continuum, or of the real number system 9t.
Since 9? contains the rational number system R, we have
0o
3. The cardinal number of the irrational or of the transcendental
numbers in any interval 21 is also c.
For the nonirrational numbers in 51 are the rational which are
enumerable ; and the nontranscendental numbers in 91 are the
algebraic which are also enumerable.
THE CARDINAL c 289
4. The cardinal number of the Cantor set S of I, 272 is c.
For each point a of & has the representation in the triadic
system n
J a = a 1 a 2 a a , a = 0, 2.
But if we read these numbers in the dyadic system, replacing
each a n = 2 by the value 1, we get all the points in the interval
(0, 1). As there is a uniform correspondence between these two
sets of points, the theorem is established.
245. An enumerable set 91 is not perfect, and conversely a perfect
set is not enumerable.
For suppose the enumerable set
91 = !, a 2 (1
were perfect. In D^^a^) lies an infinite partial set 91 j of 31,
since by hypothesis 91 is perfect. Let a mt be the point of lowest
index in 9l x . Let us take r 2 <r l such that D rs (a m2 ) lies in
D r * (a x ). In Z> r f(a m2 ) lies an infinite partial set 9^ of 9l r Let
a ms be the point of lowest index in 91 2 , etc.
Consider now the sequence
<*i i <V < a m s
It converges to a point a by I, 127, 2. But lies in 91, since this
is perfect. Thus a is some point of 1), say a = a.. But this
leads to a contradiction. For a, lies in every D r * n (am n ); on the
other hand, no point in this domain has an index as low as m n
which == oo, as n == oo. Thus 91 cannot be perfect.
Conversely, suppose the perfect set 91 were enumerable. This
is impossible, for we have just seen that when 91 is enumerable it
cannot be perfect.
246. Let 91 be the union of an enumerable set of aggregate* 9l n
each having the cardinal number c. Then Card 91 = c.
For let 48 n denote the elements of 9l n not in 91 1 ,91 2 9l n _j.
Let S n denote the interval (n 1, w*). Then the cardinal
number of gj 4 S 2 + is c.
290 AGGREGATES
But Card n < Cardg n .
Hence Card 21 < c , by 231, 6. (1
On the other hand,
Card 21 > Card 2l x = c. (2
From 1), 2) we have the theorem, by 240, 2.
247. 1. As already stated, the complex x = (x^ x%, # n ) de
notes a point in wway space. Let x, # 2 , denote an infinite
enumerable set. We may also say that the complex
x= (#!, # 2 , in inf.)
denotes a point in oo way space 9?^.
2. Let 21 denote a point set in 9t n , n finite or infinite. Then
Card 21 < c. (1
For let us first consider the unit cube (5 whose coordinates x m
range over 33 = (0*, 1*). Let ) denote the diagonal of . Then
c = Card ) < Card 6. (2
On the other hand we show Card ( < c.
For let us express each coordinate x m as a decimal in normal
form. Then __
x l ' a ll a 12 a 13 a !4 "
Let us now form the number
obtained by reading the above table diagonally. Let?) denote the
set of #'s so obtained as the #'s range over their values. Then
For the point y, for example, in which a ln = 0, n = 1, 2, lies
in 53 but not in 2) as otherwise x l = 0. Let us now set x ~ y.
Then g ^ 2) and hence Card g ^ (g
From 2), 3) we have Card g = c.
THE CARDINAL c 291
Let us now complete by adding its faces, obtaining the set C.
By a transformation of similitude T we can bring O T within &.
Hence Card > Card (7.
On the other hand, is a part of (7, hence
Card 6 < Card (7.
Thus Card = c. The rest of the theorem follows now easily.
248. Let 3 = S/i denote the aggregate of onevalued continuous
functions over a unit cube in 9? n .
Then
Let C denote the rational points of , i.e. the points all of
whose coordinates are rational. Then any / is known when its
values over C are known. For if is an irrational point of ,
we can approach it over a sequence of rational points a a , a 2 === .
But f being continuous, /(a) = lim/(a n ), and f is known at .
On the other hand, being enumerable, we can arrange its points
in a sequence n
^ C=c l , c%, ...
Let now 9?^ be a space of an infinite enumerable number of
dimensions, and let y = (y x , y 2 ) denote any one of its points.
Let f have the value t] 1 at <?j, the value ?; 2 at <? 2 an( i s n f r
the points of 0. Then the complex 7?j, ?; 2 ,  completely deter
mines / in (. But this complex also determines the point
?7 = (rjj, 7/2 ..) in 9?^. We now associate/ with ?;. Thus
Card $< Card SR = c.
But obviously Card $ > c, for among the elements of $ there
is an/ which takes on any given value in the interval (0, 1), at
a given point of @.
249. There exist aggregates whose cardinal number is greater
than any given cardinal number.
Let 33= \b\ be an aggregate whose cardinal number b is given.
Let a be a symbol so related to S3 that it has arbitrarily either
the value 1 or 2 corresponding to each b of $&. Let 21 denote the
292 AGGREGATES
aggregate formed of all possible #'s of this kind, and let a be its
cardinal number.
Let /3 be an arbitrary element of 33. Let us associate with /3
that a which has the value 1 for b = and the value 2 for all
other 6's. This establishes a correspondence between S3 and a
part of 91. Hence
a>b.
Suppose a = b. Then there exists a correspondence which
associates with each b some one a and conversely. This is
impossible.
For call a b that element of 31 which is associated with b. Then
a b has the value 1 or 2 for each j3 of S3. There exists, however,
in 21 an element a' which for each /9 of S3 has just the other
determination than the one a b has. But a f is by hypothesis
associated with some element of S3, say that
a' = a b >.
Then for b = b f , a' must have that one of the two values 1, 2
which a b ' has. But it has not, hence the contradiction.
250. The aggregate of limited integrable functions $ defined over
31 = (0, 1) has a cardinal number f > c.
For let f(x) = in 31 except at the points of the discrete
Cantor set of I, 272, and 229, Ex. 4. At each point of let /
have the value 1 or 2 at pleasure. The aggregate formed of
all possible such functions has a cardinal number > c, as the
reasoning of 249 shows. But each f is continuous except in S,
which is discrete. Hence / is integrable. But rj > . Hence
f>c.
Arithmetic Operations with Cardinals
251. Addition of Cardinals. Let 31, S3 be two aggregates with
out common element, whose cardinal numbers are a, b. We define
the sum of a and b to be
Card (31, )=a + b.
ARITHMETIC OPERATIONS WITH CARDINALS 293
We have now the following obvious relations :
K 6 + n = K , n a positive integer. (1
o+4K = , n terms. (2
KO + **o + " == NO * ^ w infinite enumerable set of terms. (3
//* the cardinal numbers of 31, 33, G are a, b, c,
af b = b + a.
The first relation states that addition is associative, the second
that it is commutative.
252. Multiplication.
1. Let 21 = ja!, 33 = J&5 have the cardinal numbers a, b. The
union of all the pairs (a, b) forms a set called the product oftyt and
53. It is denoted by 31 33. We agree that (a, b) shall be the
same as (5, a). Then
. = .
We define the product of a and b to be
Card 91 . 33 = Card 932I = a.b = b.a.
2. We have obviously the following formal relations as in finite
cardinal numbers : /* \ / \\
a(u c) = (a u)c,
a b = b  a,
which express respectively the associative, commutative, and dis
tripulative properties of cardinal numbers.
Example 1. Let 9l=Jaj, 53 = ji denote the points on two
indefinite right lines. Then
2l.$ = K^)5
If we take a, b to be the coordinates of a point in a plane 9? 2 ,
* The reader should note that here, as in the immediately following articles, c is
simply the cardinal number of ( which is any set, like 51, 53
294 AGGREGATES
Example 2. Let 21 = \a\ denote the family of circles
Let 33 = j6j denote a set of segments of length b. We can
interpret (#, 5) to be the points on a cylinder whose base is 1)
and whose height is 5. Then 2( S3 is the aggregate of these
cylinders.
253 1 K = n OT tit = c. (1
For let m , x
9e = (rt 1 , a a , ... a n ),
(5 = (^, ^ 2 in inf.)
Then SR . g = (a O Cd Ca e ) 
==^ + (52+ ... +C.
The cardinal number of the set on the left is ?iN , while the
cardinal number of the set on the right is K .
2. ec = c. ' (2
For let ( = \c\ denote the points on a right line, and @ = (1, 2,
3,.).
Then $,= \(n>c)\
may be regarded as the points on a right line l n . Obviously,
Card JU=c
Hence
ec = Card (gg = c.
254. Exponents. Before defining this notion let us recall a
problem in the theory of combinations, treated in elementary
algebra.
Suppose that there are 7 compartments
^1' ^2' "* ^Y>
and that we have k classes of objects
J^i* Sji ... K k .
ARITHMETIC OPERATIONS WITH CARDINALS 295
Let us place an object from any one of these classes in C v an
object from any one of these classes in (7 2 and so on, for each
compartment. The result is a certain distribution of the objects
from these k classes K, among the y compartments C.
The number of distributions of objects from k classes among y
compartments is &.
For in O l we may put an object from any one of the k classes.
Thus O l may be filled in k ways. Similarly (7 2 may be filled in
k ways. Thus the compartments O l , <7 2 may be filled in & 2 ways.
Similarly (7j, (7 2 , (7 3 may be filled in A 3 ways, etc.
255. 1. The totality of distributions of objects from k classes
K among the y compartments C form an aggregate which may be
denoted by j^c
We call it the distribution of K over C. The number of distri
bution of this kind may be called the cardinal number of the set,
and we have then ' K c = fr.
2. What we have here set forth for finite (7 and TTmay be ex
tended to any aggregates, 21 = \a\, 93 = \b\ whose cardinal num
bers we call a, b. Thus the totality of distributions of the a's
among the 6's, or the distribution of^[ over 33, is denoted by
a,
and its cardinal number is taken to be the definition of the symbol
a *' Thus > Card 21* = a*.
256. Example 1. Let
x n + a^ 71 " 1 + ' 4 a n = (1
have rational number coefficients. Each coefficient a 9 can range
over the enumerable set of elements in the rational number
system R = \r\, whose cardinal number is K . The n coefficients
form a set 51 = (a t , a n ) = \a\. To the totality of equations 1)
corresponds a distribution of the r's among the a's, or the set
B
whose cardinal number is
296 AGGREGATES
As Card *= Ho = e
we have the relation :
%w __ /% />n _ A
NO NO > or e c
/or any integer n.
On the other hand, the equations 1) may be associated with
the complex
0*i, <Oi
and the totality of equations 1) is associated with
= JOi> <*)}
But Ki*a2)!={i! *!
{(!, a 2 , a 3 ) = {(a x , a 2 )} \a s l , etc.
Hence ff __ , , , , 5 >
^ = Ji! \ a <n J^ni
us Card = eee , n ^Ws as factor.
But Card 6 = Card JZ
since each of these sets is associated uniformly with the equations
'* s e n = e e e , w ft'wes as factor.
257. Example 2. Any point # in wway space 9t m depends on
m coordinates 2^, a; 2 , a; m , each of which may range over the set
of real numbers SR, whose cardinal number is c. The m coordi
nates x l x m form a finite set
Thus to 9? m = {a;j corresponds the distribution of the numbers in
9t, among the m elements of X, or the set
K*
whose cardinal number is
c w .
As Card 9J X = c
we have
c m = c for any integer m. (1
As in Example 1 we show
c m = cC'"c , m times as factor.
ARITHMETIC OPERATIONS WITH CARDINALS 297
258. a b+c = a 6 a c . (1
To prove this we have only to show that
9l 8+e and . 31 S
can be put in 11 correspondence. But this is obvious. For
the set on the left is the totality of all the distributions of the
elements of 31 among the sets formed of 33 and S. On the other
hand, the set on the right is formed of a combination of a distri
bution of the elements of 91 among the 33, and among the . Hut
such a distribution may be regarded as the distribution first con
sidered.
259. (a*y = aK (1
We have only to show that we can put in 11 correspondence
the elements of
(31 V and a?'*. (2
Let 31= jaj, 33= 6j, = \c\. We note that is a union of
distributions of the a's among the 6's, and that the left side of 2)
is formed of the distributions of these sets among the c's. These
are obviously associated uniformly with the distributions of the
a's among the elements of 33 &.
260. 1 . c w == (mO n = nc = mC = c (1
where m, n are positive integers.
For each number in the interval S = (0, 1*) can be represented
in normal form once and once only by
afya z in the wadic system, (2
where the < a s < m. [I, 145] .
Now the set of numbers 2) is the distribution of 90? = (0, 1, 2,
m 1) over (g = (a x , # 2 , a 3 ), or
whose cardinal number is
9?? e .
On the other hand, the cardinal number g is c.
298 AGGREGATES
Hence, m e = c.
As n* = e, we have 1), using 1) in 257.
2. The result obtained in 247, 2 may be stated:
c' = c. (3
3. ec = c. (4
For obviously n* < e c < c e .
But by 3), c e = c and by 1) n e = c.
261. 1. 27*0 cardinal number t #/" a/ functions f (x l # m ) which
take on but two values in the domain of definition 21, of cardinal num
ber o, is 2 .
Moreover, 2 ^ > a.
This follows at once from the reasoning of 249.
2. Zrtf f Je the cardinal number of the class of all functions de
fined over a domain 21 whose cardinal number is c. Then
For the class of functions which have but two values in 21 is by
On the other hand, obviously
But
c< = (2')S by 260, 1)
= 2ec, by 259, 1)
= 2^ by 253, 2).
Thus, c c = 2 c.
That f > c
follows from 250, since the class of functions there considered lies
in the class here considered.
3. The cardinal number \ of the class of limited integrable func
tions in the interval 21 is = f, the cardinal number of all limited
functions defined over 21.
NUMBERS OF LIOUVILLE 299
For let D be a Cantor set in 21 [I, 272]. Being discrete, any
limited function defined over J) is integrable. But by 229, Ex. 4,
the points of 21 may be set in uniform correspondence with the
points of D.
4. The set of ail functions
which are the sum of convergent series, and whose terms are continu
ous in 21, has the cardinal number c.
For the set $ ^ continuous functions in 21 has the cardinal
number c by 248. These functions are to be distributed among
the enumerable set @ of terms in 2). Hence the set of these
functions is ^
$ >
whose cardinal number is
c e = c.
Remark. Not every integrable function can be represented by
the series 2).
For the class of integrable functions has a cardinal number > c,
by 250.
5. The cardinal number of all enumerable sets in an mway space
5R m is c.
For it is obviously the cardinal number of the distribution of
9t m over an enumerable set (, or
Card ffi = c c = c.
Number* of Liouville
262. In I, 200 we have defined algebraic numbers as roots of
equations of the type
where the coefficients a are integers. All other numbers in 9? we
said were transcendental. We did not take up the question
whether there are any transcendental numbers, whether in fact,
not all numbers in 9t are roots of equations of the type 1).
300 AGGREGATES
The first to actually show the existence of transcendental num
bers was Liouville. He showed how to form an infinity of such
numbers. At present we have practical means of deciding
whether a given number is algebraic or not. It was one of the
signal achievements of Hermite to have shown that e = 2.71818
is transcendental.
Shortly after Lindemann, adapting Hermite's methods, proved
that ?r = 3.14159 is also transcendental. Thereby that famous
problem the Quadrature of the Circle was answered in the negative.
The researches of Hermite and Lindemann enable us also to form
an infinity of transcendental numbers. It is, however, not our pur
pose to give an account of these famous results. We shall limit
our considerations to certain numbers which we call the numbers
of Liouville.
In passing let us note that the existence of transcendental num
bers follows at once from 235, 2 and 244, 2.
For the cardinal number of the set of real algebraic number is
e, and that of the set of all real numbers is c, and c > e.
263. In algebra it is shown that any algebraic number a is a
root of an irreducible equation,
f(x) = a x m 4 ap" 1 + 4 a m = (1
whose coefficients are integers without common divisor. We say
the order of a is m.
We prove now the theorem
Let
P
r n = , p n , q n relatively prime,
<}n
= a, an algebraic number of order m, as n = ao. Then
\ a  r \>^{ ' n>v  ( 2
!/
For let a be a root of 1). We may take 8>0 so small that
f(x)= in Z> 5 *(0> and 8 so large that r n lies in D 5 (), for n>8.
NUMBERS OF LIOUVILLE 301
for n > s, since the numerator of the middle member is an integer,
and hence >1.
On the other hand, by the Law of the Mean [I, 397],
where /8 lies in Z>s(). Now /()=0 and /'(/9)< some M.
Hence
on using 3). But however large M is, there exists a v, such that
q n > M) for any n>v. This in 4) gives 2).
264. 1. The numbers
= J?i+JSL + *IL+... (1
10 1! 10 21 10 3! ^
w^ere a n < 10 n , anrf rc0 aH of them vanish after a certain index, are
transcendental.
For if L is algebraic, let its order be m. Then if L n denotes
the sum of the first n terms of 1), there exists a v such that
But
(2
i/ being taken sufficiently large. But 3) contradicts 2).
The numbers 1) we call the numbers of Liouville.
2. The set of Liouville numbers has the cardinal number c.
For all real numbers in the interval (0*, 1) can be represented
"& + & + & +  ' ^'
where not all the 6's vanish after a certain index. The numbers
_
10 21 10 8!
can obviously be put in uniform correspondence with the set {/8j.
Thus Card \\\ =c. But \L\ > {X{, hence Card {} > c. On the
other hand, the numbers \L\ form only a part of the numbers in
(0*, 1). Hence Card \L\<t.
CHAPTER IX
ORDINAL NUMBERS
Ordered Sets
265. An aggregate 31 is ordered, when a, b being any two of
its elements, either a precedes ft, or a succeeds ft, according to some
law ; such that if a precedes 6, and b precedes e% then a shall pre
cede c. The fact that a precedes b may be indicated by
a<b.
Then a>b
states that a succeeds b.
Example 1. The aggregates
1, 2, 3, ...
2, 4, 6, ...
3, 2, 1,0, 1, 2,3, ..
are ordered.
Example 2. The rational number system R can be ordered in
an infinite variety of ways. For, being enumerable, they can be
arranged in a sequence ..
' i * r a ' r a ' " * r n '
Now interchange r T with r n . In this way we obtain an infinity
of sequences.
Example 3. The points of the circumference of a circle may be
ordered in an infinite variety of ways.
For example, let two of its points P l , P 2 make the angles af^,
f# 2 with a given radius, the angle varying from to 2 TT.
Let P l precede P 2 when l < # 2 .
302
ORDERED SETS 303
Example 4 The positive integers $ may be ordered in an infi
nite variety of ways besides their natural order. For instance, we
may write them in the order
1, 3, 5, ... 2, 4, 6, ...
so that the odd numbers precede the even. Or in the order
1, 4, 7, 10, ... 2, 5, 8, 11, ... 3, 6, 9, 12, ...
and so on. We may go farther and arrange them in an infinity
of sets. Thus in the first set put all primes ; in the second set
the products of two primes ; in the third set the products of
three primes; etc., allowing repetitions of the factors. Let any
number in set m precede all the numbers in set n >m. The num
bers in each set may be arranged in order of magnitude.
Example 5. The points of the plane 9t 2 may be ordered in an
infinite variety of ways. Let L y denote the right line parallel to
the a>axis at a distance y from this axis, taking account of the sign
of y. We order now the points of 3J 2 by stipulating that any
point on L v , precedes the points on any L^ when y 1 <y", while
points on any L v shall have the order they already possess on that
line due to their position.
266. Similar Sets. Let 31, 53 be ordered and equivalent. Let
a ~ b, a ~ ft. If when a < a in 21, b < ft in 53, we say 21 is similar
to 53, and write OT ~
zl 3o.
Thus the two ordered and equivalent aggregates are similar
when corresponding elements in the two sets occur in the same
relative order.
Example 1. Let
1 9
1, <,
In the correspondence 21 ~~ 53, let n be associated with a n . Then
^.
Example 2. Let M 100
VI = 1, A o,
53= a l 2 a m , b^ b^ J 3
304 ORDINAL NUMBERS
In the correspondence 21 ~ 33, let a r ~r for r = 1, 2, m; also
let 6 n ~ m + n, 7i = 1, 2  Then 21 ^ S3.
Example 3. Let
*
Let the correspondence between 21 and S3 be the same as in
Ex. 2. Then 21 is not similar to S3. For 1 is the first element in
21 while its associated element a l is not first in S3.
Example A. Let or 100
r 21 = 1, A o, ...
S3 = tfj, a a ij, 5 2
Let a n ~ 2 n, b n ~ 2 n  1. Then 21  S3 but 21 is not a* S3
267. i^2l^S3, S3^S. Then %**<$,.
For let a i, a 1 ~V in 21 S3. Let 6  <?, J'^^' in S3 ~ g. Let
us establish a correspondence 21 (5 by setting a ~ c, a f ~c f . Then
if a <a f in 21, c< c r in S. Hence 21 ^ S.
Eutactic Sets
268. Let 21 be any ordered aggregate, and S3 a part of 21, the
elements of S3 being kept in the same relative order as in 21. If 21
and each S3 both have a first element, we say that 21 is well ordered,
or eutactic.
Example 1. 21 = 2, 3, f>00 is well ordered. For it has a first
element 2. Moreover any part of 21 as 6, 15, 25, 496 also has a
first element.
Example 2. 21 = 12, 13, 14, in inf. is well ordered. For it
has a first element 12, and any part S3 of 21 whose elements pre
serve the same relative order as in 21, has a first element, viz.
the least number in 53.
The condition that the elements of S3 should keep the same rel
ative order as in 21 is necessary. For B = 28, 26, 24, 22, 20,
21, 23, 25, 27, ... is a partial aggregate having no first element.
But the elements of B do not have the order they have in 21.
EUTACTIC SETS 305
Example 8. Let 21 = rational numbers in the interval (0, 1)
arranged in their order of magnitude. Then 21 is ordered. It
also has a first element, viz. 0. It is not well ordered however.
For the partial set 33 consisting of the positive rational numbers of
21 has no first element.
Example 4 An ordered set which is not well ordered may some
times be made so by ordering its elements according to another
law.
Thus in Ex. 3, let us arrange 21 in a manner similar to 283.
Obviously 21 is now well ordered.
Example 5. 21 = a^ # 2 4j, J 2 is well ordered. For a is the
first element of 21 ; and any part of 21 as
has a first element.
269. 1. Every partial set 33 of a wellordered aggregate 21 is well
ordered.
For 33 has a first element, since it is a part of 21 which is well
ordered. If & is a part of 33, it is also a part of 21, and hence has
a first element.
2. If a is not the last element of a wellordered aggregate 21, there
is an element ofty. immediately following a.
For let 33 be the part of 21 formed of the elements after a. It
has a first element b since 21 is well ordered. Suppose now
a< c < b.
Then b is not the first element of 33, as c < b is in 33.
3. When convenient the element immediately succeeding a may
be denoted by
a + L
Similarly we may denote the element immediately preceding a,
when it exists, by
306 ORDINAL NUMBERS
For example, let
21 = a^a% ^1^2 ' ' "
Then a n + 1 = a n+1 , b m + 1 = b n+1
<*n  1 = nl i *m  1 = *mr
There is, however, no b l 1.
270. 1. .7/31 i Wtf/Z ordered, it is impossible to pick out an in
finite sequence of the type
a x > rt a > a 3 > ... (1
For m
1O = ' #3, # 2 ' j
is a part of 21 whose elements occur in the same relative order as
in 21, and $ has no first element.
2. A sequence as 1) may be called a decreasing sequence, while
a 1 < a 2 < a 3 ...
may be called increasing.
In every infinite well ordered aggregate there exist increasing
sequences.
3. Let 21, 48, (5, be a well ordered set. Let 21= \ a\ be well
ordered in the a #, 53 = 5 6 i ^ we# ordered in the 6's, ec.
U = , , CS 
ordered with regard to the little letters a, b ...
For U has a first element in the little letters, viz. the first ele
ment of 21. Moreover, any part of U, as 55, has a first element in
the little letters. For if it has not, there exists in 33 an infinite
decreasing sequence
t> s>r> '
This, however, is impossible, as siuli a sequence would deter
mine a similar sequence in U as
3T > @ > 5R > ..
which is impossible as U is well ordered with regard to 21, 33
4. Let 9l<<e< (1
Let each element of 21 precede each element of 53, etc.
SECTIONS 307
Let each 21, $J, be well ordered.
Let #=3l+S,
nen
CM
is a well ordered set* @ preserving the relative order of elements
intact.
For <3 has a first element, viz. the first element of 21. Any
part S of @ has a first element. F"or, if not, there exists in @
an infinite decreasing sequence
r > q> p > (2
Now r lies in some set of 1) as 9t. Hence <?, JP, also lie in
JR. But in 5R there is no sequence as 2).
5. Let 21, 33, S, be an ordered set of well ordered aggre
gates, no two of which have an element in common. The reader
must guard against assuming that 21 f .SB + & f > keeping the
relative order intact, is necessarily well ordered.
For let us modify Ex. 5 in 265 by taking instead of all the
points on each L v only a well ordered set which we denote by Sl y .
Then the sum ^ __ y^
has a definite meaning. The elements of 21 we supposed arranged
as in Ex. 5 of 265.
Obviously 21 is not well ordered.
Sections
271. We now introduce a notion which in the theory of well
ordered sets plays a part analogous to Dedekind's partitions in
the theory of the real number system 9{. Cf. I, 128.
Let 21 be a well ordered set. The elements preceding a given
element a of 21 form a partial set called the section of 21 generated
by a. We may denote it by
Sa,
or by the corresponding small letter a.
308 ORDINAL NUMBERS
Example!. Let 91 = 1 2 3
Then
100 = 1, 2, .99
is the section of 21 generated by the element 100.
Example 2. Let
21 = 6^, a 2 . b l , 6 2 ..
Then
*S* 6 = a 1 a a . ^626364
is the section generated by b^.
Sb l = a^
that generated by 6j, etc.
272. 1. Every section of a well ordered aggregate is well ordered.
For each section of 21 is a partial aggregate of 21, and hence
well ordered by 269, 1.
2. In the well ordered set 21, let a<b. Then Sa is a section
ofSb.
3. Let @ denote the aggregate of sections of an infinite well
ordered set ty. If we order @ such that Sa < Sb in @ when a<b in
21, @ is 0cK ordered.
For the correspondence between 21 and @ is uniform and similar.
273. Let 21, 58 be well ordered and 21^93. If a~~b, then
Sa ^ Sb.
For in 21 let a ff <a f >a. Let b'~a' and b n ^a rr . Since
21 2 93, we have
6"<6'<6;
hence the theorem.
274. If 21 is well ordered, 21 is not similar to any one of its
sections.
For if 21 ^$a, to a in 21 corresponds an element a l <a in Sa.
To a l in 21 corresponds an element a 2 in Sa> etc. In this way we
obtain an infinite decreasing sequence
a> a l
which is impossible by 270, 1.
SECTIONS 309
275 Let 21, 93 be well ordered and 21 ^ 93. 2%ew to Sa in 21
7&0 correspond two sections Sb, /SyS each 2* Sa.
For let b < fr and a ^ 6, #a ^ $8. Then
Sb ^ #/3, by 267. (1
But 1) contradicts 274.
276. Let 21, S3 be two well ordered aggregates. It is impossible
to establish a uniform and similar correspondence between 21 and S3
in more than one way.
For say Sa ^ Sb in one correspondence, and Sa ^ S/3 in an
other, b, /3 being different elements of S3 Then
#6 ^ tf/3, by 267.
This contradicts 275.
277. 1. We can now prove the following theorem, which is
the converse of 273.
Let 21, 93 be well ordered. If to each section of 21 corresponds one
similar section of 93, and conversely, then 93 21.
Let us first show that 21 ~ 93. Since to any Sa of 21 corre
sponds a similar section Sb in 93, let us set a ~ b. No other
a 1 ~ 6, and no other b' ~ a, as then Sa 1 ^ Sb or Sb' ^ Sa, which
contradicts 274. Let the first element of 21 correspond to the
first of 93. Thus the correspondence we have set up between 21
and 93 is uniform and 21 ~ 93.
We show now that this correspondence is similar. For let
a ~ b and a' ~ b r , a! < a.
Then b r < b. For a f lies in Sa ^ Sb and b f a r lies in Sb.
2. From 1 and 273 we have now the fundamental theorem :
In order that two wellordered sets 21, 93 be similar, it is necessary
and sufficient that to each section of 21 corresponds a similar section
of 93, and conversely.
278. Let 21, 93 be well ordered. If to each section of 21 corre
sponds a similar section of 93> but not conversely, then 21 is similar to
a section of 93.
310 ORDINAL NUMBKKS
Let us begin by ordering the sections of 21 and 93 as in 272, 3.
Let B denote the aggregate of sections of 53 to which similar sec
tions of 21 do not correspond. Then B is well ordered and has a
first section, say Sb. Let /3 < b. Then to 8ft in 53 corresponds
by hypothesis a similar section Sa in 2t. On the other hand, to
any section Sa 1 of 21 corresponds a similar section fib' of 53 Ob
viously b'<b. Thus to any section of 21 corresponds a similar
section of Sb and conversely. Hence ?l^/S7> by 277. i.
279. Let 21, 93 be well ordered. Either 21 is similar to 93 or one
is similar to a section of the other.
For either :
1 To each section of 21 corresponds a similar section of 93
and conversely ;
or 2 To each section of one corresponds a similar section of
the other but not conversely ;
or 3 There is at least one section in both 21 and 93 to which no
similar section corresponds in the other.
If 1 holds, 21 ^93 by 277, l. If 2 holds, either 21 or 93 is similar
to a section of the other.
We conclude by showing 3 is impossible.
For let A be the set of sections of 21 to which no similar section
in 93 corresponds. Let B have the same meaning for 93. If we
suppose 21, 93 ordered as in 272, 3, A will have a first section say
8a, and B a first section fl/3.
Let a < a. Then to Sa in 21 corresponds by hypothesis a sec
tion Sb of Sft as in 278. Similarly if b r < 0, to Sb' of 93 corre
sponds a section 8a' of 8a. But then 8a^8f3 by 277, l, and this
contradicts the hypothesis.
Ordinal Numbers
280. 1. With each well ordered aggregate 21 we associate an
ittribute called its ordinal number^ which we define as follows :
1 If 2193, they have the same ordinal number.
2 If 21 a section of 93, the ordinal number of 21 is less than
that of 93.
ORDINAL NUMBERS 311
3 If a section of 21 is ^ 33, the ordinal number of 21 is greater
than that of 33.
The ordinal number of 21 may be denoted by
Orel 21,
or when no ambiguity can arise, by the corresponding small letter a.
As any two well ordered aggregates 21, 55 fall under one and only
one of the three preceding cases, any two ordinal numbers a, b
satisfy one of the three following relations, and only one, viz. :
a = b , a<b , a > b,
and if a < b, it follows that b > a
Obviously they enjoy also the following properties.
2 If
J a = b , b = c , then a = c.
For if c = Ord , the first two relations state that
But then aafg ^ by
Hence _
*^ a > b , b > c , then a > c.
281. 1. Let 21 be a finite aggregate, embracing say n elements.
Then we set Onia=n.
Thus the ordinal number of a finite aggregate has exactly similar
properties to those of Unite cardinal numbers. The ordinal num
ber of a finite aggregate is called finite, otherwise transfinite.
The ordinal number belonging to the well ordered set formed
of the positive integers c\ _ i o o
O A, *j, o,
we call a).
2. The least transfinite ordinal number is to.
For suppose a = Ord 21 < o>, is transfinite. Then 21 is ^ a
section of $. But every section of 3 is finite, hence the
contradiction.
312 ORDINAL NUMBERS
3. The cardinal number of a set 51 is independent of the order
in which the elements of $ occur. This is not so in general for
ordinal numbers.
For example, let or 1 o Q
4\ 1, 4, O,
33=1, 3, 5, .2,4, 6, .
Here Card 21 = Card S=.
But Ord 21 < Ord ,
since 21 is similar to a section of S3, viz. the set of odd numbers,
1, 3, 5, ...
282. 1. Addition of Ordinals. Let 1, S3 be well ordered sets
without common elements. Let & be the aggregate formed by
placing the elements of S3 after those of 21, leaving the order in 53
otherwise unchanged. Then the ordinal number of g is called the
sum of the ordinal numbers of 21 and 53, or
Ord g = Ord 2t + Ord 53,
or c = a + 6.
The extension of this definition to any set of wellordered aggre
gates such that the result is well ordered is obvious.
2. We note that A , t ^ A rt _j_f,^f,
a f > a, a f b > I).
For 21 is similar to a section of S, and 53 is equivalent to a part
of 6.
3. The addition of ordinal numbers is associative.
This is an immediate consequence of the definition of addition.
4. The addition of ordinal numbers is not always commutative.
Thus if
let = (ojo, ... ^6 2 ... 6 n ), Ord = c,
) = (Jj ... b n a^ .), Ord = b.
Then v ,
c = a) f ^ , b = ft + o>
ORDINAL NUMBERS 313
But 21 2* a section of S, viz. : ^ >S Y 6 a , while D ^ SI. Hence
0) < C , = b,
or
CD + n> to , wj ft > : =fc>'
5. If a > b, ^e/i c f a > c + b, awd a + c > b + c.
For let
Since a > b, we can take for 93 a section Sb of 31. Then c 4 & is
the ordinal number of (S 4. 9f (\
and c + b is the ordinal number of
S + Sb, (2
preserving the relative order of the elements.
But 2) is a section of 1), and hence c 4 a > c } b.
The proof of the rest of the theorem is obvious.
283. 1. The ordinal number immediately following a is a 4 1.
For let a = Ord 91. Let 93 be a set formed by adding after all
the elements of 91 another element b. Then
a + 1 = Ord 33 = b.
Suppose now
a<c<b , c=Ordg. (1
Then is similar to a section of 53. But the greatest section
of 93 is 8b = 91. Hence
c < a,
which contradicts 1).
2. Let a > b. Then there is one and only one ordinal number b
such that  , ,
a = u h o.
For let
a = Ord 31 , b = OrdS8
We may take 93 to be a section Sb of 31. Let ) denote the set
of elements of 91, coming after Sb. It is well ordered and has an
ordinal number b. Then
91 = 93 + ,
preserving the relative order, and hence
a = b + b.
There is no other number, as 282, 6 shows.
314 ORDINAL NUMBERS
284. 1. Multiplication of Ordinals. Let 31, 93 be wellordered
aggregates having a, b as ordinal numbers. Let us replace each
element of 21 by an aggregate ^ 33. The resulting aggregate &
we denote by $.91
As 6 is a wellordered set by 270, 3 it has an ordinal number c.
We define now the product b a to be c, and write
b a = c.
We say c is the result of multiplying a % b, and call a, b factors.
We write
a . a = a 2 , a a a = a 3 , etc.
2. Multiplication is associative.
This is an immediate consequence of the definition.
3. Multiplication is not always commutative.
For example, let
S3 = (1, 2, 3 .. in inf.).
Then .a = (W 8 .., <W* 
21 33 =(*!, <?i, & 2 6 2
Hence Qrd (93 21) = 2
4. If a < b, then ca < cb.
For 6 81 is a section of g 33.
Limitary Numbers
285. 1. Let
be an infinite increasing enumerable sequence of ordinal numbers.
There exists a first ordinal number a greater than every a n .
Let n =0rd2l n .
LIMITARY NUMBERS 315
Since n ^ 1 < n , 2l n _ 1 is similar to a section of 2l n . For simplicity
we may take 2l n _! to be a section of 2l n . Let, therefore,
Consider now w w ^ ~
Zl = a x f ^ 2 + ^3 +
keeping the relative order of the elements intact. Then 21 is well
ordered and has an ordinal number a.
As any 2l n is a section of 21,
Moreover any number /3<a is also < some a m . For if 33 has
the ordinal number /3, 33 must be similar to a section of 21. Hut
there is no last section of 21.
2. The number a we have just determined is called the limit of
the sequence 1). We write
a = Km n , or n = a.
We also say that a corresponds to the sequence V).
All numbers corresponding 1 to infinite enumerable increasing
sequences of ordinal numbers are called limitary.
3. // every a n in 1) is < /3, then a < /3.
For if /3<, a is not the least ordinal number greater than
every n .
4. If /3<, /3 is
286. jfri or^r that . . ,*
1 <0 2 < (1
/3i</ 3 2<'" (2
define the same number \ it is necessary and sufficient that each
number in either sequence is surpassed by a number in the other.
For let . Q . &
n = , fin = fl
it no /3 n is greater than a m , ft<a m < , by 285, 3, and = @.
On the other hand, if each m < some /3 n , </S. Similarly
/3<a.
316 ORDINAL NUMBERS
287. Cantors Principles of Qenerating Ordinals. We have now
two methods of generating ordinal numbers. First, by adding 1
to any ordinal number a. In this way we get
a, a+ 1, a+ 2, .
Secondly, by taking the limit of an infinite enumerable increas
ing sequence of ordinal numbers, as
! < 2 < 8 < 
Cantor calls these two methods the first and second principles
of generating ordinal numbers.
Starting with the ordinal number 1, we get by successive appli
cations of the first principle the numbers
1, 2, 3, 4, ...
The limit of this sequence is CD by 285, 1. Using the first prin
ciple alone, this number would not be attained ; to get it requires
the application of the second principle. Making use of the first
principle again, we obtain
oi + l, w + 2, o) + 3, ...
The second principle gives now the limitary number &> + ct> = &>2
by 285, 1. From this we get, using the first principle, as before,
w2 + l, <2 + 2, 0)2 + 3, .
whose limit is o>3. In this way we may obtain the numbers
o)w + n , m, n finite.
The limit of any increasing sequence of these numbers as
ft> , o>2 , o)3 , o)4,
is o) o) = o) 2 , by 285, 1.
From o) 2 we can get numbers of the type
aPl + com + n l,m,n finite.
Obviously we may proceed in this way indefinitely and obtain
all numbers of the type
where # , ^  a n are finite ordinals.
LIMITARY NUMBERS 317
But here the process does not end. For the sequence
0) < O) 2 < ft) 3 <
has a limit which we denote by a> w .
Continuing we obtain
a) w&> , ft) w&)W , etc.
288. It is interesting to see how we may obtain well ordered
sets of points whose ordinal numbers are the numbers just con
sidered.
In the unit interval 21 = (0, 1), let us take the points
J . I . i ' if a
These form a well ordered set whose ordinal number is a>.
The points 1) divided 21 into a set of intervals,
i , a , a 8  ( 2
In m of these intervals, let us take a set similar to 1). This
gives us a set whose ordinal number is com.
In each interval 2), let us take a set similar to 1). This gives
us a set whose ordinal number is o> 2 . The points of this set
divide 21 into a set of &> 2 intervals. In each of these intervals,
let us take a set of points similar to 1). This gives a set of
points whose ordinal number is o> 3 , etc.
Let us now put in 2l t a set of points SS 1 whose ordinal number
is co. In 2I 2 let us put a se ^ 33 2 whose ordinal number is o> 2 , and
so on, for the other intervals of 2).
We thus get in 21 the well ordered set
whose ordinal number is the limit of
a) , a) 4 co 2 , ft) + P 4 a> 8 1
This by 286 has the same limit as
CD , a) 2 , a) 3 , or G> W .
With this set we may now form a set whose ordinal number is
fl> wW , etc.
318 ORDINAL NUMBERS
Classes of Ordinals
289. Cantor has divided the ordinal numbers into classes.
Class 1, denoted by Z, embraces all finite ordinal numbers.
Class 2, denoted by Z 2 , embraces all transfinite ordinal numbers
corresponding to well ordered enumerable sets ; that is, to sets
whose cardinal number is N . For this reason we also write
It will be shown in 293, 1 that Z^ is not enumerable. Moreover
if we set . % ~ , ~
Kj = Card Z 2 ,
there is no cardinal number between K and Kj as will be shown in
294. We are thus justified in saying that Class 8, denoted by
Z z or ^(Kj), embraces all ordinal numbers corresponding to well
ordered sets whose cardinal number is Kj, etc.
Let /3 = Ord S3 be any ordinal number. Then all the numbers
a < /8 correspond to sections of S3. These sections form a well
ordered set by 272, 3. Therefore if we arrange the numbers
a < y8 in an order such that ' precedes a when Sa r < $, they are
well ordered. We shall call this the natural order. Then the
first number in Z l is 1, the first number of Z% is ct). The first
number in Z z is denoted by fl.
290. As the numbers in Class 1 are the positive integers, they
need no comment here. Let us therefore turn to Class 2.
If a is in Z<i , so is a 4 1
For let a = Ord 21. Let S3 be the well ordered set obtained
by placing an element b after all the elements of 21. Then
+ 1 = Ord S3.
But S3 is enumerable since 21 is.
Hence a + 1 lies in Z 2 .
291  Let 1
be an enumerable infinite set of numbers in Z^. Then a = lim n lies
in Z
CLASSES OF ORDINALS 319
For using the notation employed in the proof of 285, 1, a is the
ordinal number of
But ?lj, 33j, 33 2 '" are eacn enumerable.
Hence 21 is enumerable by 2.35, l, and a lies in Z^
292. We prove now the converse of 290 and 291.
Kach number a in /^ 2 , except <y, is obtained by adding 1 to some
number in Z%; or it is the limit of an infinite enumerable increasing/
set of numbers in Z v
For, let a = Ord 21. Suppose first, that 21 has a last element,
say a. Since 21 is enumerable, so is Sa. Hence
ft = Ord Sa
is in Z r Then = + !.
Suppose secondly, that 21 has no last element. All the numbers
ft < in Z% belong to sections of 21. Since 21 is enumerable, the
numbers ft are enumerable. Let them be arranged in a sequence
ft r flv /V" (1
Since they have no greatest, let ft[ be the first number in it
>/3 1 , let /3 2 be the first number in it >/3(, etc. We get thus the
sequence /^ < # < 2 ' < (2
whose limit is X, say.
Then \ = <*. For A, is >any number in 1), which embraces all
the numbers of Z% < a. Moreover it is the least number which
enjoys this property.
293. 1 . The numbers of Z 2 are not enumerable.
For suppose they were. Let us arrange them in the sequence
i* 2 > 3 " C 1
Then, as in 292, there exists in this sequence the infinite enu
merable sequence . , . / . /0
^ ! < J < 2 < (%
such that there are numbers in 2) greater than any given number
in 1).
320 ORDINAL NUMBERS
Let = '. Then ' lies in Z% by 291. On the other hand, by
285, a f is > any number in 2), and therefore > any number in
1). But 1) embraces all the numbers of Z 2 , by hypothesis. We
are thus led to a contradiction.
2. We set p ,
Kj = Oard Z 2 .
294. There is no cardinal number between K and ttj
For let a=Card 21 be such a number. Then 21 is ~ an infinite
partial aggregate of Z%, which without loss of generality may be
taken to be a section of Z 2 . But every such section is enumer
able. Hence 21 is enumerable and =K , which is a contradiction.
295. We have just seen that the numbers in Z 2 are not enumer
able. Let us order them so that each number is less than any
succeeding number. We shall call this the natural order.
1. The numbers of Z% when arranged in their natural order form
a well ordered set.
For Z% has a first element co. Moreover any partial set Z, the
relative order being preserved, has a first element. For if it has
not, there exists an infinite enumerable decreasing sequence
This, however, is not possible. For /3, 7, form a part of Sec
which is well ordered.
There is thus one well ordered set having Kj as cardinal num
ber " Let
Let now 21 be an enumerable well ordered set whose ordinal
number is . The set
the elements of 21 coming after Z 2 , has the cardinal number Kj by
241, 3. It is well ordered by 270, 3. It has therefore an ordinal
number which lies in Z 3 , viz. H j by 282, l. Thus Z% embraces
an infinity of numbers.
2. The least number in Z B is fl.
For to any number < li corresponds a section 21 of Z y Hence
a lies in Z.
CLASSES OF ORDINALS 321
296. 1. An aggregate formed of an Kj set of Kj sets is an ^ set.
Consider the set
A = a u ,\a l2 ,
a 21 ' a 22 '
<
Here each row is an Kj set. As there are an Kj set of rows, A
is an Kj set of Kj sets. To show that A is an ^ set, we associate
each a ilc with some number in the first two number classes.
In the first place the elements a lie where i K. < CD may be associ
ated with the numbers 1, 2, 3, < co. The elements a t<0 , a^
lying just inside the &> th square and which are characterized
by the condition that i = 1, 2, o>; K = 1, 2 < co form an
enumerable set and may therefore be associated with the ordinals
o>, to f 1, ... < ft>2. For the same reason the elements just inside
the ft) + 1 st square may be associated with the ordinals ft)2, ft)2 f 1,
... < ft)3. In this way we may continue. For when we have
arrived at the a th row and column (edge of the <* th square) we
have only used up an enumerable set of numbers in the sequence
i, 2, ... w ... < n (i
in our process of association. There are thus still an 8 X set left
in 1) to continue the process of association.
2. As a corollary of 1 we have :
The ordinal numbers
n 2 , n 3 , ft 4 , ...
lie in Z% .
297. 1. Let </3<7< ... (1
be an increasing sequence of numbers in Z z having K x as cardinal
number and such that any section of 1) has K as its cardinal.
There exists a first ordinal number \ in Z% greater than any number
in 1).
For let
322 ORDINAL NUMBERS
Since a < /3 we may take 21 to be a section of S3. Similarly
we may suppose S3 is a section of (, etc.
Letnow = + *, <+
Consider now ^ _ ^ ^ ^
keeping the relative order intact. Then is well ordered by
270, 4. Let
Since Card H = Kj, by 290, l, X lies in Z 3 .
As any 21, S3, is a section of ,
<<< X.
Moreover, any number ^ < X is also < some , /3, 7 F*or if
3ft has ordinal number /*, 3W must be similar to a section of .
But there is no last section in 8.
2. We shall call sequences of the type 1), an Sj sequence.
The number X whose existence we have just established, we shall
call the limit of]). We shall also write
< ft< 7 =X
to indicate that a, /8, is an Kj sequence whose limit is X.
298. 1. The preceding theorem gives us a third method of
generating ordinal numbers. We call it the third principle.
We have seen that the first and second principles suffice to gen
erate the numbers of the first two classes of ordinal numbers but
do not suffice to generate even the first number, viz. fi in Z B . We
prove no\v the following fundamental theorem :
2. The three principles already described are necessary and suffi
cient to generate the numbers in Z B .
For let a = Ord be any number of Z 3 . If 21 has a last element,
reasoning similar to 292, l shows that
If 21 has no last element, all the numbers of Z B <a form an K
or Kj set. In the former case
a = n + ft,
CLASSES OF ORDINALS 823
where /? lies in Z a . In the latter case, reasoning similar to 292, 1
shows that we can pick out an Kj increasing sequence
299. 1. The numbers of Z form a set whose cardinal number a
is >K r
The proof is entirely similar to 293, 1. Suppose, in fact, that
a == Hj . Let us arrange the elements of Z in the Kj sequence
19 Ojj (1
As in 292, there exists in this sequence an Kj increasing sequence
a[<a< ... = a'. (2
Then ex.' lies in Z% by 297, 1. On the other hand a! is greater than
any number in 2) and hence greater than any number in 1).
But 1) embraces all the numbers in Z 3 by hypothesis. We are
thus led to a contradiction.
2. We set 2 = CardZ 8 .
3. There is no cardinal number between Kj and K 2 .
For let a = Card 21 be such a number. Then 21 is equivalent to
a section of Z%. But every such section has the cardinal num
ber K r
300. The reasoning of the preceding paragraphs may be at
once generalized. The ordinal numbers of Z n corresponding to
well ordered sets of cardinal number K n _ 2 form a well ordered set
having a greater cardinal number a than S w _ 2 . Moreover there is
no cardinal lying between K n _ 2 and a. We may therefore ap
propriately denote a by K n _ r The K n _ 2 sequence of ordinal
numbers
lying in Z n has a limit lying in Z n , and this fact embodies the
n th principle for generating ordinal numbers. The first n prin
ciples are just adequate to generate the numbers of Z n . They do
not suffice to generate even the first number in Z n+1 .
Finally we note that an S n set of N n sets forms an K n set.
CHAPTER X
POINT SETS
Pantaxis
301. 1. (JBorel.^) Let each point of the limited or unlimited set
91 lie at the center of a cube (E. Then there exists an enumerable set
of non overlapping cubes jcj such that each c lies within some (5, and
each point of 21 lies in some c. If 21 is limited and complete^ there
is a finite set \t\ having this property.
For let jZ)j, i> 2 "* be a sequence of superposed cubical divisions
of norms === 0. Any cell of D l which lies within some and
which contains a point of 21 we call a black cell ; the other cells
of D we call white. The black cells are not further subdivided.
The division D 2 divides each white cell. Any of these subdivided
cells which lies within some & and contains a point of 21 we call a
black cell, the others are white. Continuing we get an enumer
able set of nonoverlapping cubical cells Jcj.
Each point a of 21 lies within some c. For a is the center of
some S. But when n is taken sufficiently large, a lies in a cell of
Z> n , which cell lies within g.
Let now 2X be limited and complete. Each a lies within a cube c,
or on the faces of a finite number of these c. With a we associ
ate the diagonal of the smallest of these cubes. Suppose
MinS = in 21. As 21 is complete, there is a point a in 21 such
that Min S = 0, in any F^(). This is not possible, since if ?; is
taken sufficiently small, all the points of V^ lie in a finite number
of the cubes c.
Thus Min B > 0. As the c's do not overlap, there are but a
finite number.
2. In the foregoing theorem the points of 21 are not necessarily
inner points of the cubes c. Let a be a point of 21 on the face of
one of these c. Since a lies within some S, it is obvious that the
324
PANTAXIS 325
cells of some Z) n , n sufficiently large, which surround a form a
cube (?, lying within . Thus the points of 21 lie within an
enumerable set of cells }<?{, each c lying within some (. The
cells c of course will in general overlap. Obviously also, if 21 is
complete, the points of 31 will lie within a finite number of
these c's.
302. If 21 is dense, 21' is perfect.
For, in the first place, 31' is dense. In fact, let be a point of
21'. Then in any D*() there are points of 31. Let a be such a
point. Since 31 is dense, it is a limiting point of 31 and hence is a
point of 21'. Thus in any ./)*() there are points of 31'.
Secondly, 31' is complete, by I, 266.
303. Let 33 be a complete partial set of the perfect aggregate 31.
Then g = 21  33 is dense.
For if contains the isolated point c, all the points of 31 in D r *(<?)
lie in 33, if r is taken sufficiently small. But $8 being com
plete, c must then lie in 33
Remark. We take this occasion to note that a finite set is to be
regarded as complete.
304. 1. 7/31 does not embrace all 9t n , it has at least one frontier
point in 9f n .
For let a be a point of 31, and b a point of 9? n not in 31. The
points on the join of a, b have coordinates
^ = ^ + 0(^0=^(0), 0<0<1, i = 1, 2, ...n.
Let & be the maximum of those #'s such that x(d) belongs to
31 if < 0'. Then x(0 r ) is a frontier point of 31.
2. Let 31, 33 have no point in common. If Dist (21, 33) >0, we
say 31, 33 are exterior to each other.
305. 1. Let 31 = \a\ be a limited or unlimited point set in 9? m .
We say 33 < 21 is pantactic in 31, when in each D^a) there is a
point SB.
We say 33 is apantactic in 21 when in each Z)$(a) there is a point
a of 21 such that some J?T,0*) contains no point of 33.
326 POINT SETS
Example 1. Let 21 be the unit interval (0, 1), and S3 the ra
tional points in 21. Then 93 is pantactic in 21.
Example 2. Let 21 be the interval (0, 1), and $B the Cantor set
of I, 272. Then 93 is apantactic in 21.
2. If 93 < 21 is pantactic in 21, 21 contains no isolated points not
in 93.
For let a be a point of 21 not in 93. Then by definition, in any
D 5 (a) there is a point of 93. Hence there are an infinity of points
of 93 in this domain. Hence a is a limiting point of 21.
306. Let 21 be complete. We say 93 < 21 is of the 1 category
in 21, if 93 is the union of an enumerable set of apantactic sets
in 21.
If 93 is not of the 1 category, we say it is of the 2 category.
Sets of the 1 category may be called Baire sets.
Example. Let 21 be the unit interval, and 93 the rational
points in it. Then 93 is of the 1 category.
For 93 being enumerable, let 93 = \b n \. But each b n is a single
point and is thus apantactic in 21.
The same reasoning shows that if 93 is any enumerable set in
21, then 93 is of the 1 category.
307. 1. If $8 is of the 1 category in 21, 91  93 = B is > 0.
For since 93 is of the 1 category in 21, it is the union of an
enumerable set of apantactic sets {93 n J. Then by definition there
exist points a r a 2 , in 21 such that
where D(a^) contains no point of 93j, #(#2) no P ^ of 93 2 > e ^ c
Let b be the point determined by 1). Since 21 is complete by
definition, b is a point of 21. As it is not in any 93 n , it is not
in 93. Hence S contains at least one point.
2. Let 21 be the union of an enumerable set of sets 2l n j, each 2l n
being of the 1 category in 93. Then 21 is of the 1 category in 93.
This is obvious, since the union of an enumerable set of enu
merable sets is enumerable.
PANTAXIS 327
3. Let 93 be of the 1 category in 21. Then B = 21 $8 is of the
2 category in 21.
For otherwise 33 + B would be of the 1 category in 21. But
a  ( + j?) = o,
and this violates 1.
4. It is now easy to give examples of sets of the 2 category.
For instance, the irrational points in the interval (0, 1) form a
set of the 2 category.
308. Let 21 be a set of the 1 category in the cube }. Then
A = Q 21 has the cardinal number c.
If A has an inner point, Z) 6 (a), for sufficiently small 6, lies in A.
As Card D& = c, the theorem is proved.
Suppose that A luis no inner point. Let 21 be the union of the
apantactic sets 2lj < 21 2 < in Q. Let A n = Q 2l n . Let q n be
the maximum of the sides of the cubes lying wholly in A n . Ob
viously q n = 0, since by hypothesis A has no inner points. Let Q
be a cube lying in A. As q n = 0, there exists an n such that Q
has at least two cubes lying in A ni ; call them $ , Q 1 . There ex
ists an n% > n such that $ , Q l each have two cubes in A n ^\ call
them v Q O (1 O
Vo, o ' ^0,1 ' Vi, o ' Vi, i
or more shortly $ 4t t2 .
Each of these gives rise similarly to two cubes in some A n& ,
which may be denoted by $ ll} t2 lg , where the indices as before have
the values 0, 1. In this way we may continue getting the cubes
0, , 0, 4 , Q^
Let a be a point lying in a sequence of these cubes. It obvi
ously does not lie in 21, if the indices are not, after a certain stage,
all or all 1. This point a is characterized by the sequence
which may be read as a number in the dyadic system. But these
numbers have the cardinal number c.
309. Let 21 be a complete apantactic set in a cube O. Then there
exists an enumerable set of cubical celts 5qJ such that each point of
21 lies on a face of one of these q, or is a limit point of their faces.
328 POINT SETS
For let D l > D% > be a sequence of superimposed divisions
of Q, whose norms S n = 0. Let
be the cells of D l containing no point of 21 within them. Let
^21' ^22' ^23 '" (^
denote those cells of D 2 containing no point of 21 within them and
not lying in a cell of 1). In this way we may get an infinite se
quence of cells 3D = \d mn \, where for each ra, the corresponding n
is finite, and m = oo. Each point a of A lies in some d m ^ n . For 21
being complete, Dist (a, 21) > 0. As the norms S n === 0, a must lie
in some cell of D n , for a sufficiently large n. The truth of the
theorem is now obvious.
310. Let 33 be pantactic in 21. Then there exists an enumerable
set S<. S3 which is pantactic in 21.
For let Dj >D 2 > be a set of superimposed cubical divisions
of norms c? n == 0. In any cell of D l containing within it a point
of 21, there is at least one point of 93. If the point of 21 lies on
the face of two or more cells, the foregoing statement will hold
for at least one of the cells. Let us now take one of these points
in each of these cells; this gives an enumerable set @j. The
same holds for the cells of D 2 . Let us take a point in each of
these cells, taking when possible points of (Sj. Let (5 2 denote the
points of this set not in @j. Continuing in this way, let
Then (5 is pantactic in 21, and is enumerable, since each @ n is.
Corollary. In any set 21, finite or infinite, there exists an enumer
able set (5 which is pantactic in 21.
For we have only to set 93 = 21 in the above theorem.
311. 1. The points S where the continuous function f(x rr m )
takes on a given value g in the complete set 21, form a complete set.
For let tfj, <? 2 be [joints of (5 \\hirh = c. We show c is a
point of (5. For ,, f
1 ===
PANTAXIS 329
As /is continuous, *, \ ss \
/(*)=/ 00
Hence B /oo=*
and c lies in &.
2. Letf^x^ # m ) 6e continuous in the limited or unlimited set 21.
J/* /te va/we of f is known in an enumerable pantactic set (5 in 21,
which contains all the isolated points of 21, in case there be such, the
value off is known at every point of 21.
For let a be a limiting point of 21 not in (. Since (5 is pantactic
in 21, there exists a sequence of points e l ^ e z in S which = a.
Since / is continuous, /(e n )==/00 As/ is known at each e n ,
it is known at a.
X. Let g= j/{ be the class of onevalued continuous functions
defined over a limited point set 21. Then
f = Card3 = C.
For let 9?^ be a space of an infinite enumerable number of
dimensions, and let , x
y = (yii y^ )
denote one of its points. Let/ have the value rj l at e^ the value
i/ 2 at 2 for the points of @ defined in 2. Then the complex
0?r ^2 0
completely determines /. But this complex determines also a
point 77 in 9?^ whose coordinates are tj n . We now associate/ with
77. Hence c
'
^ = c.
On the other hand, f>c, since in $ there is the function
/(.rj r m ) = $r in 21, where # is any veal number.
312. Let S3 denote the class of complete or perfect subsets lying in
the infinite set 21, which latter contains at least one complete set.
For let tfj, a 2 , == a, all these points lying in 21. Then
But for t x we may take any number in 3' 1 = (1 9 2, 3, ) ; for
we may take any number in Q 2 == (t x f 1, tj + 2, ), etc.
330 POINT SETS
Obviously the cardinal number of the class of these sequences
1) is e c = C. But (a a a a ..^
^a, c* t p ^t,^ t*i 8 )
is a complete set in 21. Hence 6>.c. On the other hand, 6<c.
Forlet A>*>2> (2
be a sequence of superimposed cubical division of norms = 0.
Each D n embraces an enumerable set of cells. Thus the set of
divisions gives an enumerable set of cells. Each cell shall have
assigned to it, for a given set in 33, the sign + or according as
S3 is exterior to this cell or not. This determines a distribution
of two things over an enumerable set of compartments.
The cardinal number of the class of these distributions is 2 e = c.
But each 93 determines a distribution. Hence b< c.
Transfinite Derivatives
313. 1. We have seen, I, 266, that
Thus y M
Let now 21 be a limited point aggregate of the second species.
It has then derivatives of every finite order. Therefore by 18,
D<2T, 21", '", ) (2
contains at least one point, and in analogy with 1), we call the
set 2) the derivative of order o> <?/* 21, and denote it by
2l (w) ,
or more shortly by
2K
Now we may reason on 2l w as on any point set. If it is infinite,
it must have at least one limiting point, and may of course have
more. In any case its derivative is denoted by
((0,4.1) or att +i B
The derivative of 21"" 1 " 1 is denoted by
51(0,42) or ^42 ^ etCt
Making use of co we can now state the theorem :
TRANSFINITE DERIVATIVES 331
In order that the point set 21 is of the first species it is necessary
and sufficient that 21 r " ; = 0.
2. We have seen in 18 that 21" is complete. The reasoning 1 of
I, 266 shows that 2l u)+1 , 2l w+2 , , when they exist, are also complete.
Then 18 shows that, if 2l w+n n = 1, 2, ... exist,
Dv($? >a w+1 >8l w+2 > ) (3
exists and is complete. The set 3) is called the derivative of order
ft> 2 and is denoted by
2((a>2) or 3 j w 2 >
Obviously we may continue in this way indefinitely until we
reach a derivative of order a containing only a finite number of
points. Then ^ +1 = Q
That this process of derivation may never stop is illustrated by
taking for 21 any limited perfect set, for then
3. We may generalize as follows : Let a denote a limitary ordi
nal number. If each 2F > 0, /3 < a, we set
when it exists.
4. If 2l a > 0, while 21 +1 = 0, we say 21 is of order a.
314. 1. Let a be a limiting point of 21. Let
5 =Card F^a).
Obviously is monotone decreasing with 8. Suppose that
there exists an a and a & > 0, such that for all < 8 < &
= Card V(a).
We shall say that a is a limiting point of rank a.
If every 6 > a, we shall say that
Rank a > .
If every a$ > , we shall say that
Rank a > a.
332 POINT SETS
2. Let 2( be a limited aggregate of cardinal number a. Then there
is at least one limiting point of 21, of rank .
The demonstration is entirely similar to I, 264. Let 8 X >
S 2 > ... ~ 0. Let us effect a cubical division of 21 of norm 8j. In
at least one cell lies an aggregate 2lj having the cardinal num
ber a. Let us effect a cubical division of 21 a of norm S 2 . In at
least one cell lies an aggregate 21 2 having the cardinal number ,
etc. These cells converge to a point a, such that
Card r a (a) = a,
however small 8 is taken.
3. 7/Card 21 > e, there exists a limiting point 0/21 of rank > e.
The demonstration is similar to that of '2.
4. If there is no limiting point o/2l of rank > e, 21 is enumerable.
This follows from 3.
5. Let Card 21 be > e. Let $ denote the limiting points of 21
whose ranks are > e. Then 33 is perfect.
For obviously 33 is complete. \Ve need therefore only to show
that it is dense. To this end let b be a point of 3J. About b h j t
us describe a sequence of concentric spheres of radii r n = 0. These
spheres determine a sequence of spherical shells \8 n \, no two of
which have a point in common. If 2l n denote the points of 21 in 8 n ,
we have y = ^ (ft) = ^ + ^ + ^ + ...
Thus if eacli 2l m were enumerable, V is enumerable and hence
Rank b is not > e. Thus there is one set 2l m which is not enu
merable, and hence by 3 there exists a point of 33 in 8 m . Hut then
there are points of 33 in any T ;r r *(ft), and b is not isolated.
6. A set 21 which contains no dense component ix enumerable.
For suppose 21 were not enumerable. Let $ denote the proper
limiting points of 91. Then ^jj contains a point whose rank is > e.
But the set of these points is dense. This contradicts the hy
pothesis of the theorem.
315. Let a lie in Z n . If 2l a > 0, it is complete.
For if a is nonlimitary, reasoning similar to I, 266 shows that
2l a is complete. Suppose then that a is limitary, and 2l a is not
TRANSFINITE DERIVATIVES 333
complete. The derivatives of 21 of order < a which are not com
plete, form a well ordered set and have therefore a first element
21^, where ft is necessarily a limitary number. Then
V = Dw(v) , 7 < ft.
But every point of 21^ lies in each 21?. Hence every limiting
point of 21^ is a limiting point of each 2l v and hence lies in 21^.
Hence 2l 3 is complete, which is a contradiction.
316. Let a be a limitary number in Z n . If 21^ > for each
yS < , 2l a exists.
For there exists an K m , m < >t 2, sequence
7< 8 <e< 7; < ... = a. (1
Let c be a point of 2l>, d a point of 2l 5 ,e a point of 21% etc.
Then the set , , 
(6% d, e, f, )
has at least one limiting point I of rank S m . Let be any number
in 1). Then I is a limiting point of rank N m of the set
o,/, )
Thus I is a limiting point of every 21^, /3 < , and hence of 2l a .
317. Let us show how we may form point sets whose order a
is any number in Z l or Z 2 .
We take the unit interval 21 = (0, 1) as the base of our con
siderations.
In 21, take the points
Obviously / = 1, $; ; = 0. Hence t is of order 1. The set
SBj divides 21 into a set of intervals
!,*,, 31 3  (2
In 2l t = (0, J) take a set of points similar to 1) which has as
single limiting point, the point . In 21 2 = Q, ) take a set of
points similar to 1) which has as single limiting point, the point
, etc. Let us cull the resulting set of points 53 2 .
334 POINT SETS
Obviously $'._. i 3 i ... .to
Hence ^ = %, = l and g,,, =
A I a
Thus 93 2 is of order 2.
In eacli of the intervals 2) we may place a set of points similar
to 33 2 , such that the righthand end point of each interval 2l n is a
limiting point of the set. The resulting set $) 8 is of order 3, etc.
This shows that we may form sets of every finite order.
Let us now place a set of order 1 in 2^, a set of order 2 in 21 2 ,
etc. The resulting set 3L is of order co. For 33^ n) has no points
in 2l p 21 2 2l n ;p while the point 1 lies in every 93<, n) .
Thus gw w) ___ j
Hence oi(a,+i) n
^u> = U >
and 33o, is of order co.
Let us now place in each 2l n a set similar to 93o,, having the
righthand end point of 2l n as limiting point. The resulting set
33(u+i * 8 ^ or( l er w f 1. In this way we may proceed to form sets
of order co { 2, co f 3, just as we did for orders 2, 3, We
may also form now a set of order o>2, as we before formed a set
of order co.
Thus we may form sets of order
co , co 2 , co 3 , o> 4
and hence of order o> 2 , etc.
318. 1. Let 21 be limited or not, and let 2l t (/3) denote the isolated
points ofW. Then
a o ^ y8=l, 2, ...<fl. (1
For r = a; + a" , a" = a/ + a /f/ 
Thus a/ = ?I , + a// + ... + a(n D + 5j ( n) .
that is, 21' is the sum of the points of 2T not in 21", of the points
of 21" not in 21'", etc. If now there are points common to every
W we have r . 2a w + a ., , w==1 , 2 , ..
TRANSFINITE DERIVATIVES 335
On 8l w we can reason as on 81', and in general for any a < 1 we
have ^,
/3<a
which gives 1).
2. #*3l = 0, 21 <md 8' are enumerable.
For not every
Hence there is a first a, call it 7, such that 8l v = 0. Then 1)
reduces to ^ ^ ^ ^ ... <<y>
ft
But the summation extends over an enumerable set of terms,
each of which is enumerable by 289. Hence 81' is enumerable.
But then 81 is also enumerable by 237, 2.
3. Conversely, if 81' is enumerable, 81 = 0.
For if 81 > 0, there is a nonenumerable set of terms in 1), if
no 3l (/3) is perfect ; and as each term contains at least one point,
81' is not enumerable. If some 8l (/3) is perfect, 81' contains a per
fect partial set and is therefore not enumerable by 245.
4. From 2, 3, we have :
For 21' to be enumerable, it is necessary and sufficient that there
exists a number a in Z 1 or Z% such that 2l a = 0.
5. If 81 is complete, it is necessary and sufficient in order that 81
be enumerable, that there exists an a in Z 1 or Z 2 such that 3l a = 0.
For 8l=3l t + 3l',
and the first term is enumerable.
6. If 810 = for some /3 < fl, we say 31 is reducible, otherwise it
is irreducible.
319. If 81 > 0, it is perfect.
By 315 it is complete. We therefore have only to show that
its isolated points SlJ 1 = 0. Suppose the contrary ; let a be an
isolated point of 31".
Let us describe a sphere 8 of radius r about a, containing no
other point of 31. Let 39 denote the points of 81' in S. Let
r >r l >r z > = 0.
POINT SETS
Let S n denote a sphere about a of radius r n . Let 93 n denote the
points of 93 lying between S n ^. 1 ^ S n , including those points which
may lie on S n ^ l . Then
33= 1 + 93 a +9} 8 + + a.
Each 93 m is enumerable. For any point of 93" is a point of
93 n = a. Hence 93" = and 93 m is enumerable by 818, 2.
Thus 93 is enumerable. This, however, is impossible since
33" = a, and is thus > 0.
320. 1. In the relation
' = 221^ + 2P 0= 1, 2, ... < H,
rw on the right is enumerable.
For let US set
alsolet r^n... =0.
Let 93 n denote the points of 93 whose distance S from 31 satis
fies the relation . ^ ^
YU > o > ?*n+i
Then the distance of any point of 93!, from 21" is > r n+1 . If $
includes all points of 93 whose distance from 31 is > r x , we have
93 = 930 + 93! 4 93 2 I 
Each 93 n is enumerable. For if not, 93jf > 0. Any point of
$3% as 6 lies in 21. Hence
Dist (J, ?I n ) = 0.
On the other hand, as b lies in 93J,., its distance from 2( is
> r n+1 , which is a contradiction.
2. If W is not enumerable, there exists a first number a in Z l or
Z 2 such that ?T is perfect.
This is a corollary of 1.
8. If 21 is complete and not enumerable, there exists a first number
a in Z^ f Z% such that 2l a is perfect.
4. If 21 is complete, Qr ^ , en
* 21 = Vs f P J
where @ is enumerable, and ^J is perfect. If 21 i* enumerable, ^ = 0.
COMPLETE SETS 337
Complete Sets
321. Let us study now some of the properties of complete point
sets. We begin by considering limited perfect rectilinear sets.
Let 21 be such a set. It has a first point a and a last point b. It
therefore lies in the interval /=(#, b). If 21 is pantactic in any
partial interval J~ (a, /3) of 7, 21 embraces all the points of J,
since 21 is perfect. Let us therefore suppose that 21 is apantactic
in /. An example of such sets is the Cantor set of I, 272.
Let D = \ 8 \ be a set of intervals no two of which have a point
in common. We say D is pantactic in an interval /, when 1 con
tains no interval which does not contain some interval 8, or at
least a part of some 8.
It is separated when no two of its intervals have a point in
common.
322. 1 . Every limited rectilinear apantactic perfect set 21 deter
mines an enumerable pantactic set of separated intervals J) = jSj,
whose end points alone lie in 21.
For let 21 lie in /=(, y8), where a, /3 are the first and last
points of 21. Let 33 = / 21. Each point b of 33 falls in some in
terval 8 whose end points lie in 21. For otherwise we could
approach b as near as \ve chose, ranging over a set of points of 21.
But then b is a point of 21, as this is perfect. Let us therefore
take these intervals as large as possible and call them 8.
The intervals 8 are pantactic in /, for otherwise 21 could not be
apantactic. They are enumerable, for but a finite set can have
lengths > I/n 4 1 and < JT/w, n = 1, 2
It is separated, since 2( contains no isolated points.
2. The set of intervals .Z) = *Sj just considered are said to be
adjoint to 21, or determined by 21, or belonging to 21.
323. Let 21 be an apantactic limited rectilinear perfect point set, to
which belongs the set of intervals D = 8$. TJien 21 is formed of the
end points E\t\ of these intervals, and their limiting points JS f .
For we have just seen that the end points e belong to 21. More
over, 21 being perfect, JS f must be a part of 21.
338 POINT SETS
21 contains no other points. For let a be a point of 31 not in E,
E' . Let a be another point of 21. In the interval (a, a) lies an
end point e of some interval of D. In the interval (a, e) lies an
other end point e r In the interval (a, e^) lies another end point
f? 2 , etc. The set of points 0, e^ #% == a. Hence a lies in E 1 ',
which is a contradiction.
324. Conversely, the end points E= \e\ and the limiting points of
the end points of a pantactic enumerable set of separated intervals
D == jgj form a perfect apantactic set 21.
For in the first place, 21 is complete, since 21 = (J?, IS'). 21 can
contain no isolated points, since the intervals S are separated.
Hence 21 is perfect. It is apantactic, since otherwise 21 would em
brace all the points of some interval, which is impossible, as D is
pantactic.
325. Since the adjoint set of intervals D = \B\ is enumerable, it
can be arranged in a 1, 2, 3, order according to size as follows.
Let S be the largest interval, or if several are equally large, one
of them. The interval 8 causes /to fall into two other intervals.
The interval to the left of 8, call I Q , that to the right of 8, call I r
The largest interval in J , call S , that in / r call S r In this way
we may continue without end, getting a sequence of intervals
8, 8 , 8 X , 00 , S 01 , S 10 , S n  (1
and a similar series of intervals
A AP A' AXP 4n ***
The lengths of the intervals in 1) form a monotone decreasing
sequence which == 0.
If v denote a complex of indices i/f/c
D=!,,} = {V..J,
and J, = J^+S F + /, 1 .
326. 1. The cardinal number of every perfect limited rectilinear
point set 21 is c.
For if 21 is not apantactic, it embraces all the points of some in
terval, and hence Card 21 = c. Let it be therefore apantactic.
COMPLETE SETS 339
Let J9= {} be its adjoint set of intervals, arranged as in 325.
Let be the Cantor set of I, 272. Let its adjoint set of intervals
be H= \i] v \, arranged also as in 325. If we set S v ^ ?;, we have
D^ff. Hence Card 21 = Card 6.
But Card 6 = c by 244, 4.
2. The cardinal number of every limited rectilinear complete set 21
is either e or c.
For we have seen, 320, 4, that
where ( is enumerable and $ is perfect,
If $ = 0, Card 21 = e.
If ^>0, Card2l = c.
For Card 21 = Card <g + Card $ = e + c = c.
327. The cardinal number of every limited complete set 21 in 9t n is
either e or c. It is c, ^J2l has a perfect component.
The proof may be made by induction.
For simplicity take m = 2. By a transformation of space [242],
we may bring 21 into a unit square S. Let us therefore suppose
21 were in S originally. Then Card 21 < c by 247, 2.
Let be the projection of 21 on one of the sides of $, and 53 the
points of 21 lying on a parallel to the other side passing through a
point of . If 83 has a perfect component, Card 3$ = c, and hence
Card 21 = c. If 53 does not have a perfect component, the cardinal
number of each 53 is e. Now S is complete by I, 717, 4. Hence
if S contains a perfect component, Card S = c, otherwise Card
g = e. In the first case Card 21 = c, in the second it is e.
328. 1. Let 21 be a complete set lying within the cube Q. Let
J) l > J9 2 > denote a set of superimposed cubical divisions of Q
of norms = 0. Let d l be the set of those cubes of D l containing
no point of 21. Let rf 2 be the set of those cubes of D 2 not in d l ,
which contain no point of 21. In this way we may continue. Let
53 = [d n ] . Then every point of A = Q  21 lies in 53. For 21 being
340 POINT SETS
complete, any point a of A is an inner point of A. Hence /> p (a)
lies in .A, for some p sufficiently small. Hence a lies in some d m .
We have thus the result :
Any limited complete set is uniquely determined by an enumerable
set of cubes \d n \, each of which is exterior to it.
We may call S3 = \d n \ the border of 31, and the cells d n , border
cells.
2. The totality of all limited perfect or complete sets has the car
dinal number c.
For any limited complete set is completely determined by its
border \d n \. The totality of such sets has a cardinal number
< c c = c. Hence Card 5J < c. Since among the sets g is a cset
of segments, Card (5 > c.
329. If 3l t denote the isolated points of 31, and 31 x its proper
limiting points, we may write
a = a t + v
Similarly we have
HA=3L+A.,
31 A = 3U + 31 AS , etc.
We thus have
31 = t + 3I At + 3U +  + 3I A i t + 31 A .
At the end of each step, certain points of 31 are sifted out. They
may be considered as adhering loosely to 31, while the part which
remains may be regarded as cohering more closely to the set. Thus
we may call 3I A i t , the n th adherent, and 3l A n the w th coherent.
If the n th coherent is 0, 31 is enumerable.
If the above process does not stop after a finite number of steps,
let 3L
If 3l w > 0, we call it the coherent of order CD.
Then obviously w
We may now sift 3L as we did 31.
COMPLETE SETS 841
If a is a limitary number, defined by
we set 2l a = Dv\$\ x an (
and call it, when it exists, the coherent of order a. Thus we can
write w vr _i_w 10^/0 /1
21 =s z 2l A a t f ?i A /3 a=l, J, <# (1
a
where j8 is a number in Z 2 .
330. 1. TT/^71 21 is enumerable,
31 = 2 A ., + 31x3 a = 1, 2, .
a
= + 1) ; (1
where $ is the sum of an enumerable set of isolated sets, and J), when
it exists, is dense.
For the adherences of different orders have no point in common
with those of any other order. They are thus distinct. Thus the
sum 3 1 can contain but an enumerable set of adherents, for other
wise 21 could not be enumerable. Thus there is a first ordinal
number /3 for which
2lA = 0.
As now in general
21^= ?U
we have ^ = ^ +1 = = _
As SI A 3 thus contains no isolated points, it is dense, when not 0,
by I, 270.
2. When 21 is not enumerable, > 0. For if not, 21 = Q, and $
is enumerable.
331. g = I'. (1
For let J9 be a cubical division of space. As usual let
denote those cells of D containing a point of 21, 21' respectively.
The cells of 2l/> not in 2l/> will be adjacent to those of 2l#, and
342 POINT SETS
these may be consolidated with the cells of jD, forming a new di
vision A of norm 8 which in general will not be cubical. Then
9? 9?' 4 9? *
^*A **A ' ^*A *
The last term is formed of cells that contain only a finite number
of points of 21. These cells may be subdivided, forming a new
division E such that in
a* = i + a** (2
the last term is < e/3 Now if 8 is sufficiently small,
Hence from 2), 3) we have 1).
332. IfK >0, Card 21 = c.
For let 33 denote the sifted set of 31 [I, 712], Then $ is per
fect. Hence Card 93 = c, hence Card 21 = c.
333. Let 21 = Jttj, where each a s's metric and not discrete. If no
two of the cCs have more than their frontiers in common, 21 is an
enumerable set in the a .s*. 21 may he unlimited.
Let us first suppose that 21 lies in a cube Q. Let a denote a on
removing its proper frontier points. Then no two of the a's have
a point in common. Let
where the first term q l = Q. There can be but a finite number of
sets , such that their contents lie between two successive ^'s.
For if ,
S . i,>&
we have  , ~ , . ^
4 + a i+ " +t n >wg',.
But the sum on the left is < Q, for any n.
/>
As n may == oo, this makes Q = oo, which is absurd.
If 21 is not limited, we may effect a cubical division of 3J m .
This in general will split some of the a's into smaller sets b. In
each cube of this division there is but an enumerable set of the b's
by what has just been proved.
CHAPTER XI
MEASURE
Upper Measure
334. 1. Let 21 be a limited point set. An enumerable set of
metric sets D= \d L \^ such that each point of 31 lies in some c? t , is
called an enclosure of 21. If each point of 21 lies within some c? t , D
is called an outer enclosure. The sets d? t are called cells. To each
enclosure corresponds the finite or infinite series
which may or may not converge. In any case the minimum of all
the numbers 1) is finite and <. 0. For let A be a cubical division
of space, 21 A is obviously an enclosure and the corresponding sum
1) is also 2lA> since we have agreed to read this last symbol either
as a point set or as its content.
We call M . v .
Mm 2a t>
with respect to the class of all possible enclosures D, the upper
measure of 21, and write
2. The minimum of the sums 1) is the same when we restrict our
selves to the class of all outer enclosures.
For let J9= \d t \ be any enclosure. For each d L , there exists a
cubical division of space such that those of its cells, call them d lK ,
containing points of d t have a content differing from d t by < .
A 1
Obviously the cells \d iK \ form an outer enclosure of 21, and
343
344 MEASURE
As e is small at pleasure, Min 2c? t over the class of outer en
closures = Min 2d t over the class of all enclosures.
3. Two metric sets whose common points lie on their frontiers
are called nonoverlapping. The enclosure D = 2rf t is called non
overlapping, when any two of its cells are nonoverlapping.
Any enclosure D may be replaced by a nonoverlapping enclosure.
For let U(d l , d^) = d l + e< 2 ,
2 d s d) = d 1 + e 2 + e 3 + e 4 , etc.
Obviously each e n is metric. For uniformity let us set d l = e r
Then E '= {e n j is a nonoverlapping enclosure of 21. As
2? n <2rf B
we see that the minimum of the sums 1) is the same, ivhen we restrict
ourselves to the class of non overlapping enclosures.
Obviously we may adjoin to any cell e n , any or all of its
improper limiting points.
4. In the enclosure H== \e n \ found in 3, no two of its cells
have a point in common. Such enclosures may be called distinct.
335. 1. Let D = jc?J, J?= \e K \ be two nonoverlapping enclosures
of 81. Let
^ K =T)v(d^e K ).
Then
A=?S tM >, ,* = !, 2, .
is a non overlapping enclosure of 31.
For 8 IIC is metric by 22, 2. Two of the S's are obviously non
overlapping. Each point of 31 lies in some d t and in some e K ,
hence a lies in S^.
2. We say A is the divisor of the enclosures D, H.
336.
For let J?= [ej be an enclosure of $}. Those of its cells cZ t con
taining a point of 31 form an enclosure D= \d t \ of 31. Now the
class of all enclosures A = ^Sj of 31 contains the class D as a sub
class.
As
we have
UPPER MEASURE 345
Min 2S t < Min 2d t < Min 2? t ,
A D E
from which 1) follows at once.
337. 7/31 is metric,
3i =t.
For let D be a cubical division of space such that
(2
Let us set 33 = 2l/). I>et J?=6 t j be an outer enclosure of S3.
Since 83 is complete, there exists a finite set of cells in E which
contain all the points of 93 by 301. The volume of this set is
obviously > 33; hence a fortiori
2>8.
Hence = ^
33>93.
But = =
31 > 33, by 336,
> = />
>3le, by 2). (3
On the other hand, __ _ _
6, by 2). (4
From 3), 4) we have 1), since e is arbitrarily small.
338. If 31 is complete, = _
For by definition
with respect to all outer enclosures D = \d t \. But 21 being com
plete, we can replace D by a finite set of cells F= \f,\ lying in D,
such that F is an enclosure of 31. Finally the enclosure F can be
replaced by a nonoverlapping enclosure Gr = \g,\ by 334, 3.
Thus
31 = Min 2# t ,
with respect to the class of enclosures Gr. But this minimum
value is also 31 by 2, 8.
346 MEASURE
339. Let the limited set H = {21J be the union of a finite or infinite
enumerable set of sets 2l n . Then
For to each 2f n corresponds an enclosure D n = \d ni ] such that
Sc?m < 2l n + ~ ^ > 0, arbitrarily small.
But the cells of all the enclosures Z> n , also form an enclosure.
Hence
This gives 1), as e is small at pleasure.
340. Let 31 lie in the metric set 9ft. Let A = 9K 21, Je
complementary set. Then
For from SR = a + 4
follows = = =
2K<a + A, by 339.
But = ^
SR^aW, by 337.
341. If 21 = 93 4 S, anc? 33, & are exterior to each other ,
1 = I + f . (1
For, if any enclosure D = \d L l of 21 embraces a cell containing
a point of S3 and (, it may be split up into two metric cells rf[,
rfj', each containing points of S3 only, or of S only. Then
Thus we may suppose the cells of D embrace only cells
D 1 = \d(\ containing no point of (, and cells D n = {d'!\ con
taining no point of S3. Then
s5 t * s5; + 2rf['. (2
UPPER MEASURE 347
By properly choosing D, we may crowd the sum on the left
down toward its minimum. Now the class of enclosures D f is
included in the class of all enclosures of 93, and a similar remark
holds for D".
Thus from 2) follows that
This with 339 gives 1).
342. If 21 = + 3R, 3R being metric,
For let D be a cubical division of norm d. Let tt denote points
of 3ft in the cells containing points of Front 3ft. Let m denote
the other points of 3ft Then m and 33 are exterior to each other,
and by 337 and 341,
As a = + m + n,
Meas (33 + m) < I by 336.
Al8 l<i + m + n by 339.
Thus 5 + in<(<5 + ft + fi . (2
Now if d is sufficiently small,
3fte<m ; n<.
Thus 2) gives, as m<3ft,
which gives 1), as e>0 is arbitrarily small.
343. 1. Let ?l lie in the metric set 93, and also in the metric set
 Let 5=5321 , tf=esi.
Then 85iff.
For let
, 6) , iB = S) + 23i , <
348 MEASURE
Thus
85*8 + !(! + 5) = 8J5
2. If 2l<33, the complement of 21 with respect to 93 will
frequently be denoted by the corresponding English letter. Thus
/I = 6X81), Mod
Lower Measure
344. 1. We are now in position to define the notion of lower
measure. Let 31 lie in a metric set 2)?. The complementary set
A = 2)? 31 has ;m upper measure A. We say now that 2ft A
is the lower measure of 31, and write
By 343 this definition is independent of the set 9M chosen.
When & M
a = n
we say 21 is measurable, and write
a = Ia.
A set whose measure is is called a null set.
2. Let .'= [e^ be an enclosure of A.
Then H=Max(aH2g t ).
wrc'tfA respect to the class of all enclosures E.
3. If (S = Je t j is an enclosure of 31, the enclosures E and ( may
obviously, without loss of generality, be restricted to metric cells
which contain no points not in 2W. If this is the case, and if @,
JPare each nonoverlapping, we shall say they are normal enclosures.
If (, g are two normal enclosures of a set 21, obviously their
divisor is also normal.
LOWER MEASURE 349
345. 1. 2[>0.
For let SI lie in the metric set 9K.
Then 2=I.
But by 330,
llonce
For let SI lie in the metric set 9)?.
rheu fi + vi>9tt by 340.
Hence ?l = SW
346. 1. For any limited set 91,
[ < a < i < S (i
For let J9= \d t \ be an enclosure of 31. Then
S = Min 2<f,,
i)
when 2) ranges over the class ^P of all finite* enclosures. On the
other hand,
D
when D ranges over the class E of all enumerable enclosures.
But the class E includes the class F. Hence S < 21.
To show that < w (2
we observe that as just shown
A>A.
Hence, _ ^ 
^  A < SIR  vt = 91. (3
Z+ = 1, by 16.
This with 3) gives 2).
350 MEASURE
2. y 21 is metric, it iv measurable, and
8=5.
This follows at once from 1).
347. Let 21 be measurable and lie in the metric set 3D?. Then A
is measurable, and ** ** ^
% + A**m. (1
For ^
A = m%. (2
a=<w3 = 5,
since 21 is measurable. This last gives
! = Z8.
This with 2) shows that J. = J. ; hence ^4 is measurable. From
2) now follows 1).
348. If 21 < 93, then 8 < 8. (1
For as usual let A, B be the complements of 21, S3 with respect
to a metric set 2ft. Since 21 < 53, A > B.
Hence, by 336, = =
A. ^_ ./>.
Thus, ^ = " ^ =
<mA<<mB,
which gives 1).
349. For 21 to be measurable, it is necessary and sufficient that
where 2ft is any metric set > 21, and A = 2W 21.
It is sufficient, for then 1) shows that
i=$mZ
But the right side is by definition 21 ; hence 21 = 21.
It is necessary as 347 shows.
350. Let 21 = \a n ] be the union of an enumerable set of non
overlapping metric sets. Then 21 is measurable, and
LOWER MEASURE 351
Let S denote the infinite series on the right of 1). As usual
let S n denote the sum of the first n terms. Let 2l n = (aj, a n ).
Then 2l n < 21 and by 336,
in = S n < S , for any n. (2
Thus S is convergent and
#<I. (3
On the other hand, by 339,
1 < 8. (4
From 3), 4) follows that
S = 1 = lira S n = lira . (5
We show now that 21 is measurable. To this end, let 3ft be a
metric set > 21, and 21 B + A n = 3ft as usual.
Then ^ ^
2l n + A n = m. (6
But A < A n , hence A < A n .
Thus 6) gives = ^ ^^
A + 2i n < a,
for any n. Hence
I + lim27 n <2Jh
or using 5), l + f<^.
Hence by 339, 1 +S ,^.
Thus by 349, 31 is measurable.
351. Let
then
f 6 < .
For let 90? be a metric set > 21. Let A, B, O be the comple
ments of 21, S3, S, with reference to 2R.
Let tf={ej , F=\f n \
be normal enclosures of B, C. Let
d mn = Dv(*
and D = {d mn  the divisor of ^?, F.
352 MEASURE
As all the points of A are in , and also in (7, they are in both
E and F, and hence in the cells of D, which thus forms a normal
enclosure of A. Let
7m = 0* ml , 'C 2 " ) > Vn = (din, &<& "')
Let us set . ft / . ?
^m = 7m + 9m , / = *7n + ^n
Then by 350, ^ 3  v^
7m = ^mn , ^7n = ^mn 
By 347, ^ .'. ? 1 T
e m = 7m + ,^m ^ /n = ^7n + /l n '
Hence ^ ^ ^, ^ ^
,
Hence adding,
(SW^)
+ 2 n + Srf mn )] . (2
Now 9W = f7j.(/ m , A., d mn \ m, n = 1, 2, ...
Thus by 339, the term in [ ] is < 0. Thus 2) gives
s /n) < ^ 2 ^n < a.
But ^
S = Max ( 9W  2^ m )
e = Max (W 2^).
Thus 3) gives 1) at once.
Measurable Sets
352. 1. e 31 = 33 + 6. // , S ar^ measurable, then 21 i*
measurable, and ^
= + 6. (1
F r + 6<a , by 351
<S<S + i , by 339.
LOWER MEASURE 353
U. Let 81 = SB + S. If 8f, 53 are measurable, so it 6 <m<2
5a. (2
For let 81 lie in the metric set 3W. Then
S  8 = SB  (8 + <.) = (SW  C)  .
Thus A = <?8;
Hence C'= + A
Thus (7 is measurable by 1. Hence S is measurable by 347,
and
a = S + s.
From this follows 2) at once.
353. 1. Let 21 = 22l n 60 ^ ,<mm of an enumerable set of measur
able sets. Then 21 is measurable and
If 21 is the sum of a finite number of sets, the theorem is obvi
ously true by 352, 1. In case 21 embraces an infinite number of
sets, the reasoning of 350 may be employed.
2. Let 31 = \9l n \ be the union of an enumerable set of null sets.
Then 31 is a null set.
Follows at once from 1.
3. Let 21= 2l n J be the union of an enumerable set of measurable
scfs whose common points two and two, form null sets. Then 21 is
measurable and
i = 22l n .
4. Let @= Je n J be a nonoverlapping enclosure 0/21. Then @ is
measurable, and
i = s? n .
5. Let 33 < 21. Those cells of (g containing a point of S3 may
be denoted by S3(g, and their measure will then be of course
**'
If S3 = 21, this will be @. This notation is analogous to that
used in volume I when treating content.
8f>4 MEASURE
6. If g= \\ n \ is another nonoverlapping enclosure of some set
then
S>
f* measurable.
For the cells of 35 are
&<
Thus S l<c is metric, and
S = sS u .
354. 1. Harnack Sets. Let 21 be an interval of length I. Let
be a positive term series whose sum X > is <_ I. As in defining
Cantor's set, I, 272, let us place a black interval of length ^ in the
middle of 31. In a similar manner let us place in each of the re
maining or white intervals, a black interval, whose total lengths
= Z 2 . Let us continue in this way; we get an enumerable set of
black intervals 93, and obviously
If we omit the end points from each of the black intervals we get
a set S3*, and obviously
The set = 91  93*
we call a Harnack set. This is complete by 324 ; and by 338, 347,
= = I  \.
When X = Z, ^> is discrete, and the set reduces to a set similar
to Cantor's set. When \ < I, we get an apantactic perfect set
whose upper content is I \ > 0, and whose lower content is 0.
2. Within each of the black intervals let us put a set of points
having the end points for its first derivative. The totality of
these points form an isolated set Q and Q r == . But by 331,
$ = $' H now $ is not discrete, $ is not. We have thus the
theorem :
There exist isolated point set* which are not discrete.
LOWER MEASURE ;55f>
3. It is easy to extend Harnack sets to 5R n . For example, in 9J 2 ,
let S be the unit square. On two of its adjacent sides let us place
congruent Harnack sets . We now draw lines through the end
points of the black intervals parallel to the sides. There results
an enumerable set of black squares @ = \8 n \. The sides of the
squares @ and their limiting points form obviously an apantactic
perfect set $.
Let a\ + 02+ ... = m .
be a series whose sum < m< 1.
We can choose $& such that the square corresponding to its larg
est black interval has the area a\ ; the four squares corresponding
to the next two largest black intervals have the total area a$, etc.
Then
= 2ai = m.
Hence i = i,=5.
355. 1. If S = \t m \ is an enclosure of 21 such that
it is called an ^enclosure. Let A be the complement of 31 with
respect to the metric set 9JJ. Let E = \e n \ be an eenclosure of A.
We call @, E complementary eenclosures belonging to 21.
2. If 21 is measurable, then each pair of complementary e/2
normal enclosures @, E, whose divisor <D = Z)#(@, E), is such that
35 < e, sma// at pleasure. (1
For let @, J57 be any pair of complementary e/2 normal enclo
sures. Then
Adding, we get Q
< + j _ ( + ) < e;
or 0<i + JZ<e. (2
But the points of 3ft fall into one of three classes : 1 the points
of 3) ; 2 those of @ not in J) ; 3 those of ^ not in 2). Thus
i + J = m + ix
This in 2) gives 1).
356 MEASURE
356. 1. Up to the present we have used only metric enclosures
of a set 21. If the cells enclosing 21 are measurable, we call the
enclosure measurable.
Let @ = \t n \ be a measurable enclosure. If the points common
to any two of its cells form a null set, we say (S is non
overlapping. The terms distinct, normal, go over without
change.
2. We prove now that
with respect to the class of nonoverlapping measurable enclosures.
For, as in 339, there exists a metric enclosure m n = \d nK \ of
each e n such that 2d nK differs from e n by < e/2 n . But the set
K
jm n ( forms a metric enclosure of 21. Thus
which establishes 1).
357. Let (5 be a distinct measurable enclosure of 21. Let f denote
those cells containing points of the complement A. If for each e >
there exists an S such that f < e, then 21 is measurable.
For let @ = e + f. Then e < 21. Hence e < 21 by 348. But
Hence
and thus
358. 1. 2%e divisor 25 0/* too measurable sets 21, 93 i* #Z*0 meas
urable.
For let (, E be a pair of complementary e/4 normal enclosures
belonging to 21 ; let , F be similar enclosures of S3. Let
e = Dt>((g, E) , f =
Then
e<e/2 , T<*A by 355, 2.
LOWER MEASURE 357
Now = Dv((, 3?) is a normal metric enclosure of 35. More
over its cells g which contain points of 35 and (7(35) lie among
the cells of e, f. Hence
Thus by 357, 35 is measurable.
2. i0 91, 53 be measurable.
Let 5D = Dt;(a, 93) , U = (31,93).
For
Hence
359. Let 31 = Z7 { St m  Je tAe union of an enumerable set of
measurable cells ; moreover let ?l be limited. Then 21 is measurable.
If we set
For S) = DyCSlj, S1 2 ) is measurable by 358.
Lefc x = 3) + ^ , 2l
Then a v a a are measurable by 352, 2.
As U = (?I 1 ,2I 2 ) =
U is measurable. As U and SSj are measurable, so is 2 In a
similar manner we show that 93 3 , 93 4 are measurable. As
21 is measurable by 353, 1, and the relation 1) holds by the same
theorem.
360. Let SIj < 2I 2 < be a set of measurable aggregates whose
union 21 is limited. Then 21 is measurable, and
358 MEASURE
For let w w g r
02 = 21221! > 8 "s ~"
For uniformity let us set a x = 21. Then
2r = 2a m .
As each o n is measurable
* lira n .
361. Let 2lj, 21 2 " 6e measurable and their union 21 limited. If
3) = Dv j2l n j > 0, i measurable.
For let 21 lie in the metric set 93? ;
let s) + D = gw,a n + ^i ll = aK
as usual.
Now 3) denoting the points common to all the 2l n , no point of
D can lie in all of the 2l n , hence it lies in some one or more of the
A n . Thus D<]A n }. (1
On the other hand, a point of \A n \ lies in some A m , hence it
does not lie in 2l m . Hence it does not lie in 33. Thus it lies in
D. Hence \A n \<D. (2
From 1), 2) we have fi= $A \
As each A n is measurable, so is D. Hence 35 is.
362. If 21 1 >2I 2 > '* fl^ enumerable set of measurable aggre
gates, their divisor 3) is measurable, and
For as usual let D, A n be the complements of 2), 2l n with respect
to some metric set 2ft.
Then
Hence by 360,
lim..
LOWER MEASURE 359
As 5> = 3ttD,
we have
363. 1. The points # = (x l z m ) such that
form a standard rectangular cell, whose edges have the lengths
e l = b l a l , , e m =b m a m .
When e l = e% = = e m , the cell is a standard cube. A normal
enclosure of the limited set 91, whose cells (S = Je n f are standard
cells, is called a standard enclosure.
k l. For each e > 0, there are standard eenclosures of any limited
set%.
For let @ = \t n \ be any ^/enclosure of SI. Then
2e n i<7;. (2
Each e n being metric, may be enclosed in the cells of a finite
standard outer enclosure F n , such that
F n tt<T,/2* , n=l, 2,.
Then ^ = S ^n5 i an enclosure of SI, and
<l + 2i;, by 2).
But the enclosure jP can be replaced by a nonoverlapping
standard enclosure = fl}, as in 834, 3. But <
Hence if 2 ?; is taken < e,
and is an eenclosure.
3. Let @ = ie m {, g={W
be two nonoverlapping enclosures of the same or of different
sets. Let e mn = Z>v(e m , f n ).
360 MEASURE
Let e m =(e m<1 , e m , 2 , e,,^. ..)+* m , (3
then e m is measurable. By this process the metric or measurable
cell c m falls into an enumerable set of nonoverlapping measur
able cells, as indicated in 3). If we suppose this decomposition to
take place for each cell of @, we shall say we have superimposed $
on @.
364. (W. ff. Young.} Let S be any complete set in limited 21.
Then
H = Max 6. (1
For let 31 lie within a cube 2K, and let A = 9W  21, (7= 9W  6
be as usual the complementary sets.
Let 93 = jbj be a border set of ( [328]. It is also a non
overlapping enclosure of 0; we may suppose it is a standard en
closure of O. Let E be a standard eenclosure of A. Let us
superimpose U on 93, getting a measurable enclosure A of both
and A. Then
tf= <7 A >A.
Hence
= <m  (7= 9  (7 A < 9W  ^ A .
Thus
6 = g, by 338
<Meas (2ft A,)
! A , by 352, 2
Hence
and thus _ 
MaxS<3l (2
On the other hand, it is easy to show that
MaxS>. (3
For let A D be an eouter enclosure of A, formed of standard
nonoverlapping cells all of which, after having discarded certain
parts, lie in 2ft.
LOWER MEASURE 361
Let $ = 9W^ Z) + & (4
where g denotes the frontier points of A D lying in 21. Obviously
$ is complete. Since each face of D is a null set, g is a null set.
Thus each set on the right of 4) is measurable, hence
= m  A D +
= mI D
= aftIe' , 0<e'<e
Thus Max < > ffi ' = I >   e,
from which follows 3), since is small at pleasure.
365. 1. If 21 i$ complete, it is measurable, and
8 = a.
For by 364,
 = 21.
On the other hand,
=2l, by 338.
2. Let S3 ie any measurable set in the limited set 2(. Then
= Max . (1
For g>93 = &
Hence, >Max. (2
But the class of measurable components of 21 embraces the
class of complete components (, since each K is measurable by 1.
Thus Maxi>Max(f. (3
From 2), 3) we have 1), on using 364.
366. Van Vleck Sets. Let 6 denote the unit interval (0, 1),
whose middle point call M. Let 3 denote the irrational points of
g. Let the division D n , n = 1, 2, divide @ into equal intervals
8 n of length l/2 n .
362 MEASURE
We throw the points 3 into two classes 21 = j#{, 33 = \b\ having
the following properties :
1 To each a corresponds a point b symmetrical with respect
to M, and conversely.
2 If a falls in the segment S of J9 n , each of the other seg
ments & of D n shall contain a point a' of 21 such that a' is situated
in S f as a is situated in 8.
3 Each 8 of D n shall contain a point a f of 21 such that it is
situated in S, as any given point a of 21 is situated in (.
4 21 shall contain a point a situated in @ as any given point
a' of 21 is in any 6 n .
The 1 condition states that 21 goes over into 33 on rotating (
about M. The 2 condition states that 21 falls into n = 1, 2, 2 2 ,
2 3 , congruent subsets. The 3 condition states that the subset
2l n of 21 in & n goes over into 21 on stretching it in the ratio 2 n : 1.
The condition 4 states that 21 goes over into 2l n on contracting it
in the ratio 1 : 2 n .
We show now that 21, and therefore 33 are not measurable. In
the first place, we note that _ _
2US,
by 1. As 3 = 21 + 33, if 21 or 33 were measurable, the other would
be, and
2I = 5B = i
Thus if we show 21 or 33 = 1, neither 21 nor 33 is measurable.
We show this by proving that if 21 = < 1, then $8 is a measurable
/Sv /
set, and 33 = 1. But when 33 is measurable, 33 =  as we saw, and
we are led to a contradiction.
Let = e l f e 2 f be a positive term series whose sum e is
small at pleasure. Let Sj = \e n \ be a nonoverlapping Cjenclosure
of 21, lying in @. Then
Let SBj = 3  (g x ; then 33 t < 33, and
LOWER MEASURE 363
Each interval e n contains one or more intervals ?; nl , 77^, of
some D,, such that
27/ wm = e n  <7 n , 0<<r n
where v
<T = Z(T n
may be taken small at pleasure.
Now each rj nm has a subset 2l nm of $1 entirely similar to 31.
Hence there exists an enclosure @ n/n of 2I n/;M whose measure nm is
such that
a ntl . ?;.
_n,n _. J^n_ ^ ()r ^ _ ^ ^
1 1
But S 2 = {S n //J is a nonoverlapping enclosure of 21, whose
measure v ~ v/ ~ >.
"
if a is taken sufficiently small.
Let 33 2 denote the irrational points in (S l @ 2 . It is a part of
33, and 33 2 has no point in common with 33 t . We have
In this way we may continue. Thus 93 contains the measurable
component 5^ + 3^+...
whose measure is
As e is small at pleasure, SB = 1.
367. (F. .ff. rbwn^.) Let
,,,, , (1
6e an infinite enumerable set of point sets whose union 21 is limited.
Let 2l n > > , w = 1, 2 ?%*w ^rfj ^^s a 8^ of points each
of which belongs to an infinity of the sets 1) and of lower measure > a.
864 MEASURE
For by 365, 2, there exists in the sets 1), measurable sets
EX , 6, , 6 8 ... (2
each of whose measures S n > . Let us consider the first n of
these sets, viz. :
&J , & 2 n  (o
The points common to any two of the sets 3) form a measurable
set J)tK by 858, 1. Hence the union S ln = { J) IK  is measurable, by
359. The difference of one of the sets 3), as Sj and Dv^ v S ln ),
is a measurable set c x which contains no point in common with the
remaining sets of 3). Moreover
In the same way we may reason with the other sets ( 2 , & 3
of 3). Thus 31 contains n measurable sets c x , C 2 c n no two of
which have a common point.
Hence
c = Cl + + c n
is a measurable set and
The first and last members give
f ln >*!.
n
Thus however small > may be, there exists a /i such that
Si,, ! (4
Let us now group the sets 2) in sets of p. These sets give rise
to a sequence of measurable sets
&I M > ^2fj, , Sa^ (5
such that the points of each set in 5) belong to at least two of the
sets J.) and such that the measure of each is > the right side of 4).
We may now reason on the sets 5) as we did on those in 2).
We would thus be led to a sequence of measurable sets
Ci, , 6^ , <*  (6
ASSOCIATE SETS 365
such that the points of each set in 6) lie in at least two of the sets
5), and hence in at least 2 2 of the sets 1), and such that their
measures are.
In this way we may continue indefinitely. Let now 93 X be the
union of all the points of 21, common to at least two of the sets 1).
Let S3 2 he the union of the points of 31 common to at least 2 2 of
the sets 1), etc. In this way we get the sequence
i>a^ '
each of which contains a measurable set whose measure is
We have now only to apply 25 and 364.
368. As corollaries of 367 we have:
1. Let jQj, Q 2 be an infinite enumerable set of nonoverlapping
cubes whose union is limited. Let each Q n > a > 0. Then there
exists a set of points b whose cardinal number is c, lying in an infin
ity of the Q n and such that b > a.
2. (Arzeld.) Let y 1 ^ y% ==17. On each line y n there exists an
enumerable set of intervals of length &, r Should the number of inter
vals v n on the lines y n be finite, let v n = GO. In any case S n > a > 0,
w = l, 2, and the projections of these intervals lie in 31 = (a, b).
Then there exists at least one point x = in 21, such that the ordinate
through is cut by an infinity of these intervals.
Associate Sets
369. 1. Let l > 2 > 3 =0. (1
Let (g n be a standard e n enclosure of 2l n . If the cells of g n+1 lie in
@ n , we write l > 2 >'~ ( 2
and call 2) a standard sequence of enclosures belonging to 1).
Obviously such sequences exist. The set
3l P = D^S@ n S
is called an outer associated set of 21. Obviously
360 MEASURE
2. Each outer associated set 21, is measurable, and
a a. lime.. a
fUEOO
For each ( w is measurable; hence 3l is measurable by 362, and
= , ase n =0.
370. 1. Let A be the complement of 21 with respect to some
cube Q containing 21. Let A f be an outer associated set of A.
Then a,c^ f
is called an inner associated set of 21. Obviously
2. The inner associated set 2l t is measurable, and
21, = 21.
For A e is measurable by 369, 2. Hence 2l t =Q A, is meas
urable. But
A.f> s= A.
by 369, 2. Hence
Separated Sets
371. Let 21, 93 be two limited point sets. If there exist
measurable enclosures @, % of 21, 93 such that 3)= 7)??((S, 5) ^ s Jl
null set, we say 21, 93 are separated.
If we superimpose g on & we R e ^ an enclosure of ( = (21, 93)
such that those cells containing points of both 21, 93 form a null
set, since these cells are precisely 35. We shall call such an en
closure of ( a null enclosure.
Let 2l = {2l n } ; we shall call this a separated division of 21 into
the subsets 2l n , if each pair 2l m , 2l n is separated. We shall also
say the 2l n are separated.
SEPARATED SETS M7
372. For 21, S3 to be separated, it is necessary and sufficient that
= Dv(%,, S3.)
t a null set.
It is sufficient. For let
Then <S = (a,b,S
is a measurable enclosure of &, consisting of three measurable
cells. Of these only 35 contains points of both 21, S3. But by
hypothesis 35 is a null set. Hence 21, S3 are separated.
It is necessary. For let 9JZ be a null distinct enclosure of ,
such that those of its cells 9i, containing points of 21, 93 form a
null set. Let us superimpose 9Dt on the enclosure @ above, get
ting an enclosure 5 of 21.
The cells of arising from a contain no point of S3 ; similarly
the cells arising from b contain no point of 21. On the other
hand, the cells arising from 35, split up into three classes
The first contains no point of S3, the second no point of 21, the
cells of the last contain both points of 21, S3. As 35 a ,6^ %
$.. = 0. (1
On the other hand,
a. = a + a>>a;
hence a + 3> a + *>*>.
Thus n +>>!, (2
byl). Also i=a + ^=fl by 369, 2.
This with 2) gives ^ ^ ^ ^
a + T) a > / a hi).
Hence ^ = ^ (3
But 3:)_>35a + 3V
This with 3) gives %) b = 0.
A
In a similar manner we find that 35 a = 0. Hence 3) is a null
set by 3).
368 MEASURE
373. 1. 7/21, SJ are separated, then J) = Dv($l, 93) is a null set.
For SD e = J9i> (21,, 93,) is a null set by 372. But 2) < $),.
2. Let 21, 93 be the Van Vleck sets in 366. We saw there that
 = g = 1. Then by 369, 2, 21, = % e = 1. The divisor of 2I, 93 e is
not a null set. Hence by 372, 21, 53 are not separated. Thus the
condition that J) be a null set is necessary, but not sufficient.
374. 1. Let J2l n {, {93J be separated divisions of 21. Let
S IK = Di>(2l t , 53* ) ^Aera j t *j is a separated division o/2l afo0.
We have to show there exists a null enclosure of any two of the
sets @ IK , S mn . Now ( l(C lies in 2l t and 93*; also'6 mn lies in 2l m , 93 n .
By hypothesis there exists a null enclosure S of 2l t , 2l m ; and a null
enclosure $ of 93*, 93 n . Then = Z>v(@, g) is a nu ^ enclosure of
2l t , 2l m and of 93*, 93 n . Thus those cells of , call them , con
taining points of both 2l t , 2l m form a null set; and those of its cells
6 , containing points of both 93*, 93 n also form a null set.
Let #= \g\ denote the cells of that contain points of both
Sue, S mn . Then a cell g contains points of 2l t 2l m 33* 93 n . Thus g
lies in a or 6 . Thus in either case Q is a null set. Hence {S t *}
form a separated division of 21.
2. Let D be a separated division of 21 into the cells d^ d 2
Let E be another separated division of 21 into the cells e^ e 2
We have seen that JP = \f tK \ where / t *= Dv(d t , e^) is also a sepa
rated division of 21. We shall say that F is obtained by superim
posing E on D or D on .#, and write F=D + JE= U+ D.
3. Let J? be a separated division of the separated component 93
of 21, while D is a separated division of 21. If d t is a cell of D, e K
a cell of JE, and c?^ = Dv(d L , e K ), then
4 = OC <** )+*.
Thus superposing E on D causes each cell d t to fall into sepa
rated cells <2 tl , d ta S t . The union of all these cells, arising from
different d^ gives a separated division of 21 which we also denote
by D + E.
375. Let }2l n  be a separated division of 21. Let 93 < 21, and let
93 n denote the points of 93 in 2l n . Then *93 n  is a separated division
of 9.
SEPARATED SETS 369
For let 2) be a null enclosure of 2l m , 2l n . Let 3X& denote the
cells of 35 containing points of both 2l m , 2l n . Let S denote the
cells of ) containing points of 93; let @ 0<6 denote the cells con
taining points of both 93 m , 93 n . Then
As 33^ is a null set, so is (& .
376. 1. Let 21 = (93, 6) fo a separated division of 21.
S = i + 1. (1
For let j > 2 > = 0. There exist e n measurable enclosures
of 21, 93, 6 ; call them respectively A n , B n , O n . Then g n = A n +
J? n H (7 n is an n enclosure of 21, 93, simultaneously.
Since 93, are separated, there exist enclosures jB, (7 of 93,
such that those cells of D = B f containing points of both 93
and ( form a null set. Let us now superpose D on @ n getting
an e n enclosure JE n ~ le ns \ of 21, 93, S simultaneously. Let e bn
denote the cells of E n containing points of 93 alone ; e cn those
cells containing only points of ; and e^. those cells containing
points of both 93, . Then
2e ns = 2? 6n + 2*T cn + S^c . (2
5
As 2# 6c = 0, we see that as n == oo,
s; n ,=l , s; 6n =i , 2r cn =i.
Hence passing to the limit n= oo, in 2) we get 1).
2. .Le 21 = 593 n  be a separated division of limited 21. Then
l = 2S n . (1
For in the first place, the series
.B = 2i n (2
is convergent. In fact let 2l n == (S^ 93 2 93 n ).
Then 2l n < 21, and hence I n < I.
870 MEASl'KK
On the other hand, by 1
I n = S 1 + ... + i n = jE? n ,
the sum of the first n terms of the series 2). Thus
A,<i,
and hence B is convergent by 80, 4, Thus
B < n.
On the other hand, by 339,
^>I
The last two relations give 1).
CHAPTER XII
LEBESGUE INTEGRALS
General Theory
377. In the foregoing chapters we have developed a theory of
integration which rests 011 the notion of content. In this chapter
we propose to develop a theory of integration due to Lebesgue,
which rests on the notion of measure. The presentation here
given differs considerably from that of Lebesgue. As the reader
will see, the theory of Lebesgue integrals as here presented differs
from that of the theory of ordinary integrals only in employing
an infinite number of cells instead of a finite number.
378. In the following we shall suppose the field of integration
31 to be limited, as also the integrand 31 lies in 9J m and for brevity
we set f(x) f(x l x m ). Let us effect a separated division of
31 into cells Sj, S 2 . If each cell S t lies in a cube of side d> we
shall say D is a separated division of norm d.
As * Before, let
, *>, = Osc/= M t m, in S t .
the summation extending over all the cells of 31, are called the
upper and lower sums off over 31 with respect to D.
The sum v 5
Q, D f = 2fl) t 8 t
is called the oscillatory sum with respect to D.
379. If m = Min /, M = Max / in 31, then
For
m<m L <
371
372 LEBESGUE INTEGRALS
Hence 2wil t < 2w A < 2 J
Thus ai28 t <S D <S D <
But 28 t = I,
by 376, 2.
380. 1. Since /is limited in SI,
Max #/> , Min /S^
with respect to the class of all separated divisions D of 31, are
finite. We call them respectively the lower and upper Lebesgue
integrals of /over the field 21, and write
; Cf=UinS D .
JL<&
In order to distinguish these new integrals from the old ones,
we have slightly modified the old symbol I to resemble somewhat
script L, or I , in honor of the author of these integrals.
we say /is Lintegrable over 21, and denote the common value by
which we call the Lintegral.
The integrals treated of in Vol. I we will call Rintegrals, i.e.
integrals in the sense of Riemann.
2. Letf be limited over the null set 21. Then/ is Lintegrable in
21, and
This is obvious from 379.
381. Let 21 be metric or complete. Then
GENERAL THEORY 373
For let d v c? 2 be an unmixed metric or complete division of
of norm d. Let each cell d k be split up into the separated cells
tl , d a 
Then since c? 4 is complete or metric,
^ = (^ = 2^.
Hence using the customary notation,
^l.<
Thus summing over ,
m& <
Summing over i gives
2mA
Thus by definition,
Letting now d = 0, we get 1).
2. Let 21 be metric or complete. If f is Rintegrable in 21, it is
Linteqrable and
(2
3. In case that 21 is not metric or complete, the relations 1), 2)
may not hold.
Example 1. Let 21 denote the rational points in the interval
(0, 1).
Let
/ = 1, for x = , n even
n
= 2, when n is odd.
Then
while
since 21 is a null set. Thus 1) does not hold.
374 LEBESGtJE INTEGRALS
Example 2. Let/ = 1 at the rational points 2f in (0, 1). Then
j[/=: , jVo , *./<./ (=>
Let#= 1 in 21. Then
" 1 ' and
Thus in 3) the Zintegral is less than the 72integral, while in
4) it is greater.
Example, 3. Let /= 1 at the irrational points 21 in (0, 1).
Then
although 21 is neither metric nor complete.
382. Let Z), A be separated divisions of 91. Let
#
Then
For any cell d^ of D splits up into d^ d lt on superimposing
A, and = =
^. = s<..
But = '
M,A.
and =
383. 1. Extremal Sequences. There exists a sequence of sepa
rated divisions n n n ^i
X/ 1 , /> 2 , ^3 '" C A
each D n +! being obtained from D n by superposition, such that
^>,>^> = Cf, (2
ta
^<^< = C f. (3
~ 4r
GENERAL THEORY 375
For let j > e 2 > ^=0. For each n , there exists a division
E n such that
Let E Z + D, = I) Z , ^ 3 + D 2 =
and for uniformity set 7^ = D r Then by 382,
$i> n+l <8j> n , #/>,.< tf*,..
Hence
Letting n == oo we get 2).
Thus there exists a sequence \D' n } of the type 1) for 2), and a
sequence {!) j of the same type for 3). Let now D n = D' n + D".
Obviously 2), 3) liold simultaneously for the sequence \D n j.
2. The sequence 1) is called an extremal sequence.
3. Let \D n ] be an extremal sequence, and E any separated divi
sion of 21. Let E n = D n + E. Then E r E 2 is an extremal
sequence also.
384. Let f be Lintegrable in 21. Then for any extremal sequence
JD.I,
are the cells of /), awrf 4 aw//
Hence ^
Passing to the limit we get 1).
385. 1. Let m Min/, df =* Max /in
This follows at once from 379 and 383, 1.
376 LEBESGUE INTEGRALS
2. Let F = Max \f\in 21, then
This follows from 1.
386. In order that fbe Lintegrdble in 21, it is necessary that, for
each extremal sequence \ D n \ ,
and it is sufficient if there exists a sequence of superimposed separated
divisions \E n \, such that
71=00
It is necessary. For
/ = lim S D , / = lim S D .
eta " Jin
As /is jkintegrable,
= /  /= lim (S Dn  ^ n ) = lim Q Dn f.
o^SC 55^21
It is sufficient. For _
Both \S En \, {Sg n \ are limited monotone sequences. Their
limits therefore exist. Hence
= lim fl E = lim S E lim S E .
Thus
387. In order that f be Lintegrable, it is necessary and sufficient
that for each e > 0, there exists a separated division D of 21, for
It is necessary. For by 386, there exists an extremal sequence
\D n \, such that
:< lj) n f< e , for any n > some m.
Thus we may take D m for D.
GENERAL THEORY 377
It is sufficient. For let e 1 >e 2 > = 0. Let \D n \ be an
extremal sequence for which
Let A! = Dj , A 2 = A x + # 2 , A 3 = A 2 + D z Then { A n  is a
set of superimposed separated divisions, and obviously
Hence / is irintegrable by 386.
388. In order that f be Lintecjrable, it is necessary and sufficient
that, for each pair of positive numbers a), cr there exists a separated
division D of 31, such that if T/ X , ?; 2 , are those cells in which
Osc/> ft>, then
2^ t < a: (1
It is necessary. For by 387 there exists a separated division
D= \8,\ for which
Ha/ = 2o> t S t < wo. (2
If ^j, ^ 3 denote the cells of D in which Osc/ <. co,
fl^/ = 2ft) t ^ t f 2o> J t > o)2^ t . (3
This in 2) gives 1).
It is sufficient. For taking e > small at pleasure, let us then
take
<r^ , = 4=, (4
2fi 231
where II = Osc /in 31.
From 1), 3), and 4) we have, since o> t <. ^
&0/< 2Hf t  2o) t l t < crfl + 2o)l t < <rfi f ft>I = .
We now apply 387.
389. 1. Iff is Lintegrable in 31, it is in 95 < 31.
For let \D n \ be an extremal sequence of /relative to 31. Then
by 386,
0. (t
878 LEBESGUE INTEGRALS
Hut the sequence \D n \ defines a sequence of superposed sepa
rated divisions of 93, which we denote by \E n \. Obviously
Hence by 1),
ft* n /=o,
and / is .//integrable in 33 by 386.
2. Iff is Lintegrable in 21, so is \f\.
The proof is analogous to I, 507, using an extremal sequence
for /.
390. 1. Let j2l n j be a separated division of 31 into a finite or in
finite number of subsets. Letf be limited in 21. Then
f /=/'/+ f/+ a
4^i *LMi <^2i 2
For let us 1 suppose that the subsets 2l x 2l r are finite in num
ber. Let \D n \ bo an extremal sequence of/ relative to 21, and
}D mn \ an extremal sequence relative to 2l m . Let
Then \E n \ is an extremal sequence of /relative to 21, and also
relative to each 2l m .
Now _ _
%*.=!?*!. *.+ +^ r ,*n
Letting TI = oo, we get 1), for this case.
e /io^ r 6e infinite. We have
I=ln. (2
Let S3 = (2l 1 3I n ) , e n = SlS8 B .
Then S3 B , S B form a se[)aruti'd division of ?t, and
l=C+i.
If v is taken large enough, 2) shows that
in .
GENERAL THEORY 379
Thus by case 1,
fe in <L& n
= /'+ 4 f+', (3
<4^i *L%n
where by 385, 2
 '  < M^ n < e , n > v.
Thus 1) follows from 3) in this case.
2. Let 2l n J be a separated division of 21. Then
!''*'
cLft cL^ln
if f is L inte<jrable in 21, or if it i.s in each 2I n , and limited in 21.
391. 1. Letf = (j in 21 except at the points of a null set 91.
Then
(1
For let =+. Then
V (2
e
Similarly /"* ___ /' ^o
But/ = (7 in S3. Thus 2), 3) give 1).
392. 1. //<r>0; /V=* f/
X ^
The proof is similar to 3, 8, using extremal sequences.
2. Iffis Lintegrable in 21, ^o i* /*>
where c is a constant.
380 LEBESGUE INTEGRALS
393. 1. Let ^(V^/iO) + +/ n <, each f m being limited
in%. Then
n /* / n /*
z /*< f<* /. a
1 sta 0&9 * ^21
For let f/> n } be an extremal sequence common to F,/p /, In
each cell
^nl ^n 3 '
of D n we have
2 Min/ w < Min F < Max F < 2 Max/ m .
Multiplying by df na , summing over s and then letting w=oo,
gives 1).
2. Iff^x), /nCz) are ea^A Lintegrable in #, so Z
and
394. \. /I f* 7+ jn
I (/+*)</ f+ ff<
4^21 ^21 ^21 ci2l
For using the notation of 393,
Min (/+#) < Min/f Max^r < Max(/f ^)
in each cell c? n , of D n .
2. J/<7 f* Lintegrable in H,
Reasoning similar to 3, 4, using extremal sequences.
GENERAL THEORY 381
For x, / 7* / /
/ (/*> < / /+ / (.</) < / / J r;
a&9L _ / jQ aw91 rrf. '91 _/ .QI
etc.
4. Ijf/, ^r are Lintegrable in 21, so isf g, and
9*
ff)= ff~ f
ci;H ciil
395. J^/, # are Lintegrable in 21, so isfy.
Also their quotient f/g is Lintegrable provided it is limited in SL
The proof of the first part of the theorem is analogous to I,
505, using extremal sequences common to both f and g. The
proof of the second half is obvious and is left to the reader.
396. 1. Let f, g be limited in 21, andf<^g, except possibly in a
null set 31. Then ~ 7*
f< # a
4/51 ^21
Let us suppose first that/<. $r everywhere in 21.
Let \D n \ be an extremal sequence common to both/ and g*
Then s l)n f<s D ^.
Letting n = oo , we get 1).
We consider now the general case. Let 21 = S3 4 9t. Then
since
But in $8,f<ff without exception. We may therefore use the
result of case 1.
2. Letf> in 21. Then
For
r/= ff , r</=
4?%. 4^B ^a '
Jw Jaw
382 LEBESGUK INTEGRALS
397. The relations of 4 also hold for Lintegrals^ viz. ;
I/
(2
f I/I <//</!/]. (4
<X2l <^21 d'H
The proof is analogous to that employed for the 72integnils,
using extremal sequences.
398. Let 2l = 08to SM) be a separated division for each u == 0.
M = 0. Then
Urn f / = f /.
M> XB W ^a
For by 390, l,
IM
4^21 a^SBtt ^S,/
But by 385, 2, the last integral == 0, since S M = 0, and since/ is
limited.
399. Let f be limited and continuous in 31, except possibly at the
points of a null set Sft. Then f is Lintegrable in 21.
Let us first take 9i = 0. Then/ is continuous in 31. Let 21 He
in a standard cube Q. If Osc/ is not < e in 21, let us divide Q
into 2 n cubes. If in one of these cubes
Osc/< e, (1
let us call it a black cube. A cube in which 1) does not hold we
will call white. Each white cube we now divide in 2 n cubes.
These we call black or white according as 1) holds for them or
does not. In this way we continue until we reach a stage where
all cubes are black, or if not we continue indefinitely. In the
latter case, we get an infinite enumerable set of cubes
Hi* Hv <k (2
GENERAL THEORY 388
Each point a of 81 lies in at least one cube 2). For since / is
continuous at x * a,
l/0*)/<0</2 , * in r,(a).
Thus when the process of division has been carried so far that
the diagonals of the corresponding cubes are < S, the inequality
1) holds for a cube containing a. This cube is a black cube.
Thus, in either case, each point of 21 lies in a black cube.
Now the cubes 2) effect a separated division D of 31, and in
each of its cells 1) holds. Hence/ is iintegrable in 21.
Let us now suppose 31 > 0. We set
2l = + 9J.
Then /is iintegrable in & by case 1. It is iintegrable in 31
by 380, 2. Then it is iintegrable in 21 by 390, 1.
2. If / is Jvintegrable in 21, we cannot say that the points of
discontinuity of /form a null set.
Example. Let/= 1 at the irrational points ^ in 21 = (0, 1) ;
= at the other points $R, in 21.
Then each point of 21 is a point of discontinuity. But here
since 9? is a null set. Thus /is Zintegrable.
400. If f(x^ # w ) has limited variation in 21, it is Lintegrable.
For let D be a cubical division of space of norm d. Then by I,
709, there exists a fixed number V, such that
^o)4 m ~ l < V
for any D. Let a>, <r be any pair of positive numbers. We take
d such that
d<^. a
Let d( denote those cells in which Osc/>o>, and let the number
of these cells be i>. Let ?/ t denote the points of 21 in d( . Then
va>d m  1 < So)^ 1 < V.
384 LEBESGUE INTEGRALS
Hence T/
< f (2
tad" 1 " 1
Thus v = ., ,_^ Ftf m , ON
^.<w/<^ , by 2),
<r , byl).
ft)
Hence /is .Lintegrable by 388.
401. Jta <=/, w3l<#;
= 0, m JL = 53  21.
^ x,
//=/*,
<J/Sl at
a
if 1, </> is Lintegrable in 93 ; or 2, / is Lintegrable in 31, flftd 31, ^1
ar6? separated parts of S3.
On the 1 hypothesis let !S 8 j be an extremal sequence of <.
Let the cells of @ a be e x , 2 They effect a separated division
of 31 into cells d^, d% Let m t , J^ be the extremes off in c? t and
n t , JV" t the extremes of <f> in e t . Then for those cells containing at
least a point of 31,
nfr < m t J t < M& < Nfr (2
is obviously true when e t = d t . Let d^ < e t . If ra t j<, 0,
n t F <^ m t c? t , since m t = n t . (3
If m t > 0, w t = 0, and 3) holds.
If M L < 0, MI d, < N& , since N L = 0. (4
If M^ > 0, 4) still holds, since M^N C
Thus 2) holds in all these cases. Summing 2) gives
for the division (g,, since in a cell e of @, containing no point of 31,
= 0. Letting s = <x>, we get 1), since the end members
INTEGRAND SETS 385
On the 2 hypothesis,
r <*>= r </>+ r *= r <M r/>
XSB .Xa ^ JL% <JL%
since < being = in A, is Zintegrable, and we can apply 390.
402. 1. If
we call /a null function in 21.
2. If f^ f8 a null function in 21, ^e points ^} where />
a
For let 2T = 3 + $, so that/= in .
By 401,
0= //=//. d
Let l > e 2 > = 0. Let ^3 n denote the points of $ where
/ > n . Then
Each ^} n is a null set. For
f >*Jn=0.
cL^n
Hence ^ n = 0.
Then , ? = }^}= d + a + 
where ^^ C a =%*i, ^3=^3^3
As each Q n is a null set, $ is a null set.
Integrand Sets
403. Let 21 be a limited point set lying in an wway space 9J TO .
Let f(x l x m ) be a limited function defined over 21. Any
point of 21 may be represented by
380 LEBKSGUE !KTE(iKALS
The point x = (aj a^+j)
lies in an w + 1 way space 3? m + r The set of points \x\ in which
x m+\ ranges from oo to too is called an ordinate through a. If
x m+l is restricted by Q < < ^
we shall call the ordinate a positive ordinate of length I ; if it is re
stricted by _ i< Xm+1 <Q,
it is a negative ordinate. The set of ordinates through all the
points a of 21, each having a length =/(#), and taken positively
or negatively, as f(a) in ^ 0, form a point set $ in 3f mfl which
we call an integrand set. The points of <J for which # m+1 has a
fixed value x m+l = c form a section of 3, and is denoted by 3(0 r
by a
404. Let 21= jaj Je a limited point set in 9? m . Through each
point a, /e w ^r^<?^ a positive ordinate of constant length /, getting a
set ), m 3l m+l . Then g = ^
For let @j > @ 2 > form a standard sequence of enclosures of
), such that g = > (2
Let us project each section of @ n corresponding to a given value
of x n+l on SR m , and let 2l n be their divisor. Then 2l n > 21. Thus
Letting n == oo , and using 2), we get
5 = i *.
To prove Me res of 1), let be the complement of O with re
spect to some standard cube Q in 9J m+1 , of base Q in SR m .
Then, as just shown,
5 = IA , where ^1 = Q  21.
Hence
INTEGRAND SETS 387
405. Letf>0 be Lintegrable in %. Then
where Q is the integrand set corresponding to f.
For let \8 t \ be a separated division D of 31. On each cell S t
erect a cylinder & t of height M, = Max /in S t . Then by 404,
Let = JS t } ; the E t are separated. Hence, e>0 being small
at pleasure,
for a properly chosen D. Thus
3< f / (2
Xa
Similarly we find
From 2), 3) follows 1).
406. Letf>Q be Lintegrable over the measurable field 21.
e corresponding integrand set 3 /s measurable, and
3= / (1
L%
For by 2) in 405,
s< f /.
Xm
Using the notation of 405, let c n be a cylinder erected on B n of
height w n = Min / in S n . Let c = \c n \ . Then c < 3, and hence
c<3 ( 2
But 31 being measurable, each c n is measurable, by 404, Hence
c is by 359, Thus 2) gives
c<3 (3
Now for a properly chosen D,
388 LEBESGUE INTEGRALS
Hence
as e is arbitrarily small. From 2), 3), 4)
r = f*
I f<$<3< I f,
<Xa "= JL^i
from which follows 1).
Measurable Functions
407. Let /(^ # m ) be limited in the limited measurable set SI.
Let Six,* denote the points of SI at which
If each 3l Af * is measurable, we say f is measurable in 31.
We should bear in mind that when f is measurable in 31, neces
sarily 31 itself is measurable, by hypothesis.
408. 1. Iff is measurable in 31, the points & of 31, at which f = C,
form a measurable set.
For let 3l denote the points where
<n
where 1s ....
Then by hypothesis, 3l n is measurable. But & =
Hence S is measurable by 361.
2. If f is measurable in 31, the set of points where
is measurable, and conversely.
Follows from 1, and 407.
3. If the points 3{* in 31 where /> A, form a measurable set for
each X, f is measurable in 31.
For Slxpi having the same meaning as in 407,
Six/* = Six 31 M .
Each set on the right being measurable, so is Slx M
MEASURABLE FUNCTIONS 389
409. 1. . r ffis measurable in 81, it is Lintegrable.
For setting m = Min /, M = Max / in 21, let us effect a division
D of the interval g = (w, J!f ) of norm J, by interpolating a finite
number of points
r
Let us call the resulting segments, as well as their lengths,
rfj, rf 2 , <2 3
Let 2l t denote the points of 21 in which
m i\<f< m i i * = 1> 2, ; ra = 7w.
We now form the sums
?/,
Obviously
= , as ci! = 0. (2
We may now apply 387.
2. //'/ is measurable in 21
= lira Sw^a, = lim 2m t S t , (3
using the notation in 1.
This follows from 1), 2) in 1.
3. The relation 3) is taken by Lebesgue as definition of his
integrals. His theory is restricted to measurable fields and to
measurable functions. For Lebesgue's own development of his
theory the reader is referred to his paper, Intfyrale, Longueur,
Aire, Annali di Mat., Ser. 3, vol. 7 (1902) ; and to his book,
Lefons sur V Integration. Paris, 1904. He may also consult the
excellent account of it in ffobson's book, The Theory of Functions
of a Real Variable. Cambridge, England, 1907.
390 LEBESUCE INTEGRALS
SemiDivisors and QuasiDivisors
410. 1. The convergence of infinite series leads to the two
following classes of point sets.
Let F= s/.c*! ... *) = i/, + i/ = F n + F n , (i
1 n+l
each/ t being defined in 21.
Let us take > small at pleasure, and then fix it.
Let us denote by 2l n the points of 21 at which
n\ y V
Of course 2l n may not exist. We are thus led in general to the
se ^ s 21 21 21 (3
The complementary set A n = 21 2l n will denote the points
where I ^Y ^ I (k
I n\^/  ^ *' v
If now F is convergent at #, there exists a v such that this point
lies in 21 21 21 (5
The totality of the points of convergence forms a set which has
this property: corresponding to each of its points x, there exists
a v such that x lies in the set 5). A set having this property is
called the semidivixor of the sets 3), and is denoted by
Suppose now, on the other hand, that 1) does not converge at
the point x in 21. Then there exists an infinite set of indices
n i< n z < * ==
such that , ~
I^n 8 (*)>e.
Thus, the point x lies in an infinity of the sets
A l , A 2 , A B (6
The totality of points such that each lies in an infinity of the
sets 6) is called the quasidivisor of 6) and is denoted by
QdvMJ.
Obviously,
U + Q<lvMn} = 2l. (7
SEMIDIVISORS AND QUASIDIVISORS 391
We may generalize these remarks at once. Since F(x) is
nothing but
we can apply these notions to the case that the f unctions / t (#j # m )
are defined in 21, and that
lim/ t = <f>.
2. We may go still farther and proceed in the following abstract
manner.
The divisor 35 of the point sets
*i , v a
is the set of points lying in all the sets 1).
The totality of points each of which lies in an infinity of the sets
1) is called the quasi divisor and is denoted by
QdvfSU. (2
The totality of points a, to each of which correspond an index m a ,
such that a lies in
. , L.n,
forms a set called the semidivisor of 1), and is denoted by
SdvfSU (3
If we denote 2), 3) by Q and @ respectively, we have, obviously,
35 < @ < Q. (4
3. In the special case that Jlj >21 2 > we have
Q = @ = 3). (5
For denoting the complementary sets by the corresponding
Roman letters, we have
But Q has precisely the same expression.
Thus O = 3), and hence by 4), @ = 5).
392 LEBESGUE INTEGRALS
4. Let 2l n 4 A n = 93, n = 1, 2, ... Then
For each point b of S3 lies
either 1 only in a finite number of 2l n , or in none at all,
or 2 in an infinite number of 2l n .
In the 1 case, b does not lie in 21,, 2l, +1 ; hence it lies in
A t , A 8+l In the 2 case b lies obviously in Qdv J21J.
5. 7/Slj, 21 2 are measurable, and their union is limited,
are measurable.
For let ) n = Di;(2l n , 2l n+1 ) . Then @ = {SD n } .
But @ is measurable, as each S) w is. Thus Sdv } J. n j is measur
able, and hence Q is by 4.
6. Let O = Qdv 2l n ( , eac*A 3[ n 6em{/ measurable, and their union
limited. If there are an infinity of the 21 n , sat/
whose measure is > a, then ^
Q>a. (6
For let 93 n = (2t ln , 3t ln+1 .), then S n >.
(7
by 362. As Q>93 we have 6) at once, from 7).
Limit Functions
411. Let
a x ranges over 21, r finite or infinite. Let f be measurable in 21
and numerically <M,for each t near r. Then <f> is measurable in
21 also.
To prove this we show that the points 53 of 21 where
LIMIT FUNCTIONS 393
form a measurable set for each X, /*. For simplicity let T be finite.
Let t v 2 .. =T; also let 1 >e 2 > ==0. Let S n ,, denote the
points of 21 where
(2
Then for each point x of 93, there is an S Q such that 2) holds for
any if 8 > 8 Q . Let g n = Sdv \ S n ,} . Then 93 < 6 n . But the S n ,
being measurable, & n is by 410, 5. Finally $=> Dv jS n } , and hence
93 is measurable.
412. Let
or x in 81, anrf T finite or infinite. Let t 1 , t" = r.
, =/(a;, i (a) ) 6^ measurable, and numerically < Jtfl l>ef < =/ a
, denote the points where
i&i>.
TOen foreache>0, U m @. = 0. (1
*=<
For by 411, </> is measurable, hence #, is measurable in 81, hence
, is measurable.
Suppose now that 1) does not hold. Then
fim . = I > 0.
5=00
Then there are an infinity of the ,, as ^ ,, whose
measures are >X>0. Then by 410, 6, the measure of
=Qdv{ f 
is >\. But this is not so, since/, = <^>, at each point of 81.
413. 1. Let .
for x in 81, awrf T finite or infinite.
Ifeachf 8 =f(x, # a) ) is measurable, and numerically <Min Hfor
each sequence 1), then
/./.
i <f> = lini I / (x, t). (2
4,21 '= T J/2l'
394 LEBESGUE INTEGRALS
For set
, * ,
<t>=fs +#
an( l let I // I <^ AT * 1 9
IffflS^^f > s= i, ^
Then as in 412, $ and #, are measurable in SI. Then by 409,
they are iintegrable, and
/%= ff.+ /V (3
Jin <Asi <Xsi
Let 93, denote the points of SI, at which
and let 53,  B 9 = SI. Then 93 a , ^, are measurable, since g 9 is.
Thus by 390, r T T
I ff*= I 9*+ I ff*
^21 JL%, J*B.
Hence
By 412, , = 0. Thus
lim
^"
Hence passing to the limit in 3), we get 2), for the sequence
1). Since we can do this for every sequence of points t which
= T, the relation 2) holds.
2.
converge in SI. If each term f t is measurable, and each \ F K \ < M,
then F is Lintegrdble, and
Iterated Integrals
414. In Vol. I, 732, seq. we have seen that the relation,
holds when /is J2integrable in the metric field St. This result
was extended to iterable fields in 14 of the present volume. We
ITERATED INTEGRALS 395
wish now to generalize still further to the case that / is iinte
grable in the measurable field 21. The method employed is due to
Dr. W. A. Wilson,* and is essentially simpler than that employed
by Lebesgue.
1. Let x = (Zj z,) denote a point in sway space 3? a , s = m + n.
If we denote the first m coordinates by x 1  # w , and the remaining
coordinates by y^ y n , we have
The points x== (^ ... ^ ~. 0)
range over an raway space 9? m , when 2; ranges over 8?,. We call
x the projection of z on SK OT .
Let z range over a point set 21 lying in 3J a , then x will range
over a set 53 in 3? w > called tfAe projection of 21 on 9J W . The points
of 21 whose projection is x is called the section of 21 corresponding
to x. We may denote it by
2l(#), or more shortly by S.
We write = <
to denote that 21 is conceived of as formed of the sections , cor
responding to the different points of its projection 95.
2. Let O denote a standard cube containing 21, let q denote its
projection on $R m . Then S<_q. Suppose each section 21 (#) is
/>
measurable. It will be convenient to let 2l(#) denote a function
of x defined over q such that
2l(V) = Meas 2l(V) = S when # lies in S3,
= when # lies in q 93.
This function therefore is equal to the measure of the section of
21 corresponding to the point x, when such a section exists ; and
when not, the function = 0.
When each section 2l(#) is not measurable, we can introduce
the functions
* Dr. Wilson's results were obtained in August, 1909, and were presented by me
in the course of an address which I had the honor to give at the Second Decennial
Celebration of Clark University, September, 1909.
896 LEBESGUE INTEGRALS
Here the first = S when a section exists, otherwise it = 0, in q.
A similar definition holds for the other function.
3. Let us note that the sections
where 2l c , 2l t are the outer and inner associated sets belonging to 21.
are always measurable.
For 21, = .Z)tf{( n j, where each (g n is a standard enclosure, each
of whose cells e nm is rectangular. But the sections e nm (V) are
also rectangular. Hence
being the divisor of measurable sets, is measurable.
415. Let 2l c be an outer associated set of 21, both lying in the stand
/i
ard cube Q. Then 2l c (#) is Lintegrable in q, and
<. a
For let j( n S be a sequence of standard enclosures of 21, and
=Je Bm S. Then
<S* = 2e Bro (2
Now c nm being a standard cell, e nm (#) has a constant value >
for all x contained in the projection of c nm on q. It is thus con
tinuous in q except for a discrete set. It thus has an JSintegral,
and
e n m= j e nm (.
^
This in 2) gives
f 2e nm <, by 413, 2,
,(*), (4
by 3).
ITERATED INTEGRALS 397
On the other hand, @(#) is a measurable function by 411. Also
= Aim l n (V), by 413, 1. (5
aL/c(
pJr\Ti^ ^ '***
Thus this in 5) gives 1).
416. Let 21 lie in the standard cube Q. Let 2l t be an inner asso
ciated set. Then 2^0*0 Lintegrable in q, and
21= /'Hoe).
~~ <Xq
o = a. + A e .
21:
For
Hence S,(a;) is iintegrable in q, and
Ate*) = /"Sea:) 
oLq owq <
= QA , by 415,
= 5 t = by 370, 2.
417. e measurable ?l Ke tw ^A standard cube O
Hence < x( x )< .(*) ( 2
using 396, l, and 415, 416. From 2) we conclude 1) at once.
418. Let 21 = SB S fo measurable. Then ( are Lintegrable in
398 LEBESGUE INTEGRALS
For by 417,
by 401.
419. If 21 = S3 & is measurable, the points of $$ at which & is
measurable form a null set 31.
For by 418,
Hence
0= f (6).
1 oa
e^fijj
Thus < = 6  6
is a null function in 93, and by 402, 2, points where <f> > form a
null set.
420. i0 21 = 93 & be measurable. Let b denote the points of
/or which the corresponding sections are measurable. Then
For by 419, 33=b + 5tt,
and 5ft is a null set. Hence by 418,
/= /= / =
31= /<=/+/
^^3 Jb ^
421. ie^ /> in SI. ^ ^/^ integrand set Q, corresponding to f
be measurable, then f is Lintegrable in 31, and
*/
For the points of Q lying in an m f 1 way space 5R m+1 may be
denoted by <r (<u...<u ^
j x {y l *" y m ,z),
where y = (yi"ym) ranges over 5R m , in which 21 lies. Thus 21
may be regarded as the projection of Q on 5R m . To each point y
ITERATED INTEGRALS 399
of 21 corresponds a section 3(y), which for brevity may be denoted
by ft. Thus we may write
<*=:.
As ft is nothing but an ordinate through y of length /(y), we
have by 419, ^ /^
3= = /
422. .Z/e / Je Lintegrable over the measurable field 21 = S3 .
b denote those points of 93, /or wAi<?A / is Lintegrable over the
corresponding sections . 2%en
/ /
// // a
[ <Xfc*6@
Moreover $1 = $8 b is a null set.
Let us 1 suppose /> 0. Then by 406, 3 is measurable and
. (2
Let y8 denote the points of 93 for which 3 (x) is measurable.
Then by 420,
3 = f 300 (3
<Xp
By 419, the points
$ = 93  y8 (4
form a null set.
On the other hand, 3K#) is the integrand set of/, for 3l(V) = 6.
Hence by 421, for any x in /3,
(^)=r/, (5
ele
and ^8 < b. (6
From 2), 3), 5) we have
JL% JL? JL<
From 6) we have
9ft = 93b</3 = $,
a null set by 4). Let us set
b = + it
400 LEBESGUE INTEGRALS
Then n lying in the null set $, is a null set. Hence
Jut JL& JLn JL& JLb JL&
This with 7) gives 1).
Let f be now unrestricted as to sign. We take C > 0, such
that the auxiliary function
, n .
Then /, g are simultaneously Lintegrable over any section S.
Thus by case 1
f(/+tf)= /' f (/+#) (8
aLw cLb ~L&
Now
, r s*
(f+0)= f+ 0= I /+(?, (9
yi JLw i% <%
=
By 418, is iintegrable in S3, and hence in b. Thus
/ / / / /==
/ /(/+<?)= / / f+c <&
Xb <?6< <Xb <^(S 04/b
As b differs from 93 by a null set,
by 418. From 8), 9), 10), 11), 12) we have 1).
423. If f is Lintegrable over the measurable set 31 = 93 S, then
For by 422,
/* /^
=/ J . (2
Jit Jits,
As SB b = 91 is a null set,
/=o
e
ITERATED INTEGRALS 401
may be added to the right side of 2) without altering its value.
Thus
JL<& JLbJLo, JLwJL& JL<&JL<
424. 1. (TF. A. Wilson.) Iff(x l ^x m ) is Linteffrable in
measurable 21, f is measurable in 21.
Let us first suppose that/> 0. We begin by showing that the
set of points 21 A of 21 at which / > X, is measurable. Then by
408, 3, f is measurable in 21.
Now/ being iintegrable in 2t, its integrand set 3 is measur
able by 406. Let 3> A be the section of 3> corresponding to x m+l =* \.
Then the projection of 3? A on 9t w is 21 A . Since 3> is measurable, the
sections $A are measurable, except at most over a null set L of
values of X, by 419. Thus there exists a sequence
none of whose terms lies in L. Hence each ^A,, is measurable, and
hence 21 An is also.
As 2l An ^j < 2l An , each point of 21 A lies in
so that ^ < 2, (2
On the other hand, each point d of $) lies in $l x . For if not,
/(<*)< X.
There thus exists an s such that
< X. < X. (3
But then d does not lie in 21 A ,, for otherwise f (d) > X., which
contradicts 3). But not lying in 21 A ., d cannot lie in ), and this
contradicts our hypothesis. Thus
)<21 A . (4
From 2), 4) we have
5) = Six
But then from 1), 21 A is measurable.
Let the sign off be now unrestricted.
402 IMPROPER LINTEGRALS
Since /is limited, we may choose the constant (7, such that
Then g is Zintegrable, and hence, by case 1, g is measurable.
Hence/, differing only by a constant from g, is also measurable.
2. Let 21 be measurable. Iff is Lintegrable in 21, it is measur*
able in 21, and conversely.
This follows from 1 and 409, l.
3. From 2 and 409, 3, we have at once the theorem :
When the field of integration is measurable, an Lintegrable func
tion is integrable in Lebesgue's sense, and conversely; moreover, both
have the same value.
Remark. In the theory which has been developed in the fore
going pages, the reader will note that neither the field of integra
tion nor the integrand needs to be measurable. This is not so in
Lebesgue's theory. In removing this restriction, we have been
able to develop a theory entirely analogous to Riemaim's theory of
integration, and to extend this to a theory of upper and lower in
tegration. We have thus a perfect counterpart of the theory
developed in Chapter XIII of vol. I.
4. Let 21 be metric or complete. If f(x l # m ) is limited and
Rintegrable, it is a measurable function in 21.
For by 381, 2, it is .Lintegrable. Also since 21 is metric or
complete, 21 is measurable. We now apply 1.
IMPROPER LINTEGRALS
Upper and Lower Integrals
425. 1. We propose now to consider the case that the integrand
f(x l x m ~) is not limited in the limited field of integration 21 In
chapter II we have treated this case for jBintegrals. To extend
the definitions and theorems there given to .//integrals, we have
in general only to replace metric or complete sets by measurable
sets; discrete sets by null sets; unmixed sets by separated sets ;
UPPER AND LOWER INTEGRALS 408
finite divisions by separated divisions; sequences of superposed
cubical divisions by extremal sequences; etc.
As in 28 we may define an improper iintegral in any of the
three ways there given, making such changes as just indicated.
In the following we shall employ only the 3 Type of definition.
To be explicit we define as follows :
Let/(a; 1 a^) be defined for each point of the limited set 31.
Let 2l a /s denote the points of 21 at which
The limits , /r
lim / / , lira / / (2
, 0= eLnap , 0= at 2la3
in case they exist, we call the lower and upper (improper) Lin
tegrah, and denote them by
In case the two limits 2) exist and are equal, we denote their
common value by
ff
<%
and say/ is (improperly) Lintegrdble in 21, etc.
2. In order to use the demonstrations of Chapter II without too
much trouble, we introduce the term separated function. A func
tion f is such a function when the fields 2l a /s defined by 1) are
separated parts of 21.
We have defined measurable functions in 407 in the case that
/ is limited in 21. We may extend it to unlimited functions by
requiring that the fields 2l a j8 are measurable however large a, $ are
taken.
This being so, we see that measurable functions are special cases
of separated functions.
In case the field 21 of integration is measurable, 2f a /3 is a meas
urable part of 21, if it is a separated part. From this follows the
important result :
Iff is a separated function in the measurable field 21, it is Itin
tegrate in each 2l a p.
404 IMPROPER LINTEGRALS
From this follows also the theorem :
Let f be a separated function in the measurable field 21. If either
the lower or upper integral of f over 21 is convergent^ f is Lintegrable
in 2k and / /
/ /= lira / /.
Jb^a , fl^aoatjkp
426. To illustrate how the theorems on improper jRintegrals
give rise to analogous theorems on improper iintegrals, which
may be demonstrated along the same lines as used in Chapter II,
let us consider the analogue of 38, 2, viz. :
If f is a separated function such that / f converges^ so do I f.
ia v
Let { JEfJ'be an extremal sequence common to both
Let e denote the cells of E n containing a point of ty ft ; e' those
cells containing a point of typ ; S those cells containing a point of
2l but none of $: Then
= lira &MI e 4 2JK?,  e' f
In this manner we may continue using the proof of 38, and so
establish our theorem.
427. As another illustration let us prove the theorem analogous
to 46, viz. :
Let 2lj, 21 2 , 2l n form a separated division of 21. If f is a
separated function in 21, then
T f* r*
f = I /++ / /,
__ tt ^2li Ja&n
provided the integral on the left exists, or all the integrals on the
right exist.
For let 21,, denote the points of 2( a/3 in 21,. Then by 390, 1,
<8l, a/3
In this way we continue with the reasoning of 46.
LINTEGRALS 405
428. In this way we can proceed with the other theorems ; in
each case the requisite modification is quite obvious, by a con
sideration of the demonstration of the corresponding theorem in
JSintegrals given in Chapter II.
This is also true when we come to treat of iterated integrals
along the lines of 7078. We have seen, in 425, 2, that if 21 is
measurable, upper and lower integrals of separated functions do
not exist as such ; they reduce to iintegrals. We may still
have a theory analogous to iterated /^integrals, by extending the
notion of iterable fields, using the notion of upper measure. To
this end we define :
A limited point set at 21 = 33 S is submeasurable with respect
to 33, when
1= l.
1= f
J.
We do not care to urge this point at present, but prefer to pass
on at once to the much more interesting case of J>integrals over
measurable fields.
LIntegrals
429. These we may define for our purpose as follows :
'Letf(x l x m ) be defined over the limited measurable set 21.
As usual let 2l a /s denote the points of 21 at which
</<& , /3>0.
Let each 2l a/3 be measurable, and let / have a proper iintegral
in each 2l a p. Then the improper integral of /over 21 is
/= lim f/, (1
when this limit exists. We shall also say that the integral on
the left of 1) is convergent.
On this hypothesis, the reader will note at once that the dem
onstrations of Chapter II admit ready adaptation ; in fact some
of the theorems require no demonstration, as they follow easily
from results already obtained.
406 IMPROPER LINTEGRALS
430. Let us group together for reference the following theo
rems, analogous to those on improper J2integrals.
1. Iffis (improperly^) Lintegrable in 21, it is in any measurable
part of 21.
2. Ifg,h denote as usual the nonnegative functions associated
withf, then
L'kfr (1
3. If I f is convergent, so is I \f\, and conversely.
* JLw JL%*
4. When convergent,
i /* r*
//<//. (2
lota L%
t
f is convergent, then
'
e > 0, a > 0,
for any measurable 93 < 21, such that 93 < a.
6. Let 2l = (2l x , 21 2 2l n ) be a separated division of 21, each 2l t
measurable. Then
provided the integral on the left exists, or all the integrals on the
right exist.
7. Let 21 = J2l n ^ be a separated division of 21, into an enumerable
infinite set of measurable sets 2l n . TJien
provided the integral on the left exists.
8. Iff<g in 21, except possibly at a null set, then
L^L* < 6
when convergent.
LJNTEGRALS 407
431. 1. To show how simple the proofs run in the present
case, let us consider, in the first place, the theorem analogous to
38, 2, viz. :
If I f converges^ so do if and I f.
JLw JLy *Lm
The rather difficult proof of 38, 2 can be replaced by the follow
ing simpler one. Since
a
is a teparated division of 2T tt3 , we have
Hence
I/ I I//
\eiXafl tJL^aft' \Jdpfi JU$
But the left side is < e, for a sufficiently large a, and #, #' >
some @ Q . This shows that I is convergent. Similarly we show
the other integral converges.
2. This form of proof could not be used in 38, 2, since 1) in
general is not an unmixed division of 2f a/3 .
3. In a similar manner we may establish the theorem analo
gous to 39, viz. :
If I f and I f converge, so does I /.
JLy eLw t*
4. Let us look at the demonstration of the theorem analogous
to 43, 1, viz. :
f <7= ff ; f* f/,
% oLy eL% JxW
provided the integral on either side of these equations converges.
408 IMPROPER LINTEGRALS
Let us prove the first relation. Let 53^ denote the points of 81
at which /< ft. Then
, = Stt + %
is a separated division of 93^, and hence
f #= / ff+ Cff= I 9= //I etc 
i0 <X^ JLy ft a6$j9 JU$p
5. It is now obvious that the analogue of 44, l is the relation 1)
in 430.
6. The analogue of 46 is the relation 3) in 430. Its demon
stration is precisely similar to that in 46.
7. We now establish 430, 7. Let
=(i, Va.)
Then % = % m + Bm
is a separated division of 21, and we may take m so large that
B m < cr, an arbitrarily small positive number. Hence by 430, 6,
we may take m so large that
\L
f
Km
r/= r /+ c
tsi *L% m JL,B
.
From this our theorem follows at once.
Iterated Integrals
432. 1. Let us see how the reasoning of Chapter II may be
extended to this case. We will of course suppose that the field
of integration 21 = 33 is measurable. Then by 419, the points
of 33 for which the sections are not measurable form a null set.
Since the integral of any function over a null set is zero, we may
therefore in our reasoning suppose that every S is measurable.
Since ?l is measurable, there exists a sequence of complete com
ponents A m = B m O m in 21, such that the measure of A = \A m \ is 21.
ITERATED INTEGRALS 409
Since A m is complete, its projection B m is complete, by I, 717, 4.
The points of B m for which the corresponding sections O m are not
measurable form a null set v m . Hence the union \v m \ is a null
set. Thus we may suppose, without loss of generality in our
demonstrations, that 21 is such that every section in each A m is
measurable.
Now from
I= Ti f<7= f ((
<Z<B rJLB JL
<* r**.
we see that those points of 93 where S > O form a null set. We
may therefore suppose that S = everywhere. Then O is a
null set at each point ; we may thus adjoin them to C. Thus we
may suppose that = G at each point of 93, and that 93 = B is the
union of an enumerable set of complete sets B m *
As we shall suppose that
is convergent, let
!< 2 < = oo,
!<&<.. =00.
Let us look at the sets 2l an > 930 n , which we shall denote by 2l n .
These are measurable by 429. Moreover, the reasoning of 72, 2
shows that without loss of generality we may suppose that 21 is
such that 93 n = 93. We may also suppose that each & n is measur
able, as above.
2. Let us finally consider the integrals
/ (i
These may not exist at every point of 93, because / does not
admit a proper or an improper integral at this point. It will
suffice for our purpose to suppose that 1) does not exist at a null
set in 93. Then without loss of generality we may suppose in our
demonstrations that 1) converges at each point of 93.
On these assumptions let us see how the theorems 73, 74, 75,
and 76 are to be modified, in order that the proofs there given
may be adapted to the present case.
410 IMPROPER LINTEGRALS
433. 1 The first of these may be replaced by this :
Let B^ n denote the points of 33 at which F n > <r. Then
HmJ a , n = 0.
For by 419,
as by hypothesis the sections S are measurable. Moreover, by
hypothesis
S=(Sn + C n
is a separated division of 6, each set on the right being measur
able. Thus the proof in 73 applies at once.
2. The theorem of 74 becomes :
Let the integrals
be limited in the complete set 93. Let (5 n denote the points of $8 at
which
Then
lim @ n = 93.
n=oo
The proof is analogous to that in 74. Instead of a cubical
division of the space 9t p , we use a standard enclosure. The sets
93 n are now measurable, and thus
is measurable. Thus 6 n = b. The rest of the proof is as in 74.
3. The theorem of 75 becomes :
Let the integral
f
<X<
be limited in complete 53. Then
lim
ITERATED INTEGRALS 411
The proof is entirely similar to that in 75, except that we use
extremal sequences, instead of cubical divisions.
4. As a corollary of 3 we have
Let the integral
/ /2
be limited and Lintegrable in 33. Let JB = {J3 m } the union of an
enumerable set of complete sets. Then
lim f f/=0.
n =*JL%JLc n
For if S3 m = (^, B^ ... J5 m ), and 33 = 23 m + , we have
JL JLc n JL^ m *\ JL 2>, n JLt n
But for m sufficiently large, SD ro is small at pleasure. Hence
We h^v^ now only to apply 3.
434. 1. We are now in position to prove the analogue of
76, viz. :
Let ?{ = 33  S be measurable. Let I f be convergent. Let the
r .
integrals I f converge in 93, except possibly at a null set. Then
JLa
'{' (1
L2i JL<% at
provided the integral on the right is convergent.
We follow along the line of proof in 76, and begin by taking
/ > in 21 By 423, we have
/=lim f Cf. (2
^^^SB.iQL
412 IMPROPER LINTEGRALS
Now > being small at pleasure,
e+ C /'/< I Cf * f r & > some # ,
JL% aid %<,' <X<
"
*+
J*% << <^93 *tc n
Since we have seen that we may regard 33 as the union of an
enumerable set of complete sets, we see that the last term on the
right = 0, as n = oo, by 433, 4. Thus
/ / /* /" /*
/ / < lim / / = / , O
<A$ *l JL JLa n JL<
by 2). On the other hand,
From 3) and 4) we have 1), when/> 0.
The general case is now obviously true. For
21 = <$ + Sft,
where/ > in *ijj, and < in 9t. Here $ and 92 are measurable.
We have therefore only to use 1) for each of these fields and add
the results.
2. The theorem 1 states that if
i / /
/,///,
both converge, they are equal. Hobson* in a remarkable paper on
Lebesgue Integrals has shown that it is only necessary to assume
the convergence of the first integral ; the convergence of the second
follows then as a necessary consequence.
* Proceedings of the London Mathematical Society, Ser. 2, vol. 8 (1909),
p. 31.
ITERATED INTEGRALS 413
435. We close this chapter by proving a theorem due to
Lebesgue, which is of fundamental importance in the theory of
Fourier's Series.
Letf(x^) be properly or improperly Lintegrable in the interval
2l = a<6). Then
For in the first place,
J f< fV(* + *)!<** + \f\dx<2\f\dx. (2
fi/a
Next we note that
Hence

JL>a
or J t J a <J t _^ (3
From 2), 3) we have
Jt<J. + z\fg\dx. (4
Let now g ^ f fa\f\<a,
= for />(?.
Then by 4), J,<J. + * C\f  g \te
JLa
<^ + ^
where e' is small at pleasure, for Gr sufficiently large. Thus the
theorem is established, if we prove it for a limited function,
\ff(*)\<0>
Let us therefore effect a division of the interval F = ( #, #),
of norm d, by interpolating the points
<?<<?!< f a < <(?,
causing F to fall into the intervals
7r 7 2 ' 73
414 IMPROPER LINTEGRALS
Let h m = c m for those values of x for which g(x) falls in the in
terval 7 m , and = elsewhere in 51. Then
e', c f small at pleasure,
for d sufficiently small.
Thus we have reduced the demonstration of our theorem to a
function h(x) which takes on but two values in 21, say and 7.
Let @ be a <r/4 enclosure of the points where h = 7, while $ may
denote a finite number of intervals of ( such that g @ < cr/4.
Let <f> = 7 in (g, and elsewhere = ; let i/r = 7 in 5, and else
where = 0. Thus using 4),
since 7i =<^ in (a, y8), except at points of measure < <r/4. Similarly
Thus J h <J* + (r r i<Jt + ,
for cr sufficiently small.
Thus the demonstration is reduced to proving it for a i/r which
is continuous, except at a finite number of points. But for such a
function, it is obviously true.
CHAPTER XIII
FOURIER'S SERIES
Preliminary Remarks
436. 1. Let us suppose that the limited function f '(V) can be
developed into a series of the type
f(x) = a 4 ! cos x 4 # 2 cos 2 a; { 3 cos 3 a: 4
4 b 1 sin a; 4 6 2 sin 2 x 4 J 3 sin 3 a? 4 (1
which is valid in the interval 21 = ( TT, TT). If it is also known
that this series can be integrated term wise, the coefficients a n , b n
can be found at once as follows. By hypothesis
/*rr /ir
fdx = a G I dx 4 a* / cos a;cfe 4
<XT X^
4 b l I silica; 4
As the terms on the right all vanish except the first, we have
*. (2'
Let us now multiply 1) by cos nx and integrate.
f*7T /**
f\x) cos nxdx = a I cos nxdx 4 a^ I cos a; cos nxctx 4
r cXw <X^
7 T^
4 o l I sin a; cos :
I cos wa; cos nxdx =0 , TTI = /,
r
cos 2 wa;c?a; = TT,
Now
sin wa; cosna:=0.
r
416
410 FOURIER'S SERIES
Thus all the terms on the right of the last series vanish except
the one containing a n . Hence
a n =  rV(*Ocos nxdx. (2"
Finally multiplying 1) by sin nx, integrating, and using the
relations ,*
I sin mx sin nxdx = , m^n,
JL*
*

sin 2 nxclx = TT,
n
get
~
b n =  / f(x) sin nxdx. (2
TTeLV
Thus under our present hypothesis,
1 /' 1 /* 7r
f(x) =  / f(u)du H 2 cos w # I /*( w ) cos
"
H 2 s ^ n wa; / /( w ) s ^ n nu du. (3
7T 1 eX ff
The series on the right is known as Fourier s series ; the coeffi
cients 2) are called Fourier's coefficients or constants. When the
relation 3) holds for a set of points 93, we say^(a?) can be de
veloped in a Fourier's series in 33, or Fourier's development is valid
in 93.
2. Fourier thought that every continuous function in 31 could
be developed into a trigonometric series of the type 3). The
demonstration he gave is not rigorous. Later Dirichlet showed
that such a development is possible, provided the continuous
function has only a finite number of oscillations in 8. The func
tion still regarded as limited may also have a finite number of
discontinuities of the first kind, i.e. where
/O + O) , /OO) (4
exist, but one at least is ^=/(a).
At such a point a, Fourier's series converges to
PRELIMINARY REMARKS 417
Jordan has extended Dirichlet's results to functions having
limited variation in 21. Thus Fourier's development is valid in
certain cases when f has an infinite number of oscillations or
points of discontinuity. Fourier's development is also valid in
certain cases when f is not limited in 21, as we shall see in the
following sections.
We have supposed that f(x) is given in the interval
21 = ( TT, TT). This restriction was made only for convenience.
For if f(x) is given in the interval 3 = (a < 6), we have only to
change the variable by means of the relation
u ^7r(2xab)
b a
Then when x ranges over $, u will range over 21.
Suppose /is an even function in 21; its development in Fourier's
series will contain only cosine terms. For
00
/(a?) = 2(a n cos nx f b n sin nx),
o
CO
/(  x) = 2(a n cos nx b n sin nx).
o
Adding and remembering that f(x) =/( x) in 21, we get
00
/OB) = ^a n cos nx, f even.
o
Similarly if / is odd, its development in Fourier's series will
contain only sine terms ;
f(x) = ^2S n sin nx, f odd.
i
Let us note that if f(x) is given only in 93 = (0, TT), and has
limited variation in 93, we may develop f either as a sine or a
cosine series in 93. For let
#O)=/O) , a: in 93
=/(#) , zinCTr, 0).
Then g is an even function in 21 and has limited variation.
Using Jordan's result, we see g can be developed in a cosine
series valid in 21. Hence / can be developed in a cosine series
valid in 93.
418 FOURIER'S SERIES
In a similar manner, let
= /(*) , 7r<*<0.
Then h is an odd function in 21, and Fourier's development
contains only sine terms.
Unless /(0) = 0, the Fourier series will not converge to /(O)
but to 0, on account of the discontinuity at x = 0. The same is
true for X=TT.
If /can be developed in Fourier's series valid in 2l = ( TT, TT),
the series 3) will converge for all x, since its terms admit the
period 2 TT. Thus 3) will represent f(x) in SI, but will not
represent it unless f also admits the period 2 TT. The series 3)
defines a periodic function admitting 2 TT as a period.
EXAMPLES
437. We give now some examples. They may be verified by
the reader under the assumption made in 436. Their justifica
tion will be given later
1. /0*0 = x *> f r
Then
sin x sin 2x , sin
If we set x = ^, we get Leibnitz's formula,
4 = l~3~ f 5~~7* '"
Example 2. /() = a? > < x< TT
Then ' ~ ~~
f( ^ ^^^I CO8 ^ i_ COS ^ ^ 4_ COS ^ x 4. 1
If we set x = 0, we get
"s "r^" 1 "^ 4  ^ + " 
PRELIMINARY REMARKS 419
Example S. /(X) = 1 , < x < TT
= , X ac 0, 7T
= 1 , TT < # < 0.
Then
*/ ^ 4 f sin # , sin 3 x , sin 5 x .
Example 4. f(x) = x , 0<z<^
7T ^ ^
= ^r^ , 2<*<>r
By defining / as an odd function, it can be developed in a sine
series, valid in (0, TT). We find
n x sin 3 x , sin 5 x
Examples.
By defining / as an even function, we get a development in
cosines,
cos x cos 3a; cos
zW in (0, TT).
Example 6. f (x) = J(TT x) , < # < TT.
By defining / as an odd function we get a development in
sines,
f(x) = sin x f % sin 2# h ^ sin 3# f
vaZid m ( TT, TT).
Example 7. Let/(o?) =  , < x < ^
3 3
7T
27T
420 FOURIER'S SERIES
Developing/ as a sine series, we get
/., N . o , sin 4x , sin 8x ,
/(a;) = sm 2aH   + T +
valid in (0, TT).
Example 8. f(x)=*e x , in ( TT, TT).
We find
j oil it rj* sin. 2 x I  sin o x
valid for TT < # < TT.
Example 9. We find
__ 2 /u. ._ f 1 cos # , cos 2 # cos 3 #
vafoW/0r
.
cos //,# =  sin 
TT I 2
Let us set x = TT, and replace /x by x ; we get
TT . 1,1,1,1
COt 7TX =_+
' O 72 ^ ~2 _ 12 ' ~2 _ s>2 ^ ~2 ___ Q2 ^
* JU JU J. JU ^^ ~t Jl' ^^ _>
a decomposition of cot TTX into partial fractions, a result already
found in 216.
Example 10. We find
2 f , 2 cos 2 z 2 cos 4 a; 2 cos 6 x }
Oj i /y I I __ _ ______^__ * . * L
7^1 1.3 3.5 5.7 r
valid for < # < TT.
Summation of Fourier's Series
438. In order to justify the development of/(V) in Fourier's
series F, we will actually sum the F series and show that it con
verges to /(#) in certain cases. To this end let us suppose that
f(x) is given in the interval 21 = ( TT, TT), arid let us extend /by
giving it the period 2 TT. Moreover, at the points of discontinuity
of the first kind, let us suppose
SUMMATION OF FOURIER'S SERIES
Then the function
421
<K<0 =/O + 2 w) 4/(*  2 v)  2/O)
is continuous at w = 0, and has the value 0, at points of continuity,
and at points of discontinuity of 1 kind of/. Finally let us sup
pose that / is (properly or improperly) Zintegrable in 31 ; this
last condition being necessary, in order to make the Fourier co
efficients a n , b n have a sense.
Let
f <z 2 cos
4 6j sin # 4 J 2 s ^ n 2 a; 4
= ~ a 4 2(a n cos w# + 6 n sin nx),
where we will now write
1 /c
a n = 
I /(a?)
oLc
(2'
sn
Since /(a?) is periodic, the coefficients a n , & n have the same value
however c is chosen. If we make <?= TT, these integrals reduce
to those given in 436.
We may write
1 /e+2 jr co
_F= I / ()dt \ I 4 E(cos nx cos n^ h sin 712; sm
TTXc 1
f r+  cos n(*  x)\f(C)dt.
Thus
where
Provided
we may write
1
p=
P. I + 2 cos (*).
sin * Wo,
(3
(4
( 5
.
^j sin A"(c
sn
 *) +22 sin J(  a;) cos m(t  as)
422 FOURIER'S SERIES
P _ sin K2n + !)(**)
n ~ ' ^
if 5) holds. Let us see what happens when 5) does not hold.
In this case %(t x) is a multiple of TT. As both t and x lie in
(0, c h 2 TT), this is only possible for three singular values :
t = x ; = <?, x = c f 2 TT ; = <? f 2 TT, # = <?.
For these singular values 4) gives
As P n is a continuous function of , #, the expression on the
right of 6) must converge to the value 7) as #, t converge to these
singuLar values. We will therefore assign to the expression on
the right of 6) the value 7), for the above singular values. Then
in all cases . x _
Fm _ 1 p^8inK2 + l)Q g ) /(0<ft>
7T<X C 2 sin  ( 2:)
Let us set 241 = t =
Then !/**)+, sinw
^ n = I f(x f 2 w) c?w.
Kj*\(.cx) SI 111*
Let us choose c so that
c x = TT,
then m /.o /.I 1
^n=J = / +/
c^_7r X_E eC'O
2 2
Replacing w by u in the first integral on the right, it becomes
~, N sin vu 7
7(2? 2 u) au.
smu
Thus we get
7T
"1 /^2 *
7rJ,x) * sin u
Let us now introduce the term 2/(V) under the sign of inte
gration in order to replace the brace by </>(w). To this end let us
VALIDITY OF FOURIER'S DEVELOPMENT 425
where S3' is S3 r or S3 r _i 4 S3 r , depending on the parity of r. Now
9\* (2
du
I xrv r f ( TT\ 1
l^L^ {K)^ + Jjn
<
'2*l
V + n) ~
*f
oL/ 7
du.
<
\9\
(4
Thus J n = 0, if the three integrals 2), 8), 4) =^ 0. Moreover,
if these three integrals are uniformly evanescent with respect to
some point set & < 53, J n is also uniformly evanescent in &. In
particular we note the theorem
J n = 0, if g is Lintegrable in 33.
We are now in a position to draw some important conclusions
with respect to Fourier's series.
440. 1. Let f(x) be Lintegrable in (c, <?H27r). Then the
Fourier constants a n , b n = 0, as n = oo.
For
=  / 1+ /(^
TTrtLc
cos
is a special case of the 7 n integral. Af is iintegrable, we need
only apply the theorem at the close of the last article. Similar
reasoning applies to b n .
2. For a given value of x in 21 = ( TT, TT) let
sin u
be Lintecjrable in S3 = f 0, ]. Then Fourier s development is valid
at the point x.
426 FOURIER'S SERIES
For by 438, Fourier's series =/(#) at the point #, if D n (x) 0.
But D n is a special case of J n for which the g function is in
tegrable. We thus need only apply 439.
3. For a given x in 21 = ( TT, TT), let
X00 = *> (2
be Lintegrable in S3 = f 0, ^J. Then Fourier's development is valid
at the point x.
For let 8 > 0, then
<
= , as 8 = , by hypothesis.
4. .For a ^iv6?i x in 21 = ( TT, TT), Z^^
^) (3
6^ Lintegrable in 21. ^Ae/i Fourier's development is valid at the
point x.
Thus x is Jvintegrable in f 0,  ), as it is the difference of two
integrable functions.
441. (Lebesgue). For a given x in 21 = ( TT, TT) let
1 limn
n=
2 lim / '^(M f 8)  ^(w)  du =
^f
VALIDITY OF FOURIER'S DEVELOPMENT
427
Then Fourier's development is valid at the point x.
For as we have seen,
 sin vu
sin
du
du
, i *' J \<t> (y) 
_ I T. L sill VU
( 7
Jiftn IS''
where /3 n is a certain number which = , as u = oc.
Hence
tr^ consider D f . Since ()<w< , we have O<J/M<TT
~
sin ^^
sin w
,
f T
0<<7, T<
^ v*u?l+ av\i\ H W
1 M 1 *TT
6 V 4 J 6
; ^ j,
6
< i/, provided s > ^.
But this is indeed so. For
Hence
Thus
Fe
< VGH ^ t 7T
i f _i T 
> 1 >f , if i/>5.
D f <v I 0 dw = 0, by hypothesis.
~i/o
to D' r . We have
428 FOURIER'S SERIES
Now / being Lintegrable,
K)
is iintegrable in (17, ~J. Thus
lim = 0.
n=*>J^
But by condition 2, ,. /^ _ n
Thus lim IX' = 0.
5=0
Finally we consider D f ". But the integrand is an integrable
function in f /3, ^ J . Thus it = as n = x> .
442. 1. yAe validity of Fourier's development at the point x de
pends only on the nature off in a vicinity of x, of norm 8 as small as
we please.
For the conditions of the theorem in 441 depend only on the
value of /in such a vicinity.
2. Let us call a point x at which the function
0( tt ) =/( + 2 u) +f(x  2 u)  2 /(a?)
is continuous at u = 0, and has the value 0, a regular point.
In 438, we saw that if # is a point of discontinuity of the first
kind for /(#), then # is a regular point.
3. Fourier's development is valid at a regular point x, provided
for some rj
lim
3=0
For at a regular point #, <j>(u) is continuous at w = 0, and =
for u = 0. Now
lim J
A=0 A
LIMITED VARIATION 429
tr rr
Thus
C n 1 /**
n I  <(V)  dw = TT I  <f> 
Hence condition 1 of 441 is satisfied.
Limited Variation
443. 1. Before going farther we must introduce a few notions
relative to the variation of a function f(x) defined over an interval
21= (a < i). Let us effect a division D of 21 into subintervals,
by interpolating a finite number of points a 1 < a%< The sum
F, 2 /(a,) /(+,)  (1
is called the variation off in 21 for the division D. If
Max V D (2
is finite with respect to the class of all finite divisions of 21, we say
f has finite variation in 21. When 2) is finite, we denote its value by
Var/, or FJ,, or V
and call it the variation off in 21.
We shall show in 5 that finite variation means the same thing
as limited variation introduced in I, 509. We use the term finite
variation in sections 1 to 4 only for clearness.
2. A most important property of functions having finite vari
ation is brought out by the following geometric consideration.
Let us take two monotone increasing curves A, B such that one
of them crosses the other a finite or infinite number of times. If
/(#), g(x) are the continuous functions having these curves as
graphs, it is obvious that
d(x)=f(x)g(x)
is a continuous function which changes its sign, when the curves
A, B cross each other. Thus we can construct functions in infinite
variety, which oscillate infinitely often in a given interval, and
which are the difference of two monotone increasing functions.
430 FOURIER'S SERIES
For simplicity we have taken the curves A, B continuous. A
moment's reflection will show that this is not necessary.
Since d(x) is the difference of two monotone increasing functions,
its variation is obviously finite. Jordan has proved the following
fundamental theorem.
8. If f(x) has finite variation in the interval 31 = (a < 6), there
exists an infinity of limited monotone increasing functions </(#), h(x)
such that f = ffh. (1
For let D be a finite division of 21. Let
P D = sum of terms 5/(0 OT+1 ) /()! which are > 0,
Then +1 )  /> \=P D + N D . (2
Also
(i) /() ! + !/(> /Oh) ! + 
On the left the sum is telescopic, hence
W)M=PDN D . (a
From 2), 3) we have
Fi = 2 P,, +/(;/()= 2 ^+/(6) /(a). (4
Let now Mftx p^ = p ^ Max ^ = N
with respect to the class of finite divisions D.
We call them the positive and negative variation of /(#) in 3(.
Then 4) shows that
Adding these, we get j^ = p + j^ n;
From 5) we have
/(ft) _/() = PJK (7
Instead of the interval 21 = (<&), let us take the interval
(a < re), where x lies in 21. Replacing f> by a; in 7), we have
=/() + P(x)  N(x). (8
LIMITED VARIATION 431
Obviously P(v), N(x) are monotone increasing functions.
Let v*(x) be a monotone increasing function in 21. If we set
(9
we get 1) from 8) at once.
4. From 8) we have
1/00 1 <
5. We can now show that when f(x) has finite variation in the
interval 21 = (a < 6) i Aas limited variation and conversely.
For if / has finite variation in 21 we can set
where </>, ^ are monotone increasing in 21. Then if 21 is divided
into the intervals Sj, & 2 we have
Osc/ < Osc <j) \ Osc ^ , in S t .
OSC < = A< , OSC i/r = Ai/r , ill S t
since these functions are monotone. Hence summing over all the
intervals S t ,
< some ^If, for any division.
Hence / has limited variation.
If f has limited variation in 21,
A/ 1 < Osc/ , in 8 t .
Hence 2  A/ 1 < 2 Osc/ < some J!f.
Hence /has finite variation.
6. If f(x) has limited variation in the interval 21, its points of
continuity form a pantactic set in 21.
This follows from 5, and I, 508.
432 FOURIER'S SERIES
7. Let a<b<c; then iff has finite variation in (a, <?),
V a<b f+ V b<c f= V a , c f, (11
where V a% b means the variation off in the interval (a, i), etc.
F r V M f= Max V D f
with respect to the class of all linite divisions D of (a, <?). The
divisions D fall into two classes :
1 those divisions E containing the point 6,
2 the divisions F which do not.
Let A be a division obtained by interpolating one or more
points in the interval. Obviously
> V D f.
Let now Gr be obtained from a division F by adding the point
* Then v a f>v F f.
Hence Max F A >Max V F .
E F
Hence to find F^ c /, we may consider only the class E. Let
now E l be a division of (a, J), and E^ a division of (J, c). Then
j?! f E^ is a division of class J?. Conversely each division of class
E gives a division of (a, 6), (J, c?). Now
From this 11) follows at once.
444. We establish now a few simple relations concerning the
variation of two functions in an interval SI = (a < i).
For
where for brevity we set f f< \
2 ' F(6f)=t Vf. (2
For
LIMITED VARIATION
433
3. Letf, g be monotone increasing functions in 21. Then
 Vf+ Vg.
(3
4. _#V any two functions f, g having limited variation,
V(f+g)<Vf+Vg.
5. Letf,^ have limited variation in 21 = (a, 6).
For by 443, 8) we have
/=P
where
Thus
 NP
Hence by 2, 4,
Vff\<
 NAi + AP l 
^ + VPA l +...
...) , by 3
But
^(P + N+ a)(P, + ^ +
Vf=P + N , hence, etc.
(4
(5
445. Fourier's development is valid at the regular point x, if there
exists a < f < , such that in (0, f) <Ae variation V(u) o
e'w any (w, f ) z limited, and such that u V(u) ==0, u = 0.
By 442, we have only to show that
is evanescent with 8.
484 FOUKIKR'S SERIES
Let us first suppose that i/r(w) is monotone in some (0, f), say
monotone increasing. Similar reasoning will apply, if it is mono
tone decreasing. Then, taking < ?; f & < f,
^ =j\ty(u + )ty(u)\du= I Vo + S )^ / VOO**
In the second integral from the end, set v = w f 8.
/*? /Vf5
Then I ^r(w f B)du = I ty(v)dv.
Hence, ,,,,+a ,,,
^ = J ty(u)du  / ^(u)du
< T 25 ^  du+r**\ + \ du = ^ 4 %.
Thus
We will consider the integrals on the right separately. Let
<H in (S, 28).
smw
Now . A 9 A / i
sin n = u ar'u* , < o' < ^ .
Hence, ^ ^
=  + (TV , I o 1 < some Jf.
sin u u
= , ;is S = , since <f>(u) = 0,
as x is a regular point.
We turn now to ^ 2 . In (/, 77 + 8), 5, 77 sufficiently small,
sin w> w ^ w 8 > i/(l i; 2 ).
LIMITED VARIATION 435
Thus, if fa = Max j <f> \ in (17, rj + S),
with 8.
Thus, when i/r is monotone in some (0, f), Fourier's develop
ment is valid. But obviously when i/r is monotone, the condition
that uV(u)^=Q is satisfied. Our theorem is thus established in
this case.
Let us now consider the case that the variation V(u) of ^ is
limited in (u, f).
From 443, 10), we have
As before we have
By hypothesis there exists for each e > 0, a 8 > 0, such that
uV(u)<e , for any 0< u<S Q .
Hence,
V(u)<J
u
1 liUS,
Let us turn now to W^. Since V(u) is the sum of two limited
monotone decreasing functions P, ^Vin (w, ^), it is integrable.
Thus,
f7?+6 /T)+6
du + / V(u)du < 8 j I f (?) I +
aLn
is evanescent with S.
436 FOURIER'S SERIES
446. 1. Fourier s development is valid at the regular point x, if
<}>(u) has limited variation in some interval (0 < f),
For let < u < 7 < f, then
NOW
sin
Hence V uy ty <_ J V uy <j> f
But sin u being monotone,
si i) w sin u sin 7
Similarly, rr i ,
> yfty^ shl^ry = F 2 "
Now
0< u <M , in (0*, f).
sin w
The theorem now follows by 445. For we may take 7 so small
that p
Thus for any u < 7,
On the other hand, Wl being sufficiently large, and 7 chosen as
in 1) and then fixed,
F 2 <2tt.
Thus
for w < some S r . Hence
w
for < u < some S.
2. (Jordan.) Fourier's development is valid at the regular point
X, iff (x) has limited variation in some domain of x.
OTHER CRITERIA 437
For
+ 2 u)/(u)j +{/(* 2 )/()}
has limited variation also.
3. Fourier's development is valid at every point of 31 = (0, 2 TT),
iff is limited and has only a finite number of oscillations in 21.
Other Criteria
447. Let X=
If X = as & == 0, so does "SP, awe? conversely.
For x . ^ N xx .^ , 5,, sin(i6 f S)
 
where _ gm ,
Obviously X and "9 are simultaneously evanescent with
provided
Let
/TV N sin
u
Then p = ^(u) \ Z(u + 8)  Z(u) I
Now
. , x v cos v sin
Thus \Z'(v)\<Mv<M. 2.
438 FOURIER'S SERIES
Hence * i \ n/r
" ' smw
AS  * <!/( + 2 w)  + /(s  2 w)  +2
J?< 282ft M  =0 , with 8.
448. (LipschitzDini.^) At the regular point x, Fourier's devel
opment is valid, if for each e > 0, there exists a 8 > 0, such that for
each < 8 < S ,
 <f>(u + 8)  <(w)  < ^_L_, /or any w in (8, 8 ).
 log 8 
For
(u + 8) 
Now a: being a regular point, there exists an 17' such that
 <(w)  < e, for u in any (8, 77').
Thus taking ^ ,
*

I log 8 I 8 rj
< 2 6, for any 8 < 77.
Thus v A s A
X = 0, as o = 0.
Uniqueness of Fourier's Development
449. Suppose/ (x) can be developed in Fourier's series
00
/(a?) = a + 2(a n cos nz f t> n *\n nx), (1
a n =  I /(^) ^os wa:rfa? , J n = _ / " f(x) sin n^dz, (2
TjiTr 7TX7T "
UNIQUENESS OF FOURIER'S DEVELOPMENT 439
valid in 21 = ( TT, TT). We ask can/(V) be developed in a simi
f (x) = I 0Q + 2(a n ' cos nx + b^ sin nz), (3
also valid in H, where the coefficients are not Fourier's coefficients,
at least not all of them.
Suppose this were true. Subtracting 1), 3) we get
= 2 ( a O a o) + 2 I ( a n #n) COS TiZ + (6 n  ftj) Sill nx \ = 0,
c' f {tf n cos wo; 4 ^ n si* 1 n #J = 0, in 81. (4
Thus it would be possible for a trigonometric series of the type
4) to vanish without all the coefficients <? m , d m vanishing.
For a power series
Po + Pi* + Pv* + " ( 5
to vanish in an interval about the origin, however small, we know
that all the coefficients p m in 5) must = 0.
We propose to show now that a similar theorem holds for a
trigonometric series. In fact we shall prove the fundamental
Theorem 1. Suppose it is known that the series 4) converges to
for all the points of 3{ = ( TT, TT), except at a reducible set 3?.
Then the coefficients c m , d m are all 0, and the series 4) = at all the
points of 21.
From this we deduce at once as corollaries :
Theorem 2. Let 3J be a reducible set in 21. Let the series
4 2{a n cos nx 4 /3 n sin nx\ (6
converge in 21, except possibly at the points 3?. Then 6) defines a
function F(x) in 21 9?.
If the series / , v^c / , m >
J KQ + 2, \ cos nx + fin sin nx\
converges to F(jx) in 21 9i, its coefficients are respectively equal to
those in 6).
Theorem 3. If /(a?) admits a development in Fourier s series for
the set 21 3t, any other development off(x} of the tt/pe 6), valid in
9( 3? is necessarily Fourier's series, i.e. the coefficients m , j3 m have
the values given in 2).
440 FOURIER'S SERIES
In order to establish the fundamental theorem, we shall make
use of some results due to Riemann, Q. Cantor, Harnack and
Schwarz as extended by later writers. Before doing this let us
prove the easy
Theorem 4* If f(v) admits a development in Fourier's series
which is uniformly convergent in 21 = ( TT, TT), it admits no other
development of the type 3), which is also uniformly convergent in 21.
For then the corresponding series 4) is uniformly convergent
in 21, and may be integrated termwise. Thus making use of the
method employed in 436, we see that all the coefficients in 4)
vanish.
450. 1. Before attempting to prove the fundamental theorem
which states that the coefficients n , b n are 0, we will first show
that the coefficients of any trigonometric series which converges
in 21, except possibly at a point set of a certain type, must be such
that they == 0, as n = oc. We have already seen, in 440, 1, that
this is indeed so in the case of Fourier's series, whether it con
verges or not. It is not the case with every trigonometric series
as the following example shows, viz. :
2 sin n ! x. (1
i
When x = all the terms, beginning with the r I th , vanish,
r \
and hence 1) is convergent at such points. Thus 1) is conver
gent at a pantactic set of points. In this series the coefficients a n
of the cosine terms are all 0, while the coefficients of the sine
terms b n , are or 1. Thus b n does not = 0, as n = oo.
2. Before enunciating the theorem on the convergence of the
coefficients of a trigonometric series to 0, we need the notion of
divergence of a series due to Harnack.
Let A = a l a 2 \ (2
be a series of real terms. Let# n , & n be the minimum and maxi
mum of all the terms
A A
^n+1 * "n+2 i *
where as usual A n is the sum of the first n terms of 2). Obviously
UNIQUENESS OF FOURIER'S DEVELOPMENT 441
Thus the two sequences f# n }, { Q n \ are monotone, and if limited,
their terms converge to fixed values. Let us say
ff* = ff , #n = #
The difference
\> = &ff
is called the divergence of the series 2).
3. For the series 2) to converge it is necessary and sufficient that
its divergence b = 0.
For if A is convergent,
 + A< A n+p <A + e , ^ = 1,2...
Thus e + A<g n <G n <A + .
Thus the limits Q, g exist, and
##<2e ; or# = #,
as > is small at pleasure.
Suppose now b = 0. Then by hypothesis, #, g exist and are
equal. There exists, therefore, an n, such that
#*<<? n <# n <+,
or #n~<7n<2e.
Thus \A n+p A n \<2e , p=l, 2 
and A is convergent.
451, Let the series
00
2 (# n cos w# 4 b n sin w#)
6e such that for each B > 0, there exists a subinterval of
l = (7r, TT)
a^ eac?A jt>oir^ q/ ^AwA tY divergence b < S. TAett a n , i n = 0, as
n = QO.
For, as in 450, there exists for each x an m x , such that
*
cos M# + n sn
m x
442 FOURIER'S SERIES
for any point x in some interval 93 of 21. Thus if b is an inner
point of 23, x = b + /3 will lie in 33, if /3 lies in some interval
B = (p, q). Now
a n cos w ( b + /3; f b n sin n(b + @)
= (a n cos rc6 f & sin nb) cos w/3 (a n sin nb J n cos nb) sin ny8.
a n cos n(b /3) f # sin n(b yS)
= (a n cos w6 } 6 n sin nb) cos n/3 4 (#n y i n ^^ ^n cos n &) 8 ^ n w ^
Adding and subtracting these equations, and using 1) we have
 (a n cos nb f b n sin 116) cos wy8  < ,
g
 (a n sin nb 5 n cos w6) sin n/3 \ < ,
for all n>m x . Let us multiply the first of these inequalities by
cos nb sin n/3, and the second by sin 716 cos n/3, and add. We get
\a n sinn/3 l \<S , ^ = 2/3 , n > m x . (2
Again if we multiply the first inequality by sin nb sin n/3, and
the second by cos nb cos n/3, and subtract, we get
 b n sin n/3 l \ < 8 , n>m x . (3
From 2), 3), we can infer that for any e >
 a n \ < e , \ b n  < e , n > some m, (4
or what is the same, that a n , b n = 0.
For suppose that the first inequality of 4) did not hold. Then
there exists a sequence
n i < n i < = (5
such that on setting
loJ^+Si, , e8 = S'
we will have
S Wr > '. (6
If this be so, we can show that there exists a sequence
v \ < V 2 < " ^
in 5), such that for some /3 f in B,
 a, f sin vfl  > 8, (7
UNIQUENESS OF FOURIER'S DEVELOPMENT 443
which contradicts 2). To this end we note that 7 > may be
chosen so small that for any r and any  7  < 7 ,
I ,  cos 7 > (S + 8') cos 7 > S. (8
Let us take the integer i/ a so that
" + 2 y
. (9
qp
Then
Thus at least one odd integer lies in the interval determined by
the two numbers
o 9
. O"i + 7o) (?"! 7o>
7T 7T
Let mj be such an integer. Then
9 2
~O"i + 7o) < i < (?"i  7 )
7T 7T
If we set
\
' ft== v
...
(
we see that the interval ^ 1 = (jt> 1 , q^) lies in 5. Tlie length of
S l is 2 7 /J> r Then for any ^ in J5 X ,
Thus by 8),
 a Vi sin ^/3  =  a Vl  cos 7j > 8. (12
But we may reason on B l as we have on JB. We determine v^
by 9), replacing JP, ^ by p^ q l . We determine the odd integer m 2
by 10), replacing jp, ^, ^ t by p l ^ q l ^ v^. The relation 11) deter
mines the new interval J? 2 = (jt? 2 , <? 2 ), on replacing mj, ^j by w a , ^ 2 *
The length of J5 2 is 27 /^ 2 , and JS 2 lies in J5 1B For this relation
of 1/2 and for any y3 in 2 we have, similar to 12),
a v sin j//9 > 8.
In this way we may continue indefinitely. The intervals
B l > J9 2 > = to a point ', and obviously for this /8', the rela
444 FOURIER'S SERIES
tion 7) holds for any x. In a similar manner we see that if b n does
not = 0, the relation 3) cannot hold.
452. As corollaries of the last theorem we have :
1. Let the series
00
2(# n cos nx + b n sin nx) (1
be such that for each S > 0, the points in Sl = ( TT, TT) at which
the divergence of 1) is >S, form an apantactic set in 21. Then
a n , # n = 0, as ft = 00.
2. Jvg A# series 1) converge in 21, except possibly at the points of
a reducible set 5K. TA0w a n , 5 n ^=0.
For $R being reducible [318, 6], there exists in 21 an interval 93
in which 1) converges at every point. We now apply 451.
453. Let r, / . / . 7 x
Jf(x) = 2(a n cos nx + 6 n sin w#)
a^ ^Ae point* of 21 = ( TT, TT), where the series is convergent. At the
other points of 21, let F(x) have an arbitrarily assigned value, lying
between the two limits of indetermination g, Gr of the series. If F is
Rintegrable in 21, the coefficients a n , i n = 0.
For there exists a division of 21, such that the sum of those in
tervals in which Osc F > co is < a. There is therefore an interval
3 in which Osc F < G>. If $ is an inner interval of 3, the di
vergence of the above series is < o> at each point of $. We now
apply 451.
454. Riemanrfs Theorem.
00
Let F(x) = # + 2(a n cos nx f b n sin nx) = 2^4 n converge at
each point of 21 = ( TT, ?r), except possibly at the points of a redu
cible set 9Z. The series obtained by integrating this series termwise,
we denote by
 (a n cos wa; + 5 B sin r) =
Gr is continuous in 21.
UNIQUENESS OF FOURIER'S DEVELOPMENT 445
Let <E>(V) = G(x + 2 u) + G(x  2 w)  2 G(x). (1
Then at each point of S3 = 31 9?,
lim ^& = F(x) ; (2
M=O 4 u*
and at each point of 31,
w=0 16
For, in the first place, since 9? is a reducible set, # n , 5 n = 0. The
series Gr is therefore uniformly convergent in 31, and is thus a
continuous function.
Let us now compute <>. We have
a n cos n(x h 2 u) f # n cos n(# 2 M) 2 a n cos nx
= 2a n cos 7i# (cos 2nu  1)
= 4 a n cos Tta: sin 2 m^.
Also J n sin 7^(2; f 2 w) f ^ n sin n(x  2 w)  2 6 ft sin wa;
97 / ^v  v
^ Q sin ny\ cos ij wti I )
= 4 b n sin wa; sin 2 nu.
Thus (^=2^ f^iiL^
4 y? o n V nu
if we agree to give the coefficient of A$ the value 1. Let us
give x an arbitrary but fixed value in S3. Then for each > 0,
there exists an m such that
A Q f A 1 f 4 Aii = ^0*0 4 e n ,  e n  < , n>_m.
Thus ^4 n = n+1  n .
Hence ^
, S / N /sinnw\ 2
X + 2 (6 n4 .j  n )
i \ nu J
, ^ ffsin (ti l)w"l 2 rsium/l 2 ] .r,,
f 2 e n ^ }  (4
i IL (wl> J L WM J I
446 FOURIER'S SERIES
The index m being determined as above, let us take u such that
u < , so that m < ;
m u
and break S into three parts
1 m+l K+I
where K is the greatest integer < TT/U, and then consider each sum
separately, as u = 0.
Obviously lim S l = 0.
u=0
As to the second sum, the number of its terms increases indefin
itely as u = 0.
For any ?/,
.
<
fTsin wwH 2 ("si
IL mu J L
fsin mu~] 2 ^
 < ,
L mu J
since each term in the brace is positive. In fact
sin v
v
is a decreasing function of v as v ranges from to TT, and
nu<KU<_ f jr , n = m, m 4 1, tc.
Finally we consider S%. We may write the general term as
follows :
jTsin (ft 1)?/T 2 _ fsin (n l)wT
iL~ci)M J ~L "J
ML "nn "JL"^]}'
sin 2 (n l)w sin 2 rm __ sin (2 n l)w sin u
""
UNIQUENESS OF FOURIER'S DEVELOPMENT 447
Thus w f , 1 x ,
\s <T f l l } 4ivJL
' 8 ~
since
But # >. 1 , or feu > TT u.
u
Thus
< 6
I I t
(TT ?*) 2 TT wj
Hence S=S, + ^+ S t = 0, as u = 0,
which proves the limit 2), on using 4).
To prove the limit 8), we have
f. \2^ ^
nu
Let us give u a definite value and break T into three suras.
w
I
where m is chosen so that
 A n < , n > m ;
A
flry y
where X is the greatest integer such that
Xw <. 1 ;
and
A+l
Obviously for some M,
T 2  <. U\ < ,
since
/sin m,ii. \
448 FOURIER'S SERIES
As to the last sum,
<eX.l , sincel<X,
<.
Thus
455. SchwarzLuroth Theorem.
In 3( =(a < J) fo the continuous function f (x) be such that
S(x, u) =^ x + M) +/( V M > ~ 2/( ^ 0, as u = 0, (1
I/'
except possibly at an enumerable set (S i?i 21 At the points (, let
uS(x, u) = as u = 0. (2
2%e/& / i a linear function in 21.
Let us first suppose with Schwarz that (g = 0. We introduce
the auxiliary function,
g(x) = 7/i(V) \c(x a)(x  J),
where
Z(*) =/(*) /(a)  /(6) /(a) I,
?/ = 1, and c is an arbitrary constant.
The function g(x) is continuous in 21, and </() = <7(&) = 0.
Moreover
Thus for all < w < some S,
(3
From this follows that g(x)<Q in 21. For if #(V)>0, at any
point in 21, it takes on its maximum value at some point within 21.
Thus
for < ^ < 8, & being sufficiently small. Adding these two in
equalities gives #<.0, which contradicts 3). Thus#j<0 in 21.
Let us now suppose L ^= f or some x in 21. We take c so small
that T r
= sgn rjL = 77 sgn L.
UNIQUENESS OF FOURIER'S DEVELOPMENT 449
But rj is at pleasure 1, hence the supposition that L 3= is
not admissible. Hence L = in 31, or
/("O'/OOf {/(*)/() I (4
is indeed a linear function.
8 now suppose with Liiroth that (>0. We introduce the
auxiliary continuous function.
Thus A(a) = , A(6) = <?( J  a) 2 .
Suppose at some inner point of 1
Z,(0. (5
This leads to a contradiction, as we proceed to show. For then
provided
O=
We shall take c so that this inequality is satisfied, i.e. c lies in
the interval 6 = (0*, (7*). Thus
Hence A(V) takes on its maximum value at some inner point e
of 21. Hence for 8 > sufficiently small,
<S. (6
u 2
Now if ^ is a point of 31 @,
lira JI(e, u) = 2 c > 0.
M =
But this contradicts 7), which requires that
u=0
450 FOURIER'S SERIES
Hence 6 is a point of @. Hence by 2),
By 6), both terms have the same sign. Hence each term = 0.
Thus for u >
= lim A?J^J^ZZL.. .\?2 = lim S^ e u '
+ 2c(e a).
H puff
/0) = ft _^ + 6'Oa). (
Thus to each c in the interval E, corresponds an in @, at which
point the derivative of f(x) exists and has the value given on the
right of 8). On the other hand, two different c's, say c and c r , in
cannot correspond to the same e in @.
For then 8) shows that
c \e ""*"" ct j c \e ~" a i^
or as ^ , __
Thus there is a uniform correspondence between S whose cardi
nal number is c, and ( whose cardinal number is e, which is absurd.
Thus the supposition 5) is impossible. In a similar manner, the
assumption that L < at some point in 21, leads to a contradiction.
Hence L = in 21, and 4) again holds, which proves the theorem.
456. Cantor's Theorem. Let
00
^ a f 2(a n cos nx + b n sin nx) (1
i
converge to in 2l = ( TT, TT), except possibly at a reducible set 9?,
where nothing is asserted regarding its convergence. Then it con
verges to at every point in 21, and all its coefficients
111 ___ A
For by 452, 2, a n , b n = 0. Then Riemann's function
/ O) = 4 V 2  2 "^ ( a n COS Wir + J n Sin WSC)
f n ^
UNIQUENESS OF FOURIER'S DEVELOPMENT 451
satisfies the conditions of the SchwarzLiiroth theorem, 455, since 9?
is enumerable. Thus f(x) is a linear function of x in SI, and has
the form a + /3x. Hence
1
a + fix  a z 2 = 2} (a n cos w# + b n sin nz). (2
The right side admits the period 2 TT, and is therefore periodic.
Its period o> must be 0. For if <o > 0, the left side has this
period, which is absurd. Hence <w = 0, and the left side reduces
to a constant, which gives /3=0, # = 0. But in 21 9i, the right
side of 1) has the sum 0. Hence a= 0. Thus the right side of
2) vanishes in 21. As it converges uniformly in 31, we may deter
mine its coefficients as in 436. This gives
CHAPTER XIV
DISCONTINUOUS FUNCTIONS
Properties of Continuous Functions
457. 1. In Chapter VII of Volume I we have discussed some
of the elementary properties of continuous and discontinuous
functions. In the present chapter further developments will be
given, paying particular attention to discontinuous functions.
Here the results of Baire * are of foremost importance. Le
besgue f has shown how some of these may be obtained by sim
pler considerations, and we have accordingly adopted them.
2. Let us begin by observing that the definition of a continu
ous function given in I, 339, may be extended to sets having iso
lated points, if we use I, 339, 2 as definition.
Let therefore/^ # w ) be denned over 21, being either limited
or unlimited. Let a be any point of 21. If for each e > 0, there
exists a S > 0, such that
I/O) /O) I < > for any x in F 6 (a),
we say fis continuous <at a.
By the definition it follows at once that f is continuous at each .
isolated point of 21. Moreover, when a is a proper limiting point
of 21, the definition here given coincides with that given in I, 339.
If /is continuous at each point of 21, we say it is continuous in 21.
The definition of discontinuity given in I, 347, shall still hold,
except that we must regard isolated points as points of con
tinuity.
* " Sur les Functions de Variables reeles" Annali di Mat., Ser. 3, vol. 3
(1899).
Also his Lemons sur les Functions Discontinues. Paris, 1906.
t Bulletin de la Societe Mathematique de France, vol. 32 (1904), p. 229.
452
PROPERTIES OF CONTINUOUS FUNCTIONS 453
3. The reader will observe that the theorems I, 350 to 354
inclusive, are valid not only for limited perfect domains, but also
for limited complete sets.
458. 1. If f(v\ " %m) * 8 continuous in the limited set 21, and its
values are known at the points of S3 < 21, then f is known at all
points of S3' lying in 21.
For let 6 1 , # 2 , 6 3 be points of S3, whose limiting point b lies
in 21. Then
2. If f is known for a dense set S3 in 21, and is continuous in 21,
f is known throughout 21.
For 9 ,>^
3. If f(&i " %m) continuous in the complete set 21, the points
S3 in 21 where f= c, a constant, form a complete set. If 21 is an
interval, there is a first and a last point of S3.
For/= c at # = a v 2 which = ; we have therefore
459. The points of continuity S of ./(^i #) in 21 lie in a
deleted enclosure . If 21 is complete, $ .
For let e x > e 2 > == 0. For each e n , and for each point of
continuity c in 21, there exists a cube O whose center is c, such that
Osc/< e n , in O.
Thus the points of continuity of / lie in an enumerable non
overlapping set of complete metric cells, in each of which
Osc/< e n . Let O n be the inner points of this enclosure. Then
each point of the deleted enclosure
= Dv\& n \
which lies in 21 is a point of continuity of /. For such a point c
lies within each Q n .
HenCC Osc/<e, inF.O*
for S > sufficiently small and n sufficiently great.
454 DISCONTINUOUS FUNCTIONS
Oscillation
46  Let a> s = Osc/iX ... * m ) in Fa (a).
This is a monotone decreasing function of 8. Hence if co$ is
finite, for some S > 0,
co = lim o) 6
5=0
exists. We call co the oscillation off at x = a, and write
a) = Osc/.
ar=a
Should o> 5 = f oo, however small 8 > is taken, we say co = f oo.
When co = 0, / is continuous at a: = a, if a is a point in the domain
of definition of f. When co > 0, f is discontinuous at this point.
It is a measure of the discontinuity off at x = a ; we write
461. 1.
# = a.
 d  e  < Disc (f ^) < rf 4 e.
For in F fi (a),
 Osc/  Osc //  < Osc (fff)< Ow/ f Osc #.
2. Iffis continuous at x = a, while Disc ^7 = d, then
Disc (/ + g) = d.
x=a
For /being continuous at a, Disc/= 0.
Hence
< Disc ^ rf
3. If c is a constant,
Disc ((?f ) =  c?  Disc/ , at z = a.
For Osc ((?/) =1*1 Osc/ , in any F 6 (
4. When the limits
OSCILLATION 455
exist and at least one of them is different from /(#), the point x
is a discontinuity of the first kind, as we have already said.
When at least one of the above limits does not exist, the point x
is a point of discontinuity of the second kind.
462. 1. The points of infinite discontinuity 3? of f, defined over
a limited set 2l,/0rm a complete set.
For let *], * 2 be points of 3, having k as limiting point.
Then in any V(k) there are an infinity of the points * n and hence
in any V(k), OSC/= + OQ. The point k does not of course
need to lie in 21.
2. We cannot say, however, that the points of discontinuity of
a function form a complete set as is shown by the following
Example. In 21 = (0, 1), let /(V) = x when x is irrational, and
= when x is rational. Then each point of 21 is a point of dis
continuity except the point x = 0. Hence the points of disconti
nuity of /do not form a complete set.
3. Let f be limited or unlimited in the limited complete set 21.
The points & 0/21 at which Ose/>. k form a complete set.
For let a v a 2 be points of $ which = a. However small
S >0 is taken, there are an infinity of the a n lying in V^a). But
at any one of these points, Osc/_> &. Hence Osc/>.& in Fa (a) 9
and thus a lies in $.
4. Letf(x^  XM) be limited and Rintegrable in the limited set 21.
The points $ at which OSG f >_k form a discrete set.
For let D be a rectangular division of space. Let us suppose
St D > some constant c > 0, however D is chosen. In each cell 8
of D,
Osc/>*.
Hence the sum of the cells in which the oscillation is :> k can
not be made small at pleasure, since this sum is $ D > But this
contradicts I, 700, 5.
5. Let /(#! x m } be limited in the complete set 21. If the points
$ in 21 at 'which Osc/> k form a discrete set, for each k, then f is
Rintegrable in 21.
456 DISCONTINUOUS FUNCTIONS
For about each point of 21 $ as center, we can describe a cube
& of varying size, such that Osc/< 2 k in (. Let D be a cubical
division of space of norm d. We may take d so small that
$t D = 2d t is as small as we please. The points of 21 lie now within
the cubes S and the set formed of the cubes d,. By Borel's
theorem there are a finite number of cubes, say
such that all the points of 21 lie within these T/'S. If we prolong
the faces of these rfs, we effect a rectangular division such that
the sum of those cells in which the oscillation is > 2 k is as small
as we choose, since this sum is obviously < $ D . Hence by I, 700,
5, f is J2integrable.
6. Letf(x l x m ) be limited in 21; let its points of discontinuity
in 21 be 33. If f is Rintegrable, 33 is a null set. If 21 is complete
and 5) is a null set, f is .Rinte</rable.
Let / be Rintegrable. Then 3) is a null set. For let l > e 2
> ... = 0. Let J) n denote the points at which Osc/> e n . Then
33 = {33 W . But since/ is .Rintegrable, each S) n is discrete by 4.
Hence 33 is a null set.
Let 21 be complete and 3) a null set. Then each ) n is complete
by 3. Hence by 365, ) n = SD n . As ) = 0, we see 33 n is discrete.
Hence by 5, /is jRintegrable.
If 21 is not complete, / does not need to be 72integrable when
3) is a null set.
Example. Let2l 1 =(l , n= 1, 2 ... ; m< 2 n .
n
' "=1,2...; r<3'.
Let /(*) = : ' at *=f;
= 1 in a,.
Then each point of 21 is a point of discontinuity, and ?l = 3D.
But 2lj , 212 are null sets, hence 21 is a null set.
POINTWISE AND TOTAL DISCONTINUITY 457
On the other hand,
and / is not ./2integrable in 21.
Pointwise and Total Discontinuity
463. Let/Oj x m ) be defined over 31. If each point of 21 is a
point of discontinuity, we say /is totally discontinuous in 21.
We say f is pointwise discontinuous in 21, if f is not continuous
in 21= \a\, but has in any V(a) a point of continuity. If/ is
continuous or pointwise discontinuous, we may say it is at most
pointwise discontinuous.
Example 1. A function/^ # m ) having only a finite number
of points of discontinuity in 21 is pointwise discontinuous in 21.
Example 2. Let
f(x) = , for irrational x in 21 = (0, 1)
1 /. m
=  , lor x =
n n
= 1 , for x =0,1.
Obviously /is continuous at each irrational #, and discontinuous
at the other points of 21. Hence / is pointwise discontinuous
in 21.
Example 3. Let 3) be a Harnack set in the unit interval
21 = (0, 1). In the associate set of intervals, end points included,
let/(V)=l. At the other points of 21, let /= 0. As > is
apantactic in 21, /is pointwise discontinuous.
Example 4 In Ex. 3, let 35 = (g f gs where S is the set of end
points of the associate set of intervals. Let/=l/?i at the end
points of these intervals belonging to the n th stage. Let/= in
g. Here / is defined only over 35. The points g are points of
continuity in 3). Hence/ is pointwise discontinuous in 3).
Example 5. Let/(#) be Dirichlet's function, i.e. /= 0, for the
irrational points 3 in 21 = (0, 1), and = 1 for the rational points.
458 DISCONTINUOUS FUNCTIONS
As each point of 21 is a point of discontinuity,/ is totally discon
tinuous in 21. Let us remove the rational points in 21 ; the deleted
domain is 3 I n this domain/ is continuous. Thus on removing
certain points, a discontinuous function becomes a continuous
function in the remaining point set.
This is not always the case. For if in Ex. 4 we remove the
points g, retaining only the points (, we get a function which is
totally discontinuous in @, whereas before / was only pointwise
discontinuous.
464. 1. Iff(x l XM) is totally discontinuous in the infinite com
plete set 21, then the points b w where
Disc/>o> , o>>0,
form an infinite set, if a> is taken sufficiently small.
For suppose b w were finite however small co is taken. Let
tt) 1 >o) 2 >. =0. Let Dj, D 2 , be a sequence of superposed
cubical divisions of space of norms d n = 0. We shall only con
sider cells containing points of 21. Then if d l is taken sufficiently
small, D l contains a cell Sj, containing an infinite number of
points of 21, but no point at which Disc/>a> r If d? 2 is taken
sufficiently small, 7> 2 contains a cell S 2 <Sj, containing no point
at which Disc/>o> 2 . In this way we get a sequence of cells,
which == a point p. As 21 is complete, p lies in 21. But / is
obviously continuous at p. Hence / is not totally discontinuous
in 21.
2. If 21 is not complete, b w does not need to be infinite for
any o> > 0.
Example. Let 21 = j [ , n = 1, 2, and m odd and <2 n . At
1
~, let/= Then each point of 2l is a point of discontinuity.
*j Zi
But b w is finite, however small &>>0 is taken.
3. We cannot say /is not pointwise discontinuous in complete
2l, when bo, is infinite.
EXAMPLES OF DISCONTINUOUS FUNCTION'S 459
Example. At the points   t = % let / = ; at the other
\.n J
points of l = (0, 1), let/=l.
Obviously / is pointwise discontinuous in 21. But b w is an
infinite set for co < 1, as in this case it is formed of 5ft, and the
point 0.
Examples of Discontinuous Functions
465. In volume I, 330 seq. and 348 seq.> we have given ex
amples of discontinuous functions. We shall now consider a few
more.
Example 1. Riemanrfs Function.
Let (x) be the difference between x and the nearest integer;
and when x has the form n + , let (x) = 0. Obviously (x) has
the period 1.
It can be represented by Fourier's series thus :
s *. If sin 2 TTX sin 2 2 TTX , sin 3 2 TTX
Riemanri 8 function is now
This series is obviously uniformly convergent in 21 = ( oo, oo).
Since (#) has the period 1 and is continuous within (o> )>
we see that (nx) has the period , and is continuous within
n
( , ]. The points of discontinuity of (nx) are thus
\ 2 n 2 nj
, = 0, 1, 2, ...
n
Let (= S@ n J. Then at any x not in (g, each term of 2) is a con
tinuous function of x. Hence F(x) is continuous at this point.
On the other hand, F is discontinuous at any point e of @. For
F being uniformly convergent,
m^ (3
*
L lira F(x) = 2 lira  (4
x=e x=e n
460 DISCONTINUOUS FUNCTIONS
We show now that 3) has the value
^(0) _ JlL, for e = ^ l , e irreducible. (5
16w 2 2 n
and 4) the value
^> + i^ (6
Hence 2
 (7
*
To this end let us see when two of the numbers
1 , r A 1 . s
+ , and  +  m*n
2mm 2 n n
are equal. If equal, we have
2rf 1 2s + 1
m n
Thus if we take 2 s f 1 relatively prime to n, no two of the num
bers in @ n are equal. Let us do this for each n. Then no two of
the numbers in S are equal.
1 8
Let now x = e = h  Then (mx) is continuous at this point,
2 n n
unless 8) holds; i.e. unless m is a multiple of w, say m= In. Ju
this case, 8) gives
Thus I must be odd ; l^ 1, 8, 5 ... In this case (mx) = at 0,
while jRlim (T?IX)= . When w is not an odd multiple of n,
jr=e
obviously 72 lim (w#) = (me).
xe
Thus when m = /n, Z odd,
^ lim (m:r) = 1 L=:^)1I 1.
^=, m 2 2
When m is not a multiple of n,
EXAMPLES OF DISCONTINUOUS FUNCTIONS 461
Hence
l k +l + l + ...}
This establishes 5). Similarly we prove 6). Thus F(x) is
discontinuous at each point of @. As
F is limited. As the points (S form an enumerable set, F is
.Rintegrable in any finite interval.
466. Example 2. Let/(V)=0 at the points of a Cantor set
(7 = m a^  ; ra = 0, or a positive or negative integer, and the
a's = or 2. Let /(#) = 1 elsewhere. Since /(#) admits the
period l,/(3 nx) admits the period  Let O l be the points of
o n
G which fall in 21 = (0, 1). Let D l be the corresponding set of
intervals. Let (7 2 = Cj f Fj, where F x is obtained by putting a
O l set in each interval of D l . Let Z> 2 be the intervals correspond
ing to <7 2 . Let (7 3 = 6 7 2 + F 2 where F 2 is obtained by putting a <7 2
set in each interval of Z> 2 , etc.
The zeros of/(3naO are obviously the points of C Y n . Let
The zeros of F are the points of g = { C n \. Since each C n is a null
set, & is also a null set. Let A = 21 &. The points 4, S are
each pantactic in 31. Obviously F converges uniformly in 21,
since 0</(3 nx) <1. Since / n (V) is continuous at each point a
of A, F is continuous at a, and
462 DISCONTINUOUS FUNCTIONS
We show now that F is discontinuous at each point of . For
let e m be an end point of one of the intervals of D m+l but not of
D m . Then
Hence F(e) = H m = +  + 2
Jr m*
As the points A are pantactic in 31, there exists a sequence in
.4 which = e. For this sequence F = H. Hence
Similarly, if rj m is not an end point of the intervals D m+v but a
limiting point of such end points,
The function F is Jf2integrable in 21 since its points of discon
tinuity form a null set.
467. Let @ = i0 tl ...,J ^ an enumerable set of points lying in the
limited or unlimited set 31, which lies in 9t m . For any x in 31 and
any e^ in (, let x e l lie in 33. Let g(x l # m ) J^ limited in 33
continuous, except at x = 0,
= b.
(7= 2c ... converge absolutely. Then
is continuous in A = 21 (g, aradf at x= e^
Disc.F( = ^
For when a; ranges over 21, x e t remains in S3, and g is limited
in 93. Hence F is uniformly and absolutely convergent in 21.
Now each g(x e^) is continuous in A ; hence F is continuous
in A by 147, 2.
EXAMPLES OF DISCONTINUOUS FUNCTIONS 463
On the other hand, Fis discontinuous at #= e K . For
where If is the series F after removing the term on the right of
the last equation. But JET, as has just been shown, is continuous
at x = e K .
468. Example 1. Let @=j0 n j denote the rational numbers.
Let <V) = sin x
x
= , x=0.
Then F(x}=V(xe^ >1
^fJL
is continuous, except at the points S. At x = e n ,
Disc F = ^
Example 2. Let @ = J^ n } denote the rational numbers.
Let xx v nx i
 n=00 i + nx
= , ^ = 0,
which we considered in I, 331.
Then r,/ ^
is continuous, except at the rational points, and at x = e m ,
(x)= 
ml
469. In the foregoing g(x) is limited. This restriction may be
removed in many cases, as the reader will see from the following
theorem, given as an example.
Let JE = Se tl ... t J be an enumerable apantactic set in 2(. Let (g =
(j?, _Z7'). For any x in 81, and any e L in J?, let x e L lie within a
cube S3. Let g(x l x m ) be continuous in 53 except at # = 0, where
g = f QO, as x == 0. Let ^Lc^.., it be a positive term convergent series.
464 DISCONTINUOUS FUNCTIONS
Then , ~ .
is continuous in A = 21 G. <9>? /*e of Aer Aand, <m'7j, jt>0m 0/ S is a
point of infinite discontinuity.
For any given point x a of A lies at a distance >0 from (g.
Thus
as x ranges over some F^a), and e t over E.
I fPHPf* i i mm
1 u ct  #O  ^ )  < some M,
for 2* in F^a), and t in #. Thus jF 7 is uniformly convergent at
x = a. Ax each //(a? e k ) is continuous at x = a, J 7 is continuous
at a.
l/tf n^xf #= ^. Then there exists a sequence
x', x" . = e K (1
whose points lie in A. Thus the term g(x e^) = f oo as a; ranges
over 1). Hence a fortiori ^ = 400. Thus each point of J? is a
point of infinite discontinuity.
Finally any limit point of E is a point of infinite discontinuity,
by 462, l.
470. Example. Let g(x) = , a n = , a>l.
y> .
n ~^ y
Then n/ N ^ / >,
is a continuous function, except at the points
 1  1  1 ...
v, ~, ~,
a a* a 6
which are points of infinite discontinuity.
471. Let us show how to construct functions by limiting
processes, whose points of discontinuity are any given complete
limited apantactic set & in an mway space 3J m .
EXAMPLES OF DISCONTINUOUS FUNCTIONS 465
1. Let us for simplicity take m = 2, and call the coordinates of
a point a?, y.
Let Q denote the square whose center is the origin, and one of
whose vertices is the point (1, 0).
The edge of Q is given by the points x, y satisfying
\x\+\y\ = l. (1
Thus 1 f, on the edge
i
, outside
of the square Q. Hence
 (I:
Th "
2. We next show how to construct a function g which shall =
on one or more of the edges of Q. Let us call these sides e 1 ^ e%, 3 , 4 ,
as we go around the edge of Q beginning with the first quadrant.
If 6? = on e t , let us denote it by Gr t ; if Gr = on e t , e K let us
denote it by Q il( , etc. We begin by constructing Q r We observe
that
. ,. n\t\ fl, for^ = 0,
1 lim  ' = 4
n=00 1 + n 1 1 \ 10, for t ^ 0.
Now the equation of a right line I may be given the form
x cos a f y sin = p
where < a < 2 TJ% p > 0. Hence
Z( X , v) = 1  lim n l* CQS( * + ff sin y  = 1, on Z,
n= 1 h ?i  a; cos a + y sin a p \ (0, off L
If now we make I coincide with e we see that
E, (x, y) = 2Z(x,y)L (*, y} =
Hence
> n .
1 0, on j and without Q.
466 DISCONTINUOUS FUNCTIONS
In the same way,
#1234 = 9 ~ (#! + ^2 + ^S + ^4>
By introducing a constant factor we can replace Q by a square
Q e whose sides are in the ratio c : 1 to those of Q.
the ed g e of Qc,
Q (*, y) = li
*= 1 + M i + ) 10, outside.
\ <? c J '
We can replace the square Q by a similar square whose center
is a, b on replacing  a; ,  y  by  2; a ,  y J .
We have thus this result : by a limiting process, we can con
struct a function g(x, y) having the value 1 inside Q, and on any
of its edges, and = outside $, ail( i on the remaining edges.
Q has any point a, b as center, its edges have any length, and its
sides are tipped at an angle of 45 to the axes.
We may take them parallel to the axes, if we wish, by replacing
x
in our fundamental relation 1) by
\ x y\ > \* + y\*
Finally let us remark that we may pass to mway space, by re
placing 1) by
Kl + l*al +  + ^ = 1.
3. Let now Q = jq n  be a border set [328], of nonoverlapping
squares belonging to the complete apantactic set E, such that
Q  g = 3t the whole plane. We mark these squares in the
plane and note which sides q n has in common with the preceding
q's. We take the g n (xy) function so that it is = l in q n , except
on these sides, and there 0. Then
G(x, y) =
has for each point only one term ^ 0, if x, y lies in Q, and no
term = if it lies in .
0, for each point of S.
EXAMPLES OF DISCONTINUOUS FUNCTIONS 467
Since E is apantactic, each point of & is a point of disconti
nuity of the 2 kind ; each point of Q is a point of continuity.
4. Let /(#/) be a function defined over 21 which contains the
complete apantactic set S.
Then 9  *
472. 1. Let 21 = (0, 1), S8 n = the points m i n .
^ n
Then 93 n , S3,, have no points in common.
Let/ n (z) = 1 in 93 n , and = in n = 21  S3 n .
Let93={93 n J. Then
*<>*> (!:*
The function F is totally discontinuous in 93, oscillating be
tween and 1. The series F does not converge uniformly in
any subinterval of 21.
2. Keeping the notation in 1, let
At each point of 93 n , Gr= , while # = in J5.
w
The function 6r is discontinuous at the points of S3, but con
tinuous at the points JB. The series 6r converges uniformly in
21, yet an infinity of terms are discontinuous in any interval in 21.
473. Let the limited set 21 be the union of an enumerable set
of complete sets S2l n j. We show how to construct a function/,
which is discontinuous at the points of 21, but continuous else
where in an wway space.
Let us suppose first that 21 consists of but one set and is com
plete. A point all of whose coordinates are rational, let us call
rational, the other points of space we will call nonrational. If 21
has an inner rational point, let /= 1 at this point, on the frontier
of 21 let /= 1 also ; at all other points of space let /= 0. Then
each point a of 21 is a point of discontinuity. For if x is a fron
468 DISCONTINUOUS FUNCTIONS
tier or an inner rational point of 2l,/(#) = 1, while in any V(x)
there are points where /= 0. If x is not in 21, all the points of
some D(x) are also not in 21. At these points /= 0. Hence /is
continuous at such points.
We turn now to the general case. We have
81 = ^+^2 + ^3 + ...
where A 1 = $l 1 i A% = points of 21 2 not in 2lj, etc. Let/j = 1 at the
rational inner points of A, and at the frontier points of 2lj ; at all
other points let /j = 0. Let / 2 = at the rational inner points of
4 2 , and at the frontier points of A^ not in A l ; at all other points
let/ 2 = 0. At similar points of A Q let/ 3 = , and elsewhere = 0,
etc.
Consider now & *?* , \
* = VnOV'Zm)
Let x = a be a point of 21. If it is an inner point of some A t ,
it is obviously a point of discontinuity of F. If not, it is a proper
frontier point of one of the A* a. Then in any D(a) there are points
of space not in 21, or there are points of an infinite number of the
As. In either case a is a point of discontinuity. Similarly we
see F is continuous at a point not in 21.
2. We can obviously generalize the preceding problem by sup
posing 21 to lie in a complete set S3, such that each frontier point
of 21 is a limit point of A = S3 21.
For we have only to replace our mway space by 83.
Functions of Class I
474. 1. Baire has introduced an important classification of
functions as follows :
Let /(#!#) be defined over 21; /and 21 limited or unlimited.
If /is continuous in 21, we say its class is in 21, and write
Class /=0 , orCl/=0 , Mod 21.
If
each/ n being of class in 21, we say its class is 1, if/ does not lie
in class 0, mod 21.
FUNCTIONS OF CLASS 1 469
2. Let the series F(x) = TLf n (x)
converge in 21, each term/ n being continuous in 21. Since
we see F is of class 0, or class 1, according as F is continuous, or
not continuous in 31. A similar remark holds for infinite prod
ucts
3. The derivatives of a function f(x) give rise to functions of
class or 1. For let f(x) have a unilateral differential coeffi
cient g(x) at each point of 21. Both / and 21 may be unlimited.
To fix the ideas, suppose the righthand differential coefficient
exists. Let 7^ > 7^ 2 > = 0. Then
n
is a continuous function of x in 21. But
2O)=lim9 n <
flnoo
exists at each x in 21 by hypothesis.
A similar remark applies to the partial derivatives
&L, ... J.
dx 1 ' dx m
of a function /(a?j a; n ).
4  Let
each/ n being of class 1 in 21. Then we say, Cl/= 2 if /does not
lie in a lower class. In this way we may continue. It is of
course necessary to show that such functions actually exist.
475. Example 1.
Let f(x \ ^ lim __^_ = I !' for * > '
J ^ ' =. 1 + nx I 0, for x = 0.
This function was considered in I, 331. In any interval
21 = (0 < b) containing the origin x = 0, Cl/= 1 ; in any inter
val (a < i), a > 0, not containing the origin, Cl/= 0.
470 DISCONTINUOUS FUNCTIONS
Example 2.
Let /(*) = lirnM j = 0, in  (00,00).
n=oo
The class of f(x) is in 21. Although each f n is limited in 21,
the graphs of f n have peaks near x = which == oo, as n = oo.
Example 3. If we combine the two functions in Ex. 1, 2, we
get */ \ r f 1 ,11 f 1* forz^O,
f(x) = hm ^ \nx = 1 '
' v y nss30 1 l + nx e nx * I ( 0, for x=0.
Hence C\f(x) = 1 for any set 33 embracing the origin; =0
for any other set.
Example 4>
Let j?/ \ v *+r^ or ^A IN
/ (a?) = Inn are n , in 21 = (0, 1).
n=<x>
Then /(a?) = , for x =
i
= a:^ 2 , for x > 0.
We see thus that / is continuous in (0*, 1), and has a point of
infinite discontinuity at x = 0.
Hence Class /(af)= 1, in 21
= 0, in(OM).
Example 5.
Let / O) = Hm ^r in 2T = (0, oo).
w ~ '>* L
n
Then ..... = 1 fora . >0
= h oo , for x = 0.
Here lim/ n (V)
does not exist at x = 0. We cannot therefore speak of the class
of /(#) in 21 since it is not defined at the point x = 0. It is
defined in 93 = (0*, oo), and its class is obviously 0, mod 33.
FUNCTIONS OF CLASS 1
471
Example 6.
Let
f(x) = sin  , for x &
x
= a constant c , for x = 0.
We show that Cl/= 1 in 21 = ( 0, oo). For let
f nxj
\ , nx
+ ~  8in
o,
r *
lim A n
Now by Ex. 1,
while
im n a; =
0, for o;=0.
As each f n is continuous in 3, and
lim / n (*)=/(*) in ,
we see its class is <_ 1. As / is discontinuous at = 0, its class
is not in 21.
Example 7. Let , ^ ,. 1 .1
r f(x) = lim  sin 
= n x
Here the functions f n (x) under the limit sign are not defined
for x = 0. Thus /is not defined at this point. We cannot there
fore speak of the class of / with respect to any set embracing the
point =0. For any set S3 not containing this point, Cl /= 0,
since /(x) = in SB.
Let us set
. . N . 1 . A
<f>(x) =s sin  , for x *
x
= a constant c
for x 0.
Let
g(x) = lim <t>(x) = lim
472 DISCONTINUOUS FUNCTIONS
Here g is a continuous function in 21 = ( oo, oo). Its class is
thus in 31. On the other hand, the functions < n are each of
class 1 in 31.
Example 8.
is defined at all the points of (00, oo) except 0, 1, 2,
These latter are points of infinite discontinuity. In its domain
of definition, F is a continuous function. Hence Cl F(V) =
with respect to this domain.
476. 1. If 31, limited or unlimited, is the union of an enumerable
set of complete sets, we say 31 is hyper complete.
Example 1. The points S* within a unit sphere S, form a
hypercomplete set. For let S r have the same center as S, and
radius r<\. Obviously each 2 r is complete, while J2 r j = /S Y *, r
ranging over r l < r% < = 1.
Example 2. An enumerable set of points a l , a 2 form a hyper
complete set. For each a n may be regarded as a complete set,
embracing but a single point.
2. 7/31J, 31 2  are limited hypercomplete sets, so is their union
;3u = 3i.
For each 3l m is the union of an enumerable set of complete sets
Sl w , n . Thus 31 = j3l ;n , n S m, n = 1, 2 . is hypercomplete.
Let 31 be complete. If S3 is a complete part of 31, A = 31 93 is
hypercomplete.
For let O= Jq n  be a border set of 93, as in 328. The points
A n of A in each q n are complete, since 31 is complete. Thus
A=\A n \, and A is hypercomplete.
Let 21= \tyi n l be hypercomplete, each 3l n being complete. If $8 is
a complete part of 31, A = 31 93 is hypercomplete.
For let A n denote the points of 3l n not in 93 Then as above,
A n is hypercomplete. As A = \A n \, A is also hypercomplete.
FUNCTIONS OF CLASS 1 473
477. 1. @ e Sets. If the limited or unlimited set 21 is the union
of an enumerable set of limited complete sets, in each of which
Osc/<e, we shall say 21 is an (g e set. If, however small e>0 is
taken, 21 is an @ set, we shall say 21 is an (g c set, e = 0, which we
may also express by @^o'
2. Let /(x!'" XK) be continuous in the limited complete set 21.
Then 21 is an S set, e == 0.
For let e > be taken small at pleasure and fixed. By I, 353,
there exists a cubical division of space D, such that if 2l n denote
the points of 21 in one of the cells of D, Osc/< e in 2l n . As 2l n is
complete, since 21 is, 21 is an ( e set.
3. An enumerable set of points 21 = \ a n \ is an (S^ set.
For each a n may be regarded as a complete set, embracing but
a single point. But in a set embracing but one point, Osc/= 0.
4. The union of an enumerable set of @ e sets 21 = J2l m  is an (g e set.
For each 2t m is the union of an enumerable set of limited sets
2l m = J2l m , n {,n=l, 2,... and Osc/< e in each 2U
Thus a = j8U} , , rc=l, 2,..
But an enumerable set of enumerable sets is an enumerable set.
Hence 21 is an S e set.
5. Letf(x l # w ) be continuous in the complete set 21, except at the
points 3) = dj, rf 2 d a . Then 21 is an &=M> set.
For let 6>0 be taken small at pleasure and fixed. About each
point of 3) we describe a sphere of radius p. Let 2l p denote the
points of 21 not within one of these spheres. Obviously 2l p is com
plete. Let p range over r l > r 2 >  = 0. If we set 21 = A + 3),
obviously ^ = {2i r J. As/ is continuous in 2l rw , it is an ( set.
Hence 21, being the union of A and 35, is an @ set.
478. 1. Let 21 be an (:,, set. The points 3) of 21 common to the
limited complete set $8 form an @ e set.
For 21 is the union of the complete sets 2l n , in each of which
Osc/<. But the points of 2l n in 93 form a complete set 4 n , and
of course Osc/< e in A n . As 3) = \ A n \, it is an (g e set.
474 DISCONTINUOUS FUNCTIONS
2. Let 21 be a limited @ e set. Let 93 be a complete part of 21.
Then A = 21  $ is an @ e *tf.
For 21 is the union of the complete sets 2l n , in each of which
Osc /<. The points of 2l n not in 33 form a set J. n , such that
Osc /< in A n also. But A = \A n \, arid each J[ n being hyper
complete, is an S e set.
3. Let/(^ x # m ) be defined over 21, either/ or 21 being limited
or unlimited. The points of 21 at which
<*</< (1
may be denoted by
(</<) (2
If in 1) one of the equality signs is missing, it will of course be
dropped in 2).
479. 1. Letfi,/^, "be continuous in the limited complete set H.
If at each point of 21, Urn f n exists, 21 is an @ e=M) set and so is any
complete 33 < 21.
For let lirn f n (x l z m ) =f(x l # m ) in 21. Let us effect a
n=oo
division of norm e/2 of the interval ( 00, oo ) by interpolating
the points m_ 2 , m_ : , w = 0, m l , m 2
Let 2t t = (m t </< m t+2 ), then 21 = 21J.
Next let ^ TS f , 1 / ^ ^ 11
>n, P = ^ w, +</,< ^ l+2  
>P I n n)
Then 2l t =535 n , P S , ^^ = 1,2 (1
For let a be a point of 2l t , and say f(a) = a. Then
m t < a < w t + 2 .
But a e</ g (a) <af e , j>somejt?,
and we may take e and n so that
Hence a is in S) nip .
Conversely, let a be a point of {$) n , p }. Then a lies in some
nfp . Hence,
FUNCTIONS OF CLASS 1 475
But as/ n (a) ==/(#), we have
Hence if e is sufficiently small,
and thus a is in 2l t .
Thus 1) is established. But ) np is a divisor of complete sets,
and is therefore complete. Thus 21 is the union of an enumerable
set of complete sets J93 t j, in each of which Osc/<e, e small at
pleasure.
Let now 93 be any complete part of 21. Let a t = Dv J93, 93 t }.
Then a t is complete, and Osc/<e, in a t  Moreover, 93 = {a t .
Hence 93 is an @ =M) set.
2. // Class /< 1 m limited complete 21, / limited or unlimited,
21 is ^w (S se.
This is an obvious result from 1.
3. Let /(#! # m ) ^ # totally discontinuous function in the non
enumerable set 21. Then Class /is ## or 1 iw 21, i/* b = Disc/a^
^(?7i point is < k > 0.
For in any subset 93 of 21 containing the point #, Osc / > k.
Hence Osc/is not <e, in any part of SI, if e < &. Thus 21 cannot
be an ( e set.
4. J/ Class /(#! a^ m )<. 1 fw the limited complete set 21, the set
93 = (#</< 5) is a hyper complete set, a, b being arbitrary numbers.
For we have only to take a = m t , S = m l+2 . Then 93 = 2l t , which,
as in 1, is hypercomplete.
480. {Lebesgue.} Let the limited or unlimited function f (x^ # m )
be defined over the limited set 21. If 21 may be regarded as an
(Se^o set with respect to /, the class of f is < 3 .
For let o) 1 > G> 2 >.== 0. By hypothesis 21 is the union of a
sequence of complete sets
2l u , 2I 12 , 5ffi3*" ($1
in each of which Osc / <_&> r 21 is also the union of a sequence
of complete sets
u  < % (1
476 DISCONTINUOUS FUNCTIONS
in each of which Osc/< <o 2 . If we superpose the division 1) of
?I on the division S l ^ each 2l tlt will fall into an enumerable set
of complete sets, and together they will form an enumerable
sequence
2l a i * 2122 > IM*" (^2
in each of which Osc/<LG> 2 . Continuing in this way we see that
21 is the union of the complete sets
such that in each set of $ n , ()sc/< o> n , and such that each set lies
in some set of the preceding sequence S n _ lf
With each 2l w , , we associate a constant (7 njt , such that
!/(*) C^<*> n , in?l n ,, (2
and call C nt the corresponding field constant.
We show now how to define a sequence of continuous functions
/i'/2 '" which =/. To this end we effect a sequence of super
imposed divisions of space Dj, D^ of norms = 0. The vertices
of the cubes of D n we call the lattice points L n . The cells of D n
containing a given lattice point I of L n form a cube Q. Let 3l lti
be the first set of S l containing a point of Q. Let 2l 2 t 2 be the first
set of # 2 containing a point of Q lying in 3l ltl . Continuing in
this way we get
2l ltl >i2l 2l ,>...>2l nln .
To 2l nln belongs the field constant O nln ; this we associate with
the lattice point I and call it the corresponding lattice constant.
Let now S be a cell of D n containing a point of 21. It has 2 n
vertices or lattice points. Let P 9 denote any product of & differ
ent factors a; n , x r ^ x rg . We consider the polynomial
<f> = AP n + ^BP n ^ + 2 (7P n _ 2 4  4
the summation in each case extending over all the distinct
products of that type. The number of terms in </> is, by I, 96,
FUNCTIONS OF CLASS 1 477
We can thus determine the 2 n coefficients of <f> so that the values
of (f> at the lattice points of are the corresponding lattice con
stants. Thus <f> is a continuous function in , whose greatest and
least values are the greatest and least lattice constants belonging
to . Each cube containing a point of 21 has associated with it
a <f> function.
We now define /(#!#,) ^y stating that its value in any
cube of 7) n , containing a point of 21, is that of the correspond
ing <f> function. Since <f> is linear in each variable, two </>'s belong
ing to adjacent cubes have the same values along their common
points.
We show now that/ n (>) ==f(x) at any point x of 21, or that
e >0, v, /O) f n (x)  < , n > v. (3
Let co e < e/8. Let 3I ltl be the first set in /Si containing the point x,
2l 2l , the first set of S 2 lying in 2l ltl and containing x. Continuing
we get ^ > ^ > ^ ^ > ^
Let ty e be the union of the sets in &\ preceding 2l u ; of the sets in
$2 preceding 2l 2t and lying in 2l lt , and so on, finally the sets of
S e preceding 2l et , and lying in 2l e _ liV _ 1 . Their number being
finite, 8= Dist (2l eta , $*) is obviously > 0. We may therefore
take v > e so large that cubes of D v about the point x lie wholly
in !>(, rj < S.
Consider now/ n (#), n > v, and let us suppose first that x is not
a lattice point of /> n . Let it lie within the cell of D n . Then
f n (x) is a mean of the values of
where Z is any one of the 2 n vertices of , and C njn is the corre
sponding lattice constant, which we know is associated with the
**,*.
We observe now that each of the
For each set in S n is a part of some set in any of the preceding
sequences. Now 2l n?n cannot be a part of 2l 1Jk , k < ij, for none of
478 DISCONTINUOUS FUNCTIONS
these points lie in A,(X). Hence 2l n;  n is a part of 2l ltl . For the
same reason it is a part of 8l 2l2 , etc., which establishes 4).
Let now x' be a point of 21^ . Then
I c nja  t\. \<\o nia /<>')  + !/(*') o ett i
<a, B +o, e <l , by 2). (5
From this follows, since / (a;) is a mean of these C njii , that
l/n(0QJ< (6
But now
\f O) MX)  < \f O)  C njn  + G nin /<  . (7
As x lies in 8k ta ,
I/O)  C njn  < /(*)  O^ I + I (7 ft .  (7 n;n 
<.+<, (8
by 2), 5). From 6), 8) we have 3) for the present case.
The case that a; is a lattice point for some division and hence
for all following, has really been established by the foregoing
reasoning.
481. 1. Let fie defined over the limited set 21. If for arbitrary
a, 5, the sets 93 = (a </< 6) are hyper 'complete , then Class /< 1.
For let us effect a division of norm e/2 of (00, oo) as in
479,1. Then 2t=J2lJ, where as before 2l t = (m t </< ra t + 2 ).
But as Osc/<e in 2l t , and as each 2l t is hypercomplete by
hypothesis, our theorem is a corollary of 480.
2. For f(x l # m ) to be of class < 1 in the limited complete set
21, it is necessary and sufficient that the sets (a <f < 6) are hyper
complete, a, b being arbitrary.
This follows from 1 and 479, 2.
3. Let limited 21 be the union of an enumerable set of complete sets
j, such that Cl/< 1 in each 2l n , then Cl/< 1 in 21.
FUNCTIONS OF CLASS 1 479
For by 479, 1, ?l n is the union of an enumerable set of complete
sets in each of which Osc/ < e. Thus SI is also such a set, i.e. an
iS e set. We now apply 480, 1.
4. If Class/ < 1 in the limited complete set 31, its class is < 1,
in any complete part 33 of SI.
This follows from 479, 1 and 480, 1.
482. 1. Let f(x l x m ) be defined over the complete set SI, and
have only an enumerable set S of points of discontinuity in 31.
Then Class/ = 1 in 31.
For the points JE of 31 at which Osc/ > e/2 form a complete
part of 31, by 462, 3. Bat E, being a part of (, is enumerable
and is hence an @ g set by 477, 3. Let us turn to 33 = 31 E. For
each of its points b, there exists a 8 > 0, such that Osc/ < in
the set b of points of 93 lying in Z> 5 (6). As 31 is complete, so is b.
As E is complete, there is an enumerable set of these b, call them
bj, b 2 , such that 33 = \b a \. As 31 = S3 4 E, it is the union of
an enumerable set of complete sets, in each of which Osc/< e.
This is true however small e>0 is taken. We apply now 480, 1.
2. We can now construct functions of class 2.
Example. Let f n (x l x m )= 1 at the rational points in the
unit cube }, whose coordinates have denominators < n. Else
where let/ n = 0. Since f n has only a finite number of discontinu
ities in Q, Cl/ n = 1 in Q. Let now
At a nonrational point, each f n = 0, .. /=0. At a rational
point, / n = l for all w > some s. Hence at such a point /= 1.
Thus each point of Q is a point of discontinuity and Disc/= 1.
Hence Cl/ is not 1. As / is the limit of functions of class 1, its
class is 2.
483. Let f(x l x m ) be continuous with respect to each # t , at each
point of a limited set 31, each of whose points is an inner point.
Then Class /<!.
480 DISCONTINUOUS FUNCTIONS
For let 21 lie witliin a cube Q. Then A = Q. 21 is complete.
We may therefore regard 21 as a border set of A ; that is, a set of
nonoverlapping cubes }q n . We show now that C1/<1 in any
one of these cubes as q. To this end we show that the points 93 m
of q at which
a+<f<b
m ' m
form a complete set. For let b l , b% be points of 33 m , which = /8.
We wish to show that /3 lies in S3 m . Suppose first that i,, b s+1
have all their coordinates except one, say x, the same as the coordi
nates of /3. Since
* + </(. + P)<*,
m m
therefore  
m p=*> m
As/ is continuous in a^, and as only the coordinate x l varies in
< we have
m m
Hence lies in 33 m .
We suppose next that b 8 < b g+l have all their coordinates the
same as /3 except two, say x l , # 2 .
We may place each b n at the center of an interval t of length S,
parallel to the x l axis, such that
+* *</<>)<& + ,
m m
since /is uniformly continuous in x^, by I, 352. These intervals
cut an ordinate in the x, x 2 plane through y8, in a set of points
c t+p which == ft. Then as before,
m m
As is small at pleasure, /3 lies in S3 m . In this way we may
continue.
As Cl/< 1 in eacli q n , it is in 21, by 481, 3.
FUNCTIONS OF CLASS 1 481
484. (Volterra. ) Let J\,f ti be at most point wist' discontinuous
in the limited complete set 21. Then there exists a point of ?I at
which all thef n are continuous.
For if 21 contains an isolated point, the theorem is obviously
true, since every function is continuous at an isolated point. Let
us therefore suppose that 21 is perfect.
Let e 1 > 2 >=0. Let a l be a point of continuity of / r
Then Osc/^6 , insome2I 1 =F 5l (a 1 ).
In 2^ there is a point b of continuity of f r Hence Osc/j < e 2
in some F^J), and we may take b so that K,(6)<2l r But in
Fif(J) there is a point a 2 at which / 2 is continuous. Hence
Osc/! < 6 2 , O*c/ 2 < e l , in some 21 2 == F,( a a)'
and we may take <z 2 such that 21 2 < 2lj . Similarly there exists a
point a 3 in 21 2 , such that
Osc/t < 3 , Osc/ 2 < e 2 , Osc/ 3 < l , in some 21 3 = ^(^3)^
and we may take 8 so that 21 8 < 21 2 .
In this way we may continue. As the sets 2l n are obviously
complete, Dv\ty. n \ contains at least one point a of 21. But at this
point each/ m is continuous.
485. 1. Let 21 = 33 + & be complete, let 33, & be pantactic with
reference to 21. Then there exists no pair of functions /, g defined
over 21, such that if 33 are the points of discontinuity of f in 21, then
33 shall be the points of continuity of g in 21.
This is a corollary of Volterra's theorem. For in any Fi(a) of
a point of 21, there are points of 33 and of @. Hence there are
points of continuity of /and g. Hence/, g are at most pointwise
discontinuous in 21. Then by 484, there is a point in 21 where/
and g are both continuous, which contradicts the hypothesis.
2. Let 21= 33 hS be complete, and let 33, & each be pantactic with
reference to 21. If 33 is hypercomplete, is not.
For if 33, ( were the union of an enumerable set of complete
sets, 473 shows that there exists a function / defined over 21
which has 33 as its points of discontinuity ; and also a function g
482 DISCONTINUOUS FUNCTIONS
which has 6 as its points of discontinuity. But no such pair of
functions can exist by 1.
3. The nonrational points $ in any cube Q cannot be hyper
complete.
For the rational points in jQ are hypercomplete.
4. As an application of 2 we can state :
The limited function /(^ 1 .^ OT ) which is < at the irrational
points of a cube Q, and > at the other points 3 of ' Q, cannot be
of class or 1 in Q.
For if Cl/ < 1, the points of O where/ > must form a hyper
complete set, by 479, 4. But these are the points 3>.
486. 1. (Saire.) If the class off^x^'x^) is 1 in the com
plete set 21, it is at most pointwise discontinuous in any complete
<a.
If Cl/= 1 in 31, it is < 1 in any complete 53 < 21 by 481, 4 ; we
may therefore take 95 = 21. Let a be any point of 21. We shall
show that in any V= Fi(a) there is a point c of continuity of f.
Let e l > e 2 > = 0. Using the notation of 479, i, we saw that
the sets 2l l = (m l </< w t + 2 ) are hypercomplete. By 473, we can
construct a function ^>,(x l # m ), defined over the wway space
9{ w which is discontinuous at the points 2I t , and continuous else
where in 9t m . These functions c^, < 2 are not all at most point
wise discontinuous in V. For then, by 484, there exists in F a
point of continuity J, common to all the <'s. This point b must
lie in some 2l t , whose points are points of discontinuity of < t .
Let us therefore suppose that fy is not at most pointwise dis
continuous in V. Then there exists a point c l in F", and an ^
such that V^ = ^(^j) contains no point of continuity of <y.
Thus Fi<8k. But in 21, and hence in V^ Osc /<e r The
same reasoning shows that in V 1 there exists a F^= ^,(^2)' suc ^
that Osc/< 2 in F^. As 21 is complete, V l > F 2 > defines a
point <? in Fat which /is continuous.
2. If the class off(x l # m ) is 1 in the complete set 21, its points
of discontinuity Qform <t set of the first category.
FUNCTIONS OF CLASS 1 483
For by 462, 3, the points ) n of 3D at which Osc/>  form a
n
complete set. Each ) n is apantactic, since / is at most pointwise
discontinuous, and O n is complete. Hence 35 = \ O ft } is the union
of an enumerable set of apantactic sets, and is therefore of the 1
category.
487. 1. Let f be defined over the limited complete set 21. If
Class / is not < 1, there exists a perfect set 35 in 21, such that f is
totally discontinuous in ).
For if G\f is not <1 there exists, by 480, an e such that for
this e, 31 is not an S e set. Let now c be a point of 21 such that
the points a of 21 which lie within some cube q, whose center is tf,
form an (g e set. Let 93 = JaJ, 6 = \c\.
Then 93 = S. For obviously (<93, since each c is in some
a. On the other hand, 93 < S. For any point b of 93 lies within
some q. Thus b is the center of a cube q' within q. Obviously
the points of 21 within q' form an < e set.
By Borel's theorem, each point c lies within an enumerable set
of cubes {c n , such that each c lies within some q. Thus the
points a n of 21 in c n , form an @ e set. As S = }a w j, is an S e set.
Let 35 = 21  (. If 35 were 0, 21 = and 21 would be an @ e set
contrary to hypothesis. Thus 3) > 0.
3) is complete. For if I were a limiting point of 3) in 6, I must
lie in some c. But every point of 21 in c is a point of ( as we saw.
Thus I cannot lie in g.
We show finally that at any point d of 35,
Osc/>, with respect to .
If not, Osc/< with respect to the points b of 35 within
some cube q whose center is d. Then b is an S e set. Also the
points c of in q form an S e set. Thus the points b f e, that is,
the points of 21 in q form an (< set. Hence d belongs to , and
not to 3). As Osc/>e at each point of 3D, each point of 35 is a
point of discontinuity with respect to 3). Thus/ is totally discon
tinuous in 35.
This shows that 35 can contain no isolated points. Hence 35 is
perfect.
484 DISCONTINUOUS FUNCTIONS
2. Let f be defined over the limited complete set 21. If f is at
most pointwise discontinuous in any perfect 93 < 21, its class is < 1
w2l.
This is a corollary of 1. For if Class / were not 0, or 1, there
exists a perfect set 33 such that /is totally discontinuous in 33.
488. If the class of f, g < 1 in the limited complete set 21, the class
of their sum, difference, or product is < 1 . If f > in 21, the class
of <f> =
For example, let us consider the product h =fg. If Cl h is not
< 1, there exists a perfect set 33 in 21, as we saw in 487, 1, such
that A is totally discontinuous in 33. But/, g being of class <, 1,
are at most pointwise discontinuous in 33 by 486. Then by 484,
there exists a point of 33 at which/, g are both continuous. Then
h is continuous at this point, and is therefore not totally discon
tinous in 33.
Let us consider now the quotient <f>. If Cl (j> is not < 1, <f> is
totally discontinuous in some perfect set ) in 21. But since />
in ), / must also be totally discontinuous in >. This contradicts
486.
489. 1. Let F = 2/ ti ...<,(#! # TO ) converge uniformly in the com
plete set 21. Let the class of each termf L be <, 1, then Class F < 1
MI a.
For setting as usual [117],
there exists for each e > 0. a fixed rectangular cell R^ such that
 JF\  < e, as x ranges over 21. (2
As the class of each term in F K is < 1, Cl JP A < 1 in 21. Hence
21 is an @ set with respect to F^
From 1), 2) it follows that 21 is an (g e set with respect to F.
2. Let F = n/^...^^ # m ) converge uniformly in the complete
set $. If the class of eachf, is < 1, then Cl F < 1 in 21.
SEMICONTINUOUS FUNCTIONS 485
Semicontinuous Functions
490. Let /(#! Xm) be defined over 21. If a is a point of SI,
Max/ in Fi(a) exists, finite or infinite, and may be regarded as a
function of S. When finite, it is a monotone decreasing function
of 8. Thus its limit as S = exists, finite or infinite. We call
this limit the maximum off at x = a, and we denote it by
Max/.
x=a
Similar remarks apply to the minimum of /in F^(a). Its limit,
finite or infinite, as 8 == 0, we call the minimum of f at x = a, and
we denote it by
Min/
x=a
The maximum and minimum of /in F$(a) may be denoted by
Max/ , Min/.
a, a, 6
Obviously, Max ( _ /} = _ Min/,
:r=a .r=a
Min (/)= Max/
.r=a #=a
491. Example 1. ,
/(*0 = iin(l, 1) , fo
x
= , for x =0.
Then Max/= f oo , Min/=  oo.
ar=0 ^=0
Example 2. 
/(^) = sin  in (1,1) , f
x
= , for x = 0.
Then Max/=l , Min/=l.
X=0 X=Q
Examples. /(a; ) = 1 in ( 1, 1) , for**0
= 2 , for a; = 0.
Then Max/=2 , Min/=l.
486 DISCONTINUOUS FUNCTIONS
We observe that in Exs. 1 and 2,
Hm/==Max/ , lim/==Min/;
r=0 x=0 #=0 xQ
while in Ex. 3, _ _
lim/= 1 , arid hence Max/> lira/.
Also lim/= Min/.
#=0
Example 4* i
/(&) = (^ + 1) sin in (1,1) , for
=  2 , f or rr = 0.
Here =l , Min/ =2,
x=Q c=0
Examples. Lei f(x ) = x , for rational * in (0, 1)
= 1 , for irrational x.
Here Max/=l , Min/=0,
lhn/=l.
492. 1. For M to be the maximum of f at x = a< it is necessary
and sufficient that
1 e > 0, S > 0, /< < M+ e, for any x in F(a) ;
2 there exists for each e > 0, and in any V&(a), a point a such
M </().
These conditions are necessary. For M is the limit of Max/
in Fi(a), as S =^= 0. Hence
> 0, o > 0, Max/ <
a,
But for any x in Fi(a),
/O) < Max/.
a,
SEMICONTINUOUS FUNCTIONS 487
Hence f(x)<M+e , x in F 5 (a),
which is condition 1.
As to 2, we remark that for each e > 0, and in any F$(a),
there is a point a, such that
 +Max/ </().
a, 6
But M< Max/.
a, 8
Hence
which is 2.
These conditions are sufficient. For from 1 we have
Max/ < 7tf + e,
a, 6
and hence letting 8 = 0,
since e > is small at pleasure.
From 2 we have
Max/> Me,
a, 6
and hence letting 8=0,
Max/ > M. (2
From 1), 2) we have M = Max/.
2. J^or m to be the minimum of f at x = a, i is necessary and
sufficient that
1 e > 0, 8 > 0, m e </(.r), /or any x in T 5 (a) ;
2 ^a there exists for each e > 0, rtm# in any Fi(a), a point a
such that
/(a) < m f .
493. When Max/ = /(a), we say / is supracontinuous at x = a.
ar=a
When Min/ = /(a), we say / is mfracontinuous at a. When/ is
ar=a
supra (infra) continuous at each point of SI, we say / is supra
(infra) continuous in 21. When /is either supra or infracontinu
ous at a and we do not care to specify which, we say it is semi
continuous at a.
488 DISCONTINUOUS FUNCTIONS
The function which is equal to Max /at each point x of 21 we
call the maximal function of/ and denote it by a dash above, viz.
/(. Similarly the minimal function /(V) is defined as the value
of Min /at each point of 21. ~~
Obviously Ogc/ = Max/ _ Min/ = Digc/
xa xa xa x=a
We call 7
(*)=/()/()
the oscillatory function.
We have at once the theorem :
For f to be continuous at x = a, it is necessary and sufficient that
/(a) =/()= /(a).
F r Min / < /() < Max /.
a, 6 a, 6
Passing to the limit x = a, we have
Min /</(<*)< Max/,
x^a x=a
or
/oo <
But for / to be continuous at x = a, it is necessary and suffi
cient that
(V) = Osc/= 0.
x=a
494. 1. For f to be supracontinuous at x = a, it is necessary and
sufficient that for each e > 0, there exists a S > 0, such that
/(V) < /(a) f e , for any x in F 6 (a). (1
Similarly the condition for infracontinuity is
/(a) e < f(x) , for any x in F fi (a). ' (2
Let us prove 1). It is necessary. For when /is supracontinu
ous at a,
Then by 492, 1,
>0 , S>0 , f(x) < f(a) + e , for any x in F fi (a),
which is 1).
SEMICONTINUOUS FUNCTIONS 489
It is sufficient. For 1) is condition 1 of 492, i. The condition
2 is satisfied, since for a we may take the point a.
2. The maximal function f(x) is supracontinuous ; the minimal
function f(x) is infracontinuous, in 21.
To prove that /is supracontinuous we use 1, showing that
f(x) < /(#) + e > for any x in some Fa 00
Now by 492, 1,
e' > 0, 8 > , f(x) < 700 4 e' , for any x in Fi(a).
Thus if e' < e
8
/(V) < /(a) + , f or any x in V^ (a) , 17 =  .
3. The sum of two supra (infra) continuous functions in 21 is a
supra (infra) continuous function in 31.
For let/, g be supracontinuous in 21 ; let/+ y = h. Then by 1,
for any a; in some Fa (a) ; hence
This, by 1, shows that h is supracontinuous at #= a.
4. If f(x) is supra (infra) continuous at x = a, g(x)~f(x)
is infra (supra) continuous.
Let us suppose that /is supracontinuous. Then by 1,
/(#)</()+ e , for any x in some V*(a).
Hence
or g(a)e<g(x) , for any a: in
Thus bv 1, Q is infracontinuous at a.
490 DISCONTINUOUS FUNCTIONS
495. Iff\x l "x m ) is supracontinuous in the limited complete
set 31, the points 93 of 31 at which /> c an arbitrary constant form a
complete set.
For let /> c at b 1 , h 2 which = b ; we wish to show that b lies
in S3.
Since/ is supracontinuous, by 494, 1,
/(V)</(6) + e , for any x in some F 5 (5)= V.
But <?</(6 n ), by hypothesis ; and b n lies in FJ for n> some w
Hence
*</().
As e > is small at pleasure,
and 5 lies in 33.
496. 1. TA0 oscillatory function a>(x) is supracontinuous.
For by 493, ,, \ TVT ^ \f *
J ' a)(a;)= Max/ Mm/
= Max/+ Max (/).
But these two maximal functions are supracontinuous by 494, 2.
Hence by 494, 3, their sum o> is supracontinuous.
2. The oscillatory function o> is not necessarily infracon
tinuous, as is shown by the following
Example. /= 1 in (1, 1), except for x = 0, where /= 2.
Then &(x) = 0, except at x = 0, where CD = 1. Thus
Min o>(z) = , while o>(0) = 1.
x=Q
Hence w(x) is not infracontinuous at x = 0.
3. Let fc>(#) J# ^A^ oscillatory function of f(x l rr m ) in 31. #W
/ <o be at most pointwise discontinuous in 31, z{ zs necessary that
Min o> = at each point of 31. /f 31 z complete, this condition is
sufficient.
SEMICONTINUOUS FUNCTIONS 491
It is necessary. For let a be a point of 21. As f is at most
pointwise discontinuous, there exists a point of continuity in any
Hence Min <*>(x) = 0, in Fi(a). Hence Min o>(z) = 0.
It is sufficient. For let 1 >e 2 > =0. Since Min (.*:) = 0,
.r=
there exists in any Fi(a) a point j such that o>( 1 )< 1 .
Hence o>(#) < j in some Fi,( a i) < ^V ' n ^61 ^ nere exists a point
2 such that ct>(#)<e 2 in some F 6 /a) < F^, etc. Since 21 is com
plete and since we may let 8 n = 0,
Pi, > ^ a >== a point of 21,
at which f is obviously continuous. Thus in each ^(a) is a point
of continuity of/. Hence /is at most pointwise discontinuous.
497. 1 . At each point x of 21,
<f> = Min \f(x) f(x)\,and^= Min \f(x) f(x)\
are both == 0.
Let us show that <f> = at an arbitrary point a of 21. By 494,
2, f(x) is supracontinuous ; hence by 494, l,
f(x) </(a) he , for any x in some Fa(a) = F. (1
Also there exists a point a in Fsuch that
. (2
Also by definition
/()
If in 1) we replace ^ by a \\e
/()<
From 2), 3), 4) we have
or
As >0 is small at pleasure, this gives
<K<0 = o.
492 DISCONTINUOUS FUNCTIONS
2. If/is semicontinuous in the complete set ?l, it is at most point
wise discontinuous in 21.
For K*) =/(*)/(*)
/(*)] + [/GO f(*y\ (1
To fix the ideas let / be supracontinuous. Then <f> = in 51.
Hence 1) gives
Min to(x) = Min ty(x) = 0, by 1.
Thus by 496, 3, / is at most pointwise discontinuous in 21.
CHAPTER XV
DERIVATES, EXTREMES, VARIATION
Derivates
498. Suppose we have given a onevalued continuous function
f(x) spread over an interval 21= (a<6). We can state various
properties which it enjoys. For example, it is limited, it takes
on its extreme values, it is integrable. On the other hand, we
do not know 1 how it oscillates in 21, or 2 if it has a differ
ential coefficient at each point of 21. In this chapter we wish to
study the behavior of continuous functions with reference to these
last two properties. In Chapters VIII and XI of volume I this
subject was touched upon ; we wish here to develop it farther.
499. In I, 363, 364, we have defined the terms difference quo
tient, differential coefficient, derivative, right and lefthand dif
ferential coefficients and derivatives, unilateral differential coeffi
cients and derivatives. The corresponding symbols are
Lf(a) , Rf'(x) , Lf'(x).
The unilateral differential coefficient and derivative may be de
noted by
Z7/'() , Uf'(x). (1
When A ,
lim=
A=0 A#
does not exist, finite or infinite, we may introduce its upper and
lower limits. Thus
(2
always exist, finite or infinite. We call them the upper and lower
differential coefficients at the point x = a. The aggregate of values
493
494 DERIVATES, EXTREMES, VARIATION
that 2) take on define the upper and lower derivatives of ./(#), as
in I, 363.
In a similar manner we introduce the upper and lower right
and lefthand differential coefficients and derivatives,
Rf , Rf , Lf , Lf. (8
Thus, for example,
finite or infinite. Of. I, 336 seq.
is defined only in 21 = ( < /3), the points a, a f h must
lie in 31. Thus there is no upper or lower righthand differential
coefficient at x /3 ; also no upper or lower lefthand differential
coefficient at x = a. This fact must be borne in mind. We call
the functions 3) derivates to distinguish them from the deriva
tives Rf, Lf. When Rf ( )= Rf\a ), finite or infinite,
Rf (a) exists also finite or infinite, and has the same value. A
similar remark applies to the lefthand differential coefficient.
To avoid such repetition as just made, it is convenient to in
troduce the terms upper and lower unilateral differential coeffi
cients and derivatives, which may be denoted by
Vf' , V?'. (4
The symbol U should of course refer to the same side, if it is
used more than once in an investigation.
When no ambiguity can arise, we may abbreviate the symbols
3), 4) thus:
R , R , L , L , U , U.
The value of one of these derivates as R at a point x = a may
similarly be denoted by
5().
The difference quotient
/(>=/(*)
ab
may be denoted by
A(a, A).
DERIVATES 495
Example 1. f(x) = xs'm  , z=0 in ( 1, 1)
x
= , 2=0.
j
A sin 
Here for x = 0, ~^ = = = sin T .
Ax li h
Hence *fo + i .
Z/'(0)= + i ,
757/yvyyMflo7*> ^? ~F ( r r^l T' yi TI ^ / in ( 1 1 *\
JliJUViiiVUijG & J \^Ji> ) Ji> nlll ^ JL' =f \f 111 ^ A} J. J
X
= , z = 0.
A * Sln I
A/ a
Here for a: = , ~ = =
A^ A f
Hence .#/'(0)= 4 oo , J2f(0)=oo,
: + 00 , /'(0)=00.
Example 3. f(x) = 2: sin  , for < x < 1
x
= x^ sin  , for 1 < x <
x
= , f or x = 0.
Here
500. 1. Before taking up the general theory it will be well
for the reader to have a few examples in mind to show him how
complicated matters may get. In I, 367 seq., we have exhibited
functions which oscillate infinitely often about the points of a set
496 DERIVATES, EXTREMES, VARIATION
of the 1 species, and which may or may not have differential co
efficients at these points.
The following theorem enables us to construct functions which
do not possess a differential coefficient at the points of an enumer
able set.
2. Let S = \e n \ be an enumerable set lying in the interval 21. For
each x in 21, and e n in S, let x e n lie in an interval S3 containing
the origin. Let g(x) be continuous in S3. Let g 1 (x} exist and be
numerically <. M in S3, except at x = 0, where the difference quotients
are numerically < M. Let A = 2a n converge absolutely. Then
is a continuous function in 21, having a derivative in ( = 21 .
At the points of @, the difference quotient of F behaves essentially as
that of g at the origin.
For g(x) being continuous in S3, it is numerically < some con
stant in 21. Thus F converges uniformly in 21. As each term
g(x # n ) is continuous in 21, F is continuous in 21.
Let us consider its differential coefficient at a point x of S.
Since g'(x e n ) exists and is numerically < M,
^'(z)==2a n y(** n ) , by 156, 2.
Let now x = e m , a point of (,
F(x) = a m g(x  e m ) 4
The summation in 2* extends over all nj=m. Hence by what
has just been shown, Gr has a differential coefficient at x = e m .
Thus  behaves at x e m , essentially as ^ at x = 0. Hence
J
501. Example 1. Let
g(x) = ax , x >
b < < a.
x < 0,
DERIVATES 497
Then ,
is continuous in any interval 31, and has a derivative
n*)=SWo*o
it/
at the points of 21 not in (5. At the point e m ,
Let @ denote the rational points in 21. The graph of F(x) is a
continuous curve having tangents at a pantactic set of points ;
and at another pantactic set, viz. the set @, angular points (I, 366).
A simple example of a g function is
Example 2. Let g(x) = x* sin , x =
x
This function has a derivative
g f (x) = 2x sin TT cos ,
x x
= , x = Q.
Thus if 2e? n is an absolutely convergent series, and (g = \e n \ an
enumerable set in the interval 21 = (0, 1),
F(x) =2^0 O
is a continuous function whose derivative in 21 is
Thus F has a derivative which is continuous in 21 (, and at
the point # = e m
Disc F r = 2 c m 7r,
since
498 DKRIVATKS, KXTRKMKS, VARIATION
If S is the set of rational points in 21, the graph of F(x) is a
continuous curve having at each point of 21 a tangent which does
not turn continuously as the point of contact ranges over the
curve; indeed the points of abrupt change in the direction of the
tangent are pantactic in 21.
Example 8. Let g(x) = x sin log x 2 , x ^
Then #0*0 = s in 1M'  r2 4 ^ <' ( >s log # 2 , x^O.
At x = 0, = sin log A 2
which oscillates infinitely often between 1, as h = Ax == 0. Let
@ = j^ n j denote the rational points in an interval 21. The series
satisfies the condition of our theorem. Hence F(oi) is a continu
ous function in 21 which has a derivative in 21 @. At #= e m ,
Thus the graph of F is a continuous curve which has tangents at
a pantactic set of points in 21, and at another pantactic set it has
neither right nor lefthand tangents.
502. Weierstrass" Function. For a long time mathematicians
thought that a continuous function of x must have a derivative, at
least after removing certain points. The examples just given
show that these exceptional points may he pantactic. Weierstrass
called attention to a continuous function which has at no point a
differential coefficient. This celebrated function is defined by the
series
F (x) = 2 a n cos b n irx = cos irx + a cos birx + a 2 cos WTTX j (1
where < a < 1 ; J is an odd integer so chosen that
a6>l + *7r. (2
DERIVATKS 499
The series F converges absolutely and uniformly in any interval
21, since . _ ln , ^ _
' 0 n cos b n 7Tx I < a n .
Hence F is a continuous function in 21. Let us now consider
the series obtained by differentiating 1) term wise,
If ab < 1, this series also converges absolutely and uniformly,
and
by 155, 1. In this case the function has a finite derivative in 21.
Let us suppose, however, that the condition 2) holds. We have
^?= Q = V {cos ft"7r(2; + A)  cos ftTT^S = Q m + ~Q m . (8
A# Y ^
Now w _ n
~r J cos n 7r(x + i) cos
A
sin b n 7rudu.
Since
I /*^h^ > I /^+/t
I sin b n 7rudu, < I

Consider now _ < n
Q m = 2  f Jcos b n Tr(x 4 A) cos
m *
Up to the present we have taken h arbitrary. Let us now
take it as follows ; the reason for this choice will be evident in a
moment.
Let
where i m is the nearest integer to b m x. Thus
*<&.<*.
Then " >
500 DERIVATES, EXTREMES, VARIATION
We choose h so that
^m = m 4 hb m is 1, at pleasure.
Then _ *.
h = ^~r~^ === 0, as ra == oo ;
moreover sgn A  sgn r, m , and , ra  f.  < .
This established, we note that
cos b n jr(x f A) = cos b n ~ m 7r  b m (x f h) = cos # n ~ w (* m h ?/ m )7r
= cos (t w 47/ m )7r , since b is odd
= ( l) l w+i , since r; m is odd.
S COS 6 W 7T^ = COS 6 n ~ m (t m h f m)7T
Thus _ oo an
m 'I
W ' 16re ( \\ m +\
Now each } } > and in particular the first is > 0. Thus
sgn Q m = sgn ^ = sgn e m ij m ,
Thus if 2) holds,  Q m \ > \ Q m . Hence from 3),
sgn Q = sgn Q m = sgn e m r) m ,
and
Let now m = QO . Since i/ m = 1 at pleasure, we can make
Q = f QO, or to QO , or oscillate between GO, without becoming
definitely infinite. Thus F (x) has at no point a finite or infinite
differential coefficient. This does not say that the graph of F does
not have tangents; but when they exist, they must be cuspidal tangents.
DERIVATES 501
503. 1. VolterrcCs Function.
In the interval 21 = (0, 1), let <Q = \<r)\ be a Harnack set of
measure 0<A<1. Let A = {S n j be the associate set of black
intervals. In each of the intervals S n = (a < y8), we define an
auxiliary function f n as follows :
/nO) = O ) 2 sin , in (a*, 7), (1
where 7 is the largest value of x corresponding to a maximum of
the function on the right of 1), such that 7 lies to the left of the
middle point /i of S n . If the value of f n (x) at 7 is #, we now
* /" x
= # > HI (7, /i).
Filially f n (a)= 0. This defines / (#) for one half of the inter
val 8 n . We define f n (x) for the other half of S n by saying that if
x<x r are two points of S n at equal distances from the middle
point* then /(*)=/(*')
With Volterra we now define a f unction f(x) in 31 as follows:
f(x) = f n (x) , inS n , n = l, 2, .
= , in .
Obviously /(#) is continuous in 21.
At a point x of 1 not in &,f(jz) behaves as
2 a; sin  cos,
x x
as is seen from 1). Thus as x converges in 8 n toward one of its
end points a, /?, we see that f f (x) oscillates infinitely often be
tween limits which = 1. Thus
R lim/'<= + 1,7? lira/ (a?) =  1 ;
*= a 1^.
similar limits exist for the points 0.
Let us now consider the differential coefficient at a point rj of
. We have
502 DERIVATES, EXTREMES, VARIATION
If r) f k is a point of ,/0? 4 &) = 0. If not, 77 f A lies in some
interval 8 m . Let x = e be the end point of 8 m nearest 1; + A.
Then
Thus/' (97)= 0. Hence Volterra's function /(a?) has a differen
tial coefficient at each point of 21 ; moreover f (x) is limited in 21.
Each point rj of is a point of discontinuity of /'(#), and
Disc/' (a?) > 2.
Hence /'(#) is not /2integrable, as
We have seen, in I, 549, that not every limited .Rintegrable
function has a primitive. Volterra's function illustrates con
versely the remarkable fact that Not every limited derivative is
Rintegrable.
2. It is easy to show, however, that The derivative of Volterra's
function is Lintegrable.
For let 21 A denote the points of 21 at which /'(a?) >X. Then
when X>l/w, w=l, 2, 21 A consists of an enumerable set of
intervals. Hence in this case 21 A is measurable. Hence 21 A , X>0,
is measurable. Now 21 , X>0, differs from the foregoing by add
ing the points n in each S n at which/' {x) = 0, and the points ^>.
But each $ n is enumerable, and hence a null set, and j is measur
able, as it is perfect. Thus 21 A , X>0, is measurable. In the
same way we see 21 A is measurable when X is negative. Thus 21 A
is measurable for any X, and hence iintegrable.
504. 1. We turn now to general considerations and begin by
considering the upper and lower limits of the sum, difference, prod
uct, and quotient of two functions at a point x = a.
Let us note first the following theorem :
Letf(x^ Xjn) be limited or not in 21 which has x = a as a limiting
point. Let&= Max /, < 6 = Min / in V** (a) . Then
lim/=lim<^5 , lim / = lim 4>$ .
=5 5=0 #=a fi=o
This follows at once from I, 338.
DERI VAXES 508
2. Letf(x l > x m ), g(^x l x m ) be limited or not in 21 which has
x = a as limiting point.
Let lim/= , lim cj = ft
lim/ = A , lim g = B
as x^= a. Then, these limits being finite,
a + < Hm (/ + //) < A + A (1
aB< lim (/  </) < ^  ft. (2
For in any FS*(),
Min / f Min g < Min (/ 4 //) < Max (/ + </) < Max/ + Max #.
Lulling 8 = 0, we gel 1 ).
Also in Fi*(),
Min/ Max // < Min (./' #) < Max(/ ^) < Max / Min^.
Let ling 8 = 0, we gel 2).
Imi ./// < ^1 J5. (3
(4
/ ,   f ^ A ,r
<hm f :< (5
 
a<0<A , ,^(^)>
^ l"^ f ^ A
st*
(6
The relations 3), 4), 5), 6) may be proved as in 2. For exam
ple, lo prove 5), we observe Ibal in
A,. /^AT f ^ Max/
Mm ^ < Max tL. < _ _y_ .
 . _^ _
Max// /y ^ Min
504 DERIVATES, EXTREMES, VARIATION
5. + /3< lim (/ + #)< + . (7
. (8
(9
. (10
If /(*)><> , </(*)> o,
a/3 < limfff < , (11
A^<\^nfg<AB. (12
V g(x) > k > 0,
f<l^. (13
<E/<. (14
6. If lim/ exists,
lim (/ h ^) = lim / f lim g> (15
(16
If \\rng exists,
lim(/50 = lmi/limflr,
(18
f(x) > 0, ^r(rr) > 0. Let lim # ri^.
lim fg = lim / lim g, (19
lim fg = lim / lim g. (20
g(x) > k > 0,
(21
iim //gr = lim //lim #. (22
505. The preceding results can be used to obtain relations be
tween the derivates of the sum, difference, product, and quotient
of two functions as in I, 373 seq.
DERIVATES 505
1. Let w (V) = u (x) f v (V) . Then
A^y__Aw Av ,^
Ax A# Az
Thus from 504, 1), we get the theorem :
Uu f + v'U<Uw'< Uu f + Uv'. (2
// u has a unilateral derivative Uu',
Uw f = Uu r + Uv', (3
Uw' = ZTw' + Uv'. (4
We get 3), 4) from 1), using 504, 15), 16).
2. In the interval 31, M, v are continuous, u is monotone increasing,
v is > 0, awe? v' exists. Then, if w = MV,
Uiv' = uv' + vUu', (1
Uw r =uv r +vUu r . (2
For from A Ai , A
 = (WH A?/,) f w ,
"
we have ' = ^ +
which gives 1). Similarly we establish 2).
506. 1. We show now how we may generalize the Law of the
Mean, I, 393.
Let f(x) oe continuous in 21 =(#<&). Let m, M be the mini
mum and maximum of one of the four derivates off in SI. Then for
.
/3 a
To fix the ideas let us take Rf'(x) as our derivate. Suppose
now there exists a pair of points < /3 in 31, such that
506 DERIVATKS, EXTREMES, VARIATION
We introduce the auxiliary function
$(aO=/(aOCaf+e)*, (2
where 0<c<e = c+8.
Then Q8 ) QO = /()/() (Jfg)= s.
Henue
Consider now the equation
It is satisfied for # = a. If it is satisfied for any other x in the
interval (a/3), there is a last point, say x = 7, where it is satisfied,
by 458, 3.
Thus for x > 7, <O)i 8 >4>().
Hence Jty'(7)>0. (3
Now from 2) we have
Hence M is not the maximum of Rf'(x) in 21. Similarly the
other half of 1) is established. The case that m or M is infinite
is obviously true.
2. Let f(x) be defined over 91 = (a < 5). JW rtj < # 2 < < a n fe
m 31. Let m and M denote the minimum and maximum of the dif
ference quotients
AO&^ag) , A(a 2 , a 8 ) , ... A(a n _ r a n ).
Then
For let us first take three points < /3 < 7 in 21. We have iden
tically Q Q
7).
Now the coefficients of A on the right lie between and 1.
Hence 1) is true in this case. The general case is now obvious.
DERIVATES 507
507. 1. Let f(x) be continuous in 21 = (a < 6). The four deri
vates off have the same extremes in 21.
To fix the ideas let
Min L =5 m , Min R = /K, in 21.
We wish to show that m = /*. To this end we
For there exists an a in 21, such that
L^oC) < ra  e.
There exists therefore a /3 < a in 21, such that
,
a p
Now by 506, 1,
fjb = Min R<q.
Hence
as >0 is small at pleasure.
JF0 sAow wo?u ^Aa^ ^ /0
7M < /I. (2
For there exists an a in 21, such that
jR() < fJL + .
There exists therefore a /9 > a in 21, such that
C6/3
Thus by 506, 1,
w = Miu L<q.
Hence as before ra</*. From 1), 2) we have m = /it.
2. In 499, we emphasized the fact that the lefthand derivates
are not defined at the lefthand end point of an interval, and the
righthand derivates at the righthand end point of an interval
for which we are considering the values of a function. The fol
lowing example shows that our theorems may be at fault if this
fact is overlooked.
508 DERIVATES, EXTREMES, VARIATION
Example. Let/ (x) = j x \.
If we restrict x to lie in 21 = (0, 1), the four derivates = 1 when
they are defined. Thus the theorem 1 holds in this case. If,
however, we regarded the lefthand derivates as defined at x = 0,
and to have the value
=  1,
as they would have if we considered values of / to the left of 31,
the theorem 1 would no longer be true,
For then Min =  1 , Min ]B = + 1,
and the four derivates do not have the same minimum in 21.
3. Let f '(#) be continuous about the point x= c. If one of its
four derivates is continuous at x = c, all the derivates defined at this
point are continuous, and all are equal.
For their extremes in any Fi(<?) are the same. If now R is
continuous at x = <?,
R(c)  e < R(x) < JRO) + e,
for any x in some V^(ci).
4. Let f (x) be continuous about the point x c. If one of its
four derivates is continuous at x = c, the derivative exists at this
point.
This follows at once from 3.
Remark. We must guard against supposing that the derivative
is continuous at x = <?, or even exists in the vicinity of this point.
Example. Let F(x) be as in 501, Ex. 1. Let
21= (0,1) and g= ().
I n)
Let
Then
tx) = 2 xF(x) f
LH'(x) = 2 xF(x) +
Obviously both Rff' and Lff f are continuous at x = and
J5F(0) = 0. But H 1 does not exist at the points of (, and hence
DERIVATES 509
does not exist in any vicinity (0, 8) of the origin, however small
S > is taken.
9
5. If one of the derivates of the continuous function f(x) is
continuous in an interval 21, the derivative /'(#) exists, and is con
tinuous in 21.
This follows from 3.
6. If one of the four derivates of the continuous function f (x) is
= in an interval 21, f(x) = const in 21.
This follows from 3.
508. 1. If one of the derivates of the continuous function f(x) is
> in 21 = (a < J), f(x) is monotone increasing in 21.
For then m = Min Ef f > 0, in (a < x). Thus by 506, i,
2. If one of the derivates of the continuous function f(x) is _>_
in 21, f(x) is monotone decreasing.
3. If one of the derivates of the continuous function f(x) is >
in 21, without being constantly in any little interval of 21, /(a?) is
an increasing function in 21. Similarly f is a decreasing function
in 21, if one of the derivates is <C 0, without being constantly in any
little interval of 21.
The proof is analogous to I, 403.
509. 1. Letf(x) be continuous in the interval 21, and have a deriv
ative, finite or infinite, within 21. Then the points where the deriva
tive is finite form a pantactic set in 21.
For let a < ft be two points of 21. Then by the Law of the
Mean,
As the right side has a definite value, the left side must have.
Thus in any interval (a, ft) in 21, there is a point 7 where the
differential coefficient is finite.
510 DERI VAT ES, EXTREMES, VARIATION
2. Let f(x) be continuous in the interval 2l = (#<6). Then
Uf* (#) cannot be constantly + GO, or constantly oo in 21.
For consider
a
which is continuous, and vanishes for x = a, x = b. We observe
that $(V) differs from /(V) only by a linear function. If now
t/f'(V)= + ac constantly, obviously 7<'(V)= h oo also. Thus </>
is a uni variant function in 21. This is not possible, since </> has
the same value at a and b.
8. Let f(x) be continuous in 2J[ = (a< b), <mc? Aowtf a derivative,
finite or infinite, in 2(=(a*, J). TVierc
Min/ (a) < Rf(a)< Max/ (a;) , m 91.
For the Law of the Mean holds, hence
Letting now A = 0, we get the theorem.
Remark. This theorem answers the question : Can a continu
ous curve have a vertical tangent at a point x a, if the deriva
tives remain < M in V*(a) ? The answer is, No.
4. Let f(x) be continuous in 21 = (a < J), and have a derivative,
finite or infinite, in 21* = (a*, b). Iff'(a) exists, finite or infinite,
there exists a sequence j > 2 > == a in 21, such that
 , , . (2
/I
Let now A range over Aj > A 2 > = 0. If we set w n = ,, , the
relation 1) follows at once from 2), since 1 f f (a) exists by
hypothesis.
510. 1. A righthand derivate of a continuous function f(x)
cannot have a discontinuity of the 1 kind on the riyht. A similar
statement holds for the other derivates.
DEB1VATES /ill
For let R(x) be one of the righthand derivates. It it has a
discontinuity of the 1 kind on the right at or = a, there exists a
number I such that
I e <_ R(x) <_ I f c , in some ( " < # t 8).
Then by 500, i,
Hence R(a)= I,
and R(V) is continuous on the right at # = a, which is contrary
to hypothesis.
2. It can, however, have a discontinuity of the 1 kind on the
left, as is shown by the following
Example. Let/(r)==  r = + Va? , in 31 = ( 1, 1).
Here R(x)=+\ , for ?.> in 31
= I , for x < 0.
Thus at x = 0, R is continuous on the right, but has a discon
tinuity of the 1 kind on the left.
8. Let f(x) be continuous in ?I = (a, ?>), tt/tc? /m^' a derivative,
finite or infinite, in 31* =(<*, />*). Then tlie discontinuities off'(x)
in 31, if any exist, must be of the second kind.
This follows from 1.
^O in 31 = (0, 1)
Example.
/O) =
a? 2 sin  , f(
x
=
, for x =
Then
. 1
0*0 =2 2; sin cos ,
a: #
= , x = 0.
The discontinuity of ./ v (.r) at a; = 0, is in fact of the 2 kind.
4. Let f(x) be continuous in 3t = (a<6), except at x a, which
is a point of discontinuity of the 2 kind. Let f 1 (x) exist ^ finite or
infinite, in (a*, 6). Then x = a is a point of infinite discontinuity
<>//'(*)
512 DERIVATES, EXTREMES, VARIATION
For if
there exists a sequence of points 1 > 2 >... =a, such that
/(a n )=_=jt?; and another sequence ft l >/3 2 > =, such that
== ? We may suppose
>&. , ora n </3 n , rc=l, 2, 
Then the Law of the Mean gives
where 7 n lies between n , /3 n . Now the numerator = JP ^, while
the denominator = 0. Hence Q n = f oo , or oo , as we choose.
5. Let f (x) have a finite unilateral differential coefficient U at
each point of the interval ?l. Then U is at most pointwise discon
tinuous in 21.
For by 474, 3, 7is a function of class 1. Hence, by 486, l, it is
at most pointwise discontinuous in 21.
511. Let f (x) be continuous in the interval (a < J). Let R(x)
denote one of the righthand derivates of f(x). If R is not con
tinuous on the right at a, then
ujhere _____
I = R lim R(x) , m = R lim R(x) , x = a.
To fix the ideas let R be the upper righthand derivate. Let us
suppose that a = Rf f (a) were >ra. Let us choose 77, and c such
that .
wi f V < c < (2
We introduce the auxiliary function
Now if > is sufficiently small,
72/'O)< + >? , for any x in 21* = (a*, a + S).
DERI VAXES 513
Thus 2), 3), show that
R$(X)><T , <r>0.
Hence </>(#) is an increasing function in 31*. But, on the other
since a > m. Hence
Mfi (a) = c  fif' (a) = c  a < 0.
Hence < is a decreasing function at x = a. This is impossible
since <f> is continuous at a. Thus <.m.
Similarly we may show that <..
512, 1. Let f(x) be continuous in ?I = (a < 5), <mc? have a
derivative, finite or infinite. Ifa=f'(a), /9 =/'(&), then f (x)
takes on all values between a, /3, as x ranges over 81.
For let a < 7 < /3, and let
) = /<JLJI/M , A>0 .
We can take h so small that
Q(a, A)<7 , and
N W <?(i, *)
Hence (*,*)> 7
If now we fix A, ^ (#, A) is a continuous function of #. As
is < 7, for x =s a, and > 7, for # = 6 A, it takes on the value 7
for some x, say for x = , between a, 6 A. Thus
<?(?, A) = 7
But by the Law of the Mean,
(*)=/' Oi),
Thus/' (x) = 7, at x == 77 in 31.
2. ie /(#) 5e continuous in the interval 31, and admit a deriva
tive, finite or infinite. If f'(x) = in 31, except possibly at an
enumerable set @, then f = aZs0 in (
514 DER1VATES, EXTREMES, VARIATION
For if /'() = 0, and /'() = b & 0, then f'(x) ranges over all
values in (0, b), as x passes from a to ft. But this set of values
has the cardinal number c. Hence there is a set of values in
(, /3) whose cardinal number is c, where /'(#) =jt 0. This is
contrary to the hypothesis.
8. Let /(#), #(#) ^ continuous and have derivatives, finite or
infinite, in the interval 81. // in ?{ //ere i an /or which
/3 /"or
/ a 7 /0r which
/' (7) =/(?),
*(*)=/<*>*(*>
Aas a derivative, finite or infinite.
For by hypothesis
'()> , S'(/3)<0.
Hence by 1 there is a point where 8' = 0.
513. 1. If one of the four derivates of the continuous function
f(x) is limited in the interval 81, all four are, and they have the
same upper and lower Rintegrals.
The first part of the theorem is obvious from 507, 1. Let us
effect a division of 81 of norm d. Then
R = li
d=Q
lim ^Mdi , Mi = Max R, in d,.
Hut the maximum of the three other derivates in d L is also M^ by
f)07, l. Hence the last part of the theorem.
2. Let f(x) be continuous and have a limited unilateral derivate
as R in 81 = (a < 6). Then
For let a < aj < a 2 < ... < ft determine a division of 31, of norm d.
DERIVATES 515
Then by 506, 1,
Min R < ^lAz/^n). < Max R,
tfm+l  m
in the interval (a m , a mJrl ) = d m .
Hence
2d m Min R </6) /a < 2<k Max 5.
Letting d == 0, we get 1).
3. If f(x) is continuous, and Uf is limited and Rintegrable in
, then
514. 1. Letf(x) be limited in 21 = (a < 5), and
F(x)= Cfdx , a<a;<6.
\/a
Whew _ _
 ^ 71 U\\mf<UF(u)<U\imf, (1
a^=u *="
or any u within 31.
To fix the ideas let us take a righthand derivate &tz = u. Then
A Min/ < f/ds < h Max/ , in (w*, u + A), A > 0.
_ M
Thus
Letting A = 0, we get
R lim / < RF' (u) < R lim /,
*= ^ M
which is 1) for this case.
2. i# /(^) ^^ limited in the interval 21 = (a < 6). If f(x + 0)
JS derivative I fdx=f(x f 0) ;
^a
and iff(x 0) exists, a<x<b
L derivative ( fdx f(x 0).
Ja
5ir> DKRIVATES, KXTRKMKS, VARIATION
3. Let f(x) be limited and Rintegrable in 2l=(a<S). The
points where
F<= Cfdx , a<x<b
v/O
does not have a differential coefficient in 31 1 form
F r J*(:r)=/Cr) by 1,537, 1,
when / is continuous at #. But by 462, 6, the points where / is
not continuous form a null set.
515. In I, 400, we proved the theorem :
Let /(#) be continuous in 21 = (a < ft), and let its derivative
= within 21. Then /is a constant in 31. This theorem we have
extended in 507, 6, to a derivate of f(x). It can be extended still
farther as follows :
1. (L. Scheefer). ///(#) is continuous in 31 = (#</>), and if
one of its derivates = in 21 except possibly at thr points of an
enumerable set (, then f = constant in 21.
If /is a constant, the theorem is of course true. We show that
the contrary case leads to an absurdity, by showing that Card (5
would = c, the cardinal number of an interval.
For if / is not a constant, there is a point c in 21 where
jp=/(6') /(a) is =^0. To fix the ideas let jt?>0; also let us
suppose the given derivate is R = Rf'(x).
Let g(z,f)=f(x)f(a)t(xa) , * > 0.
Obviously  g \ is the distance/ is above or below the secant line,
Thus in particular for any ,
g(a, 0=0 ,
Let q > be an arbitrary but fixed number < p. Then
#0, t)  q = p  q  t(v  a )
\it<T, where
r.
a
DER1VATES 517
Hence a* > ,
for any t in the interval Z = (T, 7), < T < J 7 . We note that
Card X = c.
Since for any in I, #(a, = 0, and #O, > 9, let x = e ( be
the maximum of the points < c where g(x, ) = ? Then e < c,
and for any h such that e f 7t lies in (e, 6'),
Thus for any t in I, 0< lies in @. As ranges over , let ^
range over @ x < S. To each point e of (S^ corresponds but one
point of Z. For
O^Cu, 0^,0 = (*O().
Hence * = f , as >a.
Thus Card I = Card (gj < Card @,
which is absurd.
2. Let f (x) be continuous in < $t = (a<b). Let S denote the
points of 31 wAere o?i^ of the derivates has one sign. If S exists,
Card (5 = c, A# cardinal number of the continuum.
The proof is entirely similar to that in 1. For let c be a point
of 6. Then there exists a d > c such that
We now introduce the function
g(z, Q =/GO/(<0 '(**) * *>0,
and reason on this as we did on the corresponding g in 1, using
here the interval (c, d) instead of (a, ). We get
Card g 1= =Card = c.
3. Letf(x), g(x) be continuous in the interval 31. Let a pair of
corresponding derivates as Rf, Eg 1 be finite and equal, except pos
sibly at an enumerable set g. Then f=g + C, in 31, where C is a
constant.
518 DERI VAXES, EXTREMES, VALUATION
Then in
But if .72<// < at one point in 31, it is < at a set of points 95
whose cardinal number is c. But 93 lies in @. Hence R<f> is
never < 0, in 21. The same holds for i/r. Hence, by 508, <f> and
*fr are both monotone increasing. This is impossible unless
<f> = a constant.
516. The preceding theorem states that the continuous function
/(#) in the interval 21 is known in 21, aside from a constant, when
f (x) is finite and known in 21, aside from an enumerable set.
Thus f(x) is known in 21 when f 1 is finite and known at each
irrational point of 21.
This is not the case when/' is finite and known at each rational
point only in SI.
For the rational points in 21 being enumerable, let them be
r n r v r 3 0
Let J=* 1 + J 2 +J 8 +
be a positive term series whose sum I is < 21. Let us place r^
within an interval S l of length < ? x . Let r t be the first number
in 1) not in S r Let us place it within a nonoverlapping interval
S 2 of length < ? 2 , etc.
We now define a function /(a?) in 21 such that the value of /at
any x is the length of all the intervals and part of an interval
lying to the left of x. Obviously f(x) is a continuous function of
x in 21. At each rational point /' (x) = 1. But f(x) is not de
termined aside from a constant. For 28 n < I. Therefore when
I is small enough we may vary the position and lengths of the
Sintervals, so that the resulting /'s do not differ from each other
only by a constant,
517. 1. Let f(x) be continuous in 21 = (a < J) and have a finite
derivate, say Ef\ at each point of 21. Let S denote the points of 21
where R has one sign, say > 0. If S exists, it cannot be a null set.
DERI VAXES 519
For let c be a point of g, then there exists a point d > c such
that
Let S n denote the points of where
nl<Rf'<n. (2
Then g = 6 X 4 S 2 4 Let < q < p. We take the positive
constants ft, q% such that
If now & is a null set, each @ m is also. Hence the points of 6 m
can be inclosed within a set of intervals S mn such that 2S mn < q m .
n
Let now q m (#) be the sum of the intervals and parts of intervals
^m, m n == 1> ^ which lie in the interval (a < #). Let
Obviously #(^) is a monotone increasing function, and
0<<?(aO<9 ( 3
Consider now
P(^)=/(^
We have at a point of 21 S,
Hence at such a point
EP f < Rf < 0.
But at a point x of (, /ZP' < also. For x must lie in some
S m , and hence within some S mn . Thus q m (x*) increases by at least
A# when x is increased to x 4 Aa?. Hence mq m (x), and thus
#(a; is increased at least wA#. Thus
Aa;
ThUS
' < Rf>  m < 0, by 2),
520 DERI VAXES, EXTREMES, VARIATION
since x lies in m . Thus RP' < at any point of 31. Thus P is
a monotone decreasing function in 2, by 508, 2. Hence
Hence > 0,
or using 1), 3)
jt>?<0,
which is not so, as p is > q.
2. (Lebesgue*) Let f(x)i g (&) be continuous in the interval 81,
#W6? have a pair of corresponding derivates as Hf 1 * Rg 1 which are
finite at each point of 21, and also equal, the equality holding except
possibly at a null set. Thenf(x) g(x) constant in 31.
The proof is entirely similar to that of 515, 3, the enumerable
set ( being here replaced by a null set. We then make use of 1.
518. Letf'(x) be continuous in some interval A = (u 8, u 4 8).
Letf"(x) exist, finite or infinite, in A, but he finite at the point x=u.
A=0
where
Let us first suppose that/"(M) = 0. We have for < h < rj <
_
"Al " A X
= ^ {/'(a/) /'(" )j , <*'< + /* , uh<x"<u
= [(a,') {/() + e'  ^"^{/"do I e"{J,
/fc
where e',  e /;  are < e/2 for T; sufficiently small.
Now x' u^* \x n ^l <^i
~Y~~" ~T~^
while /"(ti) = , by hypothesis.
Hence  <?/!<* , for
and 1 ) holds in this case.
MAXIMA AND MINIMA 521
Suppose now that f"(u) = a = 0. Let
y(. x ) =/O)  ?O) where f/(.r) =  ax 2 f fo + c.
Since ^"(w) = a , </"( = 0.
Thus we are in the preceding case, and lim Qg = 0.
But Q9=QfQq.
Hence lim Qf= a.
Maxima and Minima
519. 1. In I, 466 and 476, we have defined the terms /(V) as
a maximum or a minimum at a point. Let us extend these terms
as follows. Let/(^ .r m ) be defined over 91, and let x= a be an
inner point of 21.
We say f has a maximum at x = a if 1, / (a) / (#) > 0, for any
x in some V(a), and 2,/(#) J\x) >0 for some x in any F(a).
If the sign ^ can be replaced by > in 1, we will say f has a
proper maximum at a, when we wish to emphasize this fact; and
when > cannot be replaced by >, we will say / has an improper
maximum. A similar extension of the old definition holds for
the minimum. A common term for maximum and minimum is
extreme.
2. If f(x) is a constant in some segment 33, lying in the inter
val 21, 33 is called a segment of invariability, or a constant segment
of /in 21.
Example. Let/(#) be continuous in 21 = (0, 1*).
Let /1
x = a^a^a^ (I
be the expression of a point of 2l in the normal form in the dyadic
system. Let fc
J = 
be expressed in the triadic system, where n = a n , when a n = 0,
and =2 when a n = l. The points = jj form a Cantor set,
I, 272. Let j3 n j be the adjoint set of intervals. We associate
522 DERIVATES, EXTREMES, VARIATION
now the point 1) with the point 2), which we indicate as usual by
x~ f . We define now a function g(x) as follows :
#() =/O) > when x ~ .
This defines ^ for all the points of . In the interval 3n let 9
have a constant value. Obviously g is continuous, and has a
pantactic set of intervals in each of which g is constant.
3. We have given criteria for maxima and minima in I, 468
seq., to which we may add the following :
Let f(x) be continuous in (a , a + ). If Rf 1 '(#) > and
0, finite or infinite, f (x^) has a minimum at x = a.
)< and Lf'(a) > 0, finite or infinite, f(x) has a maxi
mum at x = a.
For on the 1 hypothesis, let us take a such that J2a
Then there exists a 8' >0 such that
h
Hence /(a + *)>/() , a + A in (a, a + 8')
Similarly if /S is chosen so that L 4 /? < 0, there exists a S" > 0,
such that /.. 7x /rx >.
h
Hence /(aA)>/(a) , a + A hi (a  8", a*).
520. Example 1. Let/(a;) oscillate between the #axis and the
two lines y = x and y = x, similar to
In any interval about the origin, y oscillates infinitely often, hav
ing an infinite number of proper maxima and minima. At the
point # = 0,/has an improper minimum.
Example 2. Let us take two parabolas P l , P 2 defined by y = # 2 ,
y = 2 # 2 . Through the points x= , ^ let us erect ordi
nates, and join the points of intersection with P x , P 2 , alternately
by straight lines, getting a broken line oscillating between the
MAXIMA AND MINIMA 523
parabolas P l , P 2 . The resulting graph defines a continuous func
tion f(x) which has proper extremes at the points @ = j  1
i n )
However, unlike Ex. 1, the limit point x of these extremes is
also a point at which f(x) has a proper extreme.
Example 3. Let jSj be a set of intervals which determine a
Harnack set lying in 21 = (0, 1). Over each interval B = (a, /3)
belonging to the n ih stage, let us erect a curve, like a segment of
a sine curve, of height h n = 0, as n == oo, and having horizontal
tangents at a, /3, and at 7, the middle point of the interval 8. At
the points \%\ of 21 not in any interval S, let/" (x) = 0. The func
tion/ is now defined in 21 and is obviously continuous. At the
points \y\if has a proper maximum ; at points of the type a, /3,
f,/has an improper minimum. These latter points form the set
whose cardinal number is c. The function is increasing in each
interval (a, 7), and decreasing in each (7, /3). It oscillates in
finitely often in the vicinity of any point of .
We note that while the points where / has a proper extreme
form an enumerable set, the points of improper extreme may form
a set whose cardinal number is c.
Example 4. We use the same set of intervals jSj but change
the curve over S, so that it has a constant segment 77 = (X, /A) in its
middle portion. As before /=0, at the points not in the
intervals 8.
The f unction /(V) has now no proper extremes. At the points
of ^p, / has an improper minimum ; at the points of the type X, ^, it
has an improper maximum.
Example 5. Weierstrass" Function. Let S denote the points in
an interval 21 of the type
x = ~ , r, s, positive integers.
For such an x we have, using the notation of 502,
b m x = t m 4 m = b m ~'r.
Hence  m = , for m>_8.
Thus e m = ( 1> +1 = ( l) r+1 .
524 DER1VATES, EXTREMES, VARIATION
Hence sgn = sgn Q = sgu e m y m = sgn (  1 yh
if r is even, and reversed if r is odd. Thus at the points @, the
curve has a vertical cusp. By 519, 3, F has a maximum at the
points S, when r is odd, and a minimum when r is even. The
points ( are pan tactic in 31.
Weierstrass' function has no eonstant segment fi, for then
f'(x) = in S. Hut F 1 does not exist at any point.
521. 1. Let f (JL\ jc m ) be continuous in the limited <>r unlimited
set 21. Let ($ denote the points of ?{ where f has a proper extreme.
Then ( is enumerable.
Let us first suppose that 21 is limited. Let S > be a fixed
positive number. There can be but a finite number of points in
31 such that
For if there were an infinity of such points, let ft be a limiting
point and 77 <  8. Then in V^(ff) there exist points ', " such
that Fi(cc'), V s (a rf ) overlap. Thus in one case
/(')>/(">,
and in the other
/(')</(">,
which contradicts the first.
Let now Sj > 8 3 > =0. There are but a finite number of
points a for which 1) holds for 8 = Sj, only a finite number for
S = S 2 , etc. Hence ( is enumerable. The case that 21 is unlim
ited follows now easily.
2. We have seen that Weierstrass' function lias a pantactic set
of proper extremes. However, according to 1, they must be
enumerable. In Ex. 3, the function has a minimum at each point
of the nonenumerable set ; but these minima are improper. On
the other hand, the function has a proper maximum at the points
}7J, but these form an enumerable set.
MAXIMA AND MINIMA 525
522. 1. Let f(x) be continuous in the interval 21. Let f have a
proper maximum at x = a, and .*; = /3 in 21. Then there is a point 7
between a, ft where f has a minimum, which need not however be a
proper minimum.
For say a < 13. In the vicinity of a, f(x) is </() ; also in
the vicinity of /3, /(#) is </(/?). Thus there are points S3 in
(a, /3) where /is < either /() or/(/8). Let /A be the minimum
of the values of /(#), as # ranges over S3. There is a least value
of x in (, /3) for which /(V) = /x. We may take this as the
point in question. Obviously 7 is neither nor /3.
2. That at the point 7, / docs not need to have a proper mini
mum is illustrated by Exs. 1, or 3.
3. In 21 = (#, 6) /<' /'(.r) esist, finite or infinite. The points
within 2( fl which f 1m* an extreme proper or improper, lie among
the zeros off'(x).
This follows from the proof used in I, 408, 2, if we replace there
< 0, by <: 0, and > 0, by > 0.
4. Let /'OO be continuous in the interval 21, and let f(x) have
no constant segments in 21. The points (5 of 21 where f has an ex
treme, form an apantactic set in 21. Let denote the zeros of f (x)
in ty. If 33 = Jb n ( is the border set of intervals lying in 21 corre
sponding to S,f(%) is univariant in each b n .
For by 3, the points (5 lie in $ As f\x) is continuous, is
complete and determines the border set 33. Within each b n ,
/'(#) lias one sign. Hence /(a;) is univariant in b n .
5. Letf(x) be a continuous function having no constant segment
in the interval 21. If the points ( where f has an extreme form a
pantarfic set in 21, then the points 53 where /'(#) does not exist or is
discontinuous, form also a pantactic set in 21.
For if 93 is not pantactic in 21, there is an interval & in 21
containing no point of 33. Thus /'(#) is continuous in @. But
the points of (S in S form an apantactic set in S by 4. This,
however, contradicts our hypothesis.
Example. Weierstrass' function satisfies the condition of the
theorem 5. Hence the points where F f (x) does not exist or is
526 DERIVATES, EXTREMES, VARIATION
discontinuous form a pantactic set. This is indeed true, since
F' exists at no point.
6. Let f(x) be continuous and have no constant segment in the
interval 21. Let f'(x) exist, finite or infinite. The points where
/'(X) i^ finite and is = form a pantactic set in 21.
For let a < & be any two points in 21. If /() =/(y8), there is
a point a < 7 < /3 such that /() ^/(Y), since / has no constant
segment in 21. Then the Law of the Mean gives
ay
Thus in the arbitrary interval (a, /3) there is a point f, where
f 1 (x) exists and is = 0.
7. Let f \x) be continuous in the interval 21. Then any interval
S3 in 21 which is not a constant segment contains a segment @ in which
f is univariant.
For since f is not constant in S3, there are two points a, b in S3
at which f has different values. Then by the Law of the Mean
/()/(*)=( *)AO . eill 
Hence f f (c) = 0. As f r (x) is continuous, it keeps its sign in
some interval (c 8, c + ), and/ is therefore univariant.
523. Letf(x) be continuous in the interval 21, and have in any in
terval in 21 a constant segment or a point at which f has an extreme.
If f(x) exists, finite or infinite, it is discontinuous infinitely often in
any interval in 21, not a constant segment. At a point of continuity
of the derivative, /' (#) = 0.
For if f(x) were continuous in an interval S3, not a constant
segment, / would be univariant in some interval S:<S3, by 522, 7.
But this contradicts the hypothesis, which requires that any inter
val as has a constant segment. Hence /'(#) is discontinuous
in any interval, however small.
Let now x = c be a point of continuity. Then if c lies in a con
stant segment, /'(V) = obviously. If not, there is a sequence of
points e 19 e% = e such that /(#) ^ as & n extreme at e n . But then
/'(e n )=0, by 522,3. As f(x) is continuous at x = c, /'(c?) =
also.
MAXIMA AND MINIMA 527
524. (Kttnig.) Letf(x) be continuous in 31 and have a pantactic
set of cuspidal points . Then for any interval 93 of 21, there exists
a /3 such that f(x) = /3 at an infinite set of points in 33. Moreover,
there is a pantactic set of points \%\ in 33, such that k being taken at
For among the points & there is an infinite pantactic set c of
proper maxima, or of proper minima. To fix the ideas, suppose
the former. Let x = c be one of these points within 53. Then
there exists an interval 6:<33, containing (?, such that
Let p. = Min/(V), in b
Then there is a point x where / takes on this minimum value.
The point c divides the interval b into two intervals. Let I be
that one of these intervals which contains #, the other interval we
denote by ttl. Within tit let us take a point c l of c. Then in I
there is a point c[ such that
The point c 1 determines an interval b x , just as c determined b.
Obviously bjfCtn, and bj falls into two segments t x , ttt 1 as before
b did. Within m t we take a point of c. Then in I there is a
point c^, and in Ij a point c%, such that
In this way we may continue indefinitely. Let
be the points obtained in this way which fall in (. Let c' be a
limit point of this set. Let
/" /*" /" ...
c l 5 C 2 C 3
be the points obtained above which fall in lj, and let c 11 be a limit
point of this set. Continuing in this way we get a sequence of
limiting points c , ^ c n ^ c m ... ^2
lying respectively in I, I L , I a
f>28 DEKIVATES, EXTREMES, VARIATION
Since f is continuous,
/(O=/('")=/<y")= (3
Thus if we set /(V)= /9 we see that f(x) takes on the value y8 at
the infinite set of points 2), which lie in 33.
Let 7j, 7 2 ... be a set of points in 2) which = 7.
Then /Cl) /(_7l) == /(7)"/(7t) = . . . = o. ( 4
7 ~ 7i 7 ~ 7 2
Thus if f'(x) exists at # = 7, the equations 3) show that ^(7)
= 0. If/' does not exist at 7, they show that
/' < < /' , at x = 7
Let now k be taken at pleasure. Then
g(x)=f(x)kx
is constituted as/, and
/<>)=/<>)*.
This gives 1).
525. 1. LineoOwillatinf/ Functions. The oscillations of a con
tinuous function fall into two widely different classes, accord
ing as f(x) becomes monotone on adding ii linear function
l(x)=*ax + b, or does not.
The former are called lineooscillating functions. A continu
ous function which does not oscillate in 21, or if it does is lineo
oscillating, we say is at most a lineooscillating function.
Example 1. Let */ \ js \
r J {x) = sin x , l(x) = x.
If we set jf f ^ , 7 . .
y =/()+()
and plot the graph, we see at once that y is an increasing function.
At the point # = 77, the slope of the tangent to /(#)= sin# is
greatest negatively, i.e. sin # is decreasing here fastest. But the
angle that the tangent to sin x makes at this point is 45, while
the slope of the line l(x) is constantly 45. Thus at x = TT, y has
a point of inflection with horizontal tangent.
If we take l(x) = ax, a > 1, y is an increasing function, increas
ing still faster than before.
MAXIMA AND MINIMA f>29
All this can be verified by analysis. For setting
y = sin x f ax i # > 1,
and
Thus y is a lineooscillating function in any interval.
Example 2. /(#) = :r 2 sin  , xJ=Q
x
= , r=0.
J(#) = ax + 5 , y =/O) + ZO).
Then
u' = 2 a; sin cosfa ,
XX
a , x= 0.
Hence, if a > 1 h 2 TT, y is an increasing function in 21 = ( TT, TT).
The function /" oscillates infinitely often in 21, but is a lineooscil
lating function.
Example 8. f(x)
=
Here 11 1
^'=sin^  cos  4 a
a; ^ ^
For x=Q, y f does not exist, finitely or infinitely.
Obviously, however great a is taken, y has an infinity of oscilla
tions in any interval about x= 0. Hence/' is not a lineooscillat
ing function in such an interval.
2. If one of the four derivates of the continuous function f(x) is
limited in, the interval 21, /(>) is at most lineooscillating in 21.
For say Rf >  a in ?(. L t < < /3,
and
530 DERIVATES, EXTREMES, VARIATION
Then
/<=/3 +/'(*)> 0.
Hence g is monotone increasing by 508, 1.
3. Letf^x) be at most lineooscillating in the interval 21. If Uf f
does not exist finitely at a point x in 21, it is definitely infinite at the
point. Moreover, the sign of the GO is the same throughout 21.
For if / is monotone in 21, the theorem is obviously true. If
not ' let <,(*>=/(*)+*
be monotone. Then
Uf'=Ug'a,
and this case is reduced to the preceding.
Remark. This shows that no continuous function whose graph
has a vertical cusp can be lineooscillating. All its vertical tan
gents correspond to points of inflection, as in
Variation
526. 1. Letf(x) be continuous in the interval 21, and have limited
variation. Let D be a division of 21 of norm d. Then usin</ the no
tation 0/443,
\imVjJ=Vf , limP D f=Pf , UmN D f=Nf. (1
For there exists a division A such that
where for brevity we have dropped / after the symbol V. Let
now A divide 21 into v segments whose minimum length call X.
Let D be a division of 21 of norm d<d <\. Then not more
than one point of A, say a x , can lie in any interval as (a t , a t+1 ) of
D. Let E= D + A, the division obtained by superposing A on D.
Then fi denoting some integer < *>,
2j
K 1
VARIATION 531
If now c? is taken sufficiently small, Osc/ in any interval of D
is as small as we choose, say < . Then
But since E is got by superposing A on D,
Hence for any D of norm < rf ,
IF Fl <
which proves the first relation in 1. The other two follow at
once now from 443.
527. If f(x) is continuous and has limited variation in the in
terval 21 = O<6), then
POO , N(x) , FOO
are a?s0 continuous functions of x in 21.
Let us show that V(x) is continuous ; the rest of the theorem
follows at once by 443.
By 526, there exists a d , such that for any division D of norm
d<d Q , F(6) = F^(J) + e' , 0<e'<e/3.
Then a fortiori, for any z< b in 21,
In the division J), we may take # as one of the end points of an
interval, and x + h as the other end point. Then
F(a? + A) ^F^C*) + /(a; + A) /() + ^ , 0< 2 </3. (2
On the other hand, if d Q is taken sufficiently small,
/0*+A)/00<! , forO<A<S. (3
d
From 1), 2), 3) we have
V(x) < e , for any < A < 8. (4
532 DERIVATES, EXTREMES, VARIATION
But in the division J), x is the righthand end point of some in
terval as (x k,x). The same reasoning shows that
\V(xK)V(x)\<e , foranyO<<S. (5
From 4), 5) we see V(x) is continuous.
528. 1. If one of the derivates of the continuous function f (a?) is
numerically < Ttfm Ae interval ?l, i/if? variation Voff is < M^{.
For by definition
with respect to all divisions D= \d,\ of 81. Here
Now by 506, 1,
*
Hence
2. Letf(x) be limited and Rintegrahle in 31 = (a< 6).
F(x)= Cfdx , a< ^< 6
*^a
As limited variation in ?I.
For let D be a division of 31 into the intervals d t = (a t , l+1 ).
Then
< 2 l/ j da: < M 2
*/a t
Thus Max F,,  F <
and J 7 has limited variation.
529. 1. If f (x) has limited variation in the interval 21, the
points $ where Osc / > A, are finite in number.
For suppose they were not. Then however large Gr is taken,
we may take n so large that nk > Gr. There exists a division D
VARIATION /W3
of 31, such that there are at least n intervals, each containing a
point of & within it. Thus for the division D,
Thus the variation of f is large at pleasure, and therefore is not
limited.
2. If f has limited variation in the interval 21, its points of dis
continuity form an enumerable set.
This follows at once from 1.
530. 1. Let D 19 J9 2 be a sequence of superposed divisions, of
norms d n = 0, of the interval 21. Let L Dn be the sum of the oscilla
tions of f in the intervals of D n . If Max fl Dn is finite, f(x) has
limited variation in 21.
For suppose f does not have limited variation in 21. Then
there exists a sequence of divisions jE^, J? 2 such that if H^ n is
the sum of the oscillations of /in the intervals of IH n , then
n^< fi*,< = +QO. (i
Let us take v so large that no interval of D v contains more than
one interval of E n or at most parts of two E n intervals. Let
.F n = E n + D v . Then an interval 8 of D v is split up into at most
two intervals ', B n in F n . Let a>, a/, (*> n denote the oscillation of
f in 8, S', S" . Then the term co in D v goes over into
a> r +a>"<2co
in n^. Hence if Max fl = M,
which contradicts 1).
2. Let V Dn 2 /( t ) /(X+i) I ^ ie summation extended
over the intervals (a t , a l+1 ) of the division D n . If Max V Dn is
n
tinite with respect to a sequence of superposed divisions \D n \> we
cannot say that /has limited variation.
Example. For let/(V) = 0, at the rational points in the inter
val 21 = (0, 1), and = 1, at the irrational points. Let D n be
534 DERIVATES, EXTREMES, VARIATION
obtained by interpolating the points m f 7 in & Then /=
at the end points a t , a t+1 of the intervals of D n . Hence V Dn = 0.
On the other hand, f(x) has not limited variation in 21 as is
obvious.
531. Let F (x) = lim/(#, ), r finite or infinite, for x in the
t = T
interval 21. Let Var/(o:, t) <_M for each t near r.
Then F(x) has limited variation in 21.
To fix the ideas let r be finite. Let
Then for a division D of 21,
V D F<V D f+ V D g.
But
V D g = 2 \g(a m )  ff(a m+1 )  f
where (a m , a w+1 ) are the intervals of D.
But for some t = t' near r, each
where 8 is the number of intervals in the division D.
Thus
Hence
and J 7 has limited variation.
532. Let /(#), g(%) have limited variation in the interval 21, then
their sum, difference, and product have limited variation.
If also l#l>7>0 , in 21
thenf/g has limited variation.
Let us show, for example, that h=fg has limited variation.
Forlet Min/=m , Min</ = 7*
in the interval rf t .
Osc/=a> , Osc g = T
VARIATION 535
en / = m + o> , g = n + fir , in d t ,
< a < 1 , < < 1.
us
fg = mn h 7W/3r 4 r&aa> f a/3a>T.
Now
77M W T U ft) WT<g < 77171 + W T + U CO + ft)T.
Hence ^ = OscA<^2Jrm + a>  /i  +O>T{.
Rnf"
I m 1 1 I w I T <. some ff.
Thus ^ h < 4 JT2a> + 2 JT2r,
< some 6?,
and h has limited variation.
533. 1. Let us see what change will be introduced if we
replace the finite divisions D employed up to the present by
divisions JE, which divide the interval 21 = {a < 6) into an infinite
enumerable set of intervals (a t , # l+1 ).
and !F
for the class of finite or infinite enumerable divisions
Obviously TT> F;
hence if TFis finite, so is V.
We show thab if V is finite, so is W. For suppose W were
infinite. Then for any Q > 0, there exists a division E, and an
w, such that the sum of the first n terms in 1) is > (?, or
Ws, n >&. (2
Let now D be the finite division determined by the points a^ ,
#2 " a n+i which figure in 2).
Then v ^^
heuee ^=00, which is contrary to our hypothesis.
53(3
DERI VAXES, EXTREMES, VARIATION
We show now that V and W are equal, when finite. For let
E be so chosen that
W <W E <W.
Now
W E = W Et n + e' ,  e f < /2
if w is sufficiently large.
Let D correspond to the points a l a% in TF^ n . Then
and hence V + ' > W 4 ' W
Hence TT^<6.
We may therefore state the theorem :
2. Iff has limited variation in the interval 31 with respect to th<>
class of finite divisions D, it has with respect to the class of enumer
able divisions E, and conversely. Moreover
Max V D = Max V K .
534. Let us show that Weierstrass' function F, considered in
502, does not have limited variation in any interval 21 = (a < /3)
when ab > 1. Since F is periodic, we may suppose > 0. Let
Jm Jm frm
be the fractions of denominator b m which lie in 21.
These points effect a division D m of 31, and
> )
If I is the minimum of the terms F 3  under the 2 sign,
Now ft 1
Hence  yu f 2
^ ^ fz, ' /^
(1
(2
NOXLNTUITIONAL CURVKS 537
On the other hand, using the notation and results of 502,
and also
F(x+h) F(x)
> amftm /2 _ TT \
" \3 a6l/
Let us now take
Then _&+/ L
*T ' *
Hence from 3), p > /2 TT
^~ a U~^=
ThuB ^>^(^)(^ 2 ) ' byl),2).
As a < 1, and a6 > 1, we see that
V Dm = f QO, as m = QO .
Nonintuitional Curves
535. 1. Let /(#) be continuous in the interval 31. The graph
of/ is a continuous curve C. If / has only a finite number of os
cillations in 21, and has a tangent at each point, we would call an
ordinary or intuitional curve, it might even have a finite num
ber of angle points, i.e. points where the righthand tangent is
different from the lefthand one [cf. I, 366]. But if there were
an infinity of such points, or an infinity of points in the vicinity
of each of which / oscillates infinitely often, the curve grows less
and less clear to the intuition as these singularities increase in
number and complexity. Just where the dividing point lies be
tween curves whose peculiarities can be clearly seen by the intui
tion, and those which cannot, is hard to say. Probably different
persons would set this point at different places.
For example, one might ask : Is it possible for a continuous
curve to have tangents at a pantactic set of points, and no tangent
at another pantactic set? If one were asked to picture such a
curve to the imagination, it would probably prove an impossibility.
538 DERIVATES, EXTREMES, VARIATION
Yet such curves exist, as Ex. 3 in 501 shows. Such curves might
properly be called nonintuitional.
Again we might ask of our intuition : Is it possible for a con
tinuous curve to have a tangent at every point of an interval 21,
which moreover turns abruptly at a pantactic set of points ? Again
the answer would not be forthcoming. Such curves exist, how
ever, as was shown in Ex. 2 in 501.
We wish now to give other examples of nonintuitional curves.
Since their singularity depends on their derivatives or the nature
of their oscillations, they may be considered in this chapter.
Let us first show how to define curves, which, like Weierstrass'
curve, have a pantactic set of cusps. To effect this we will extend
the theorem of 500, 2, so as to allow g(x) to have a cusp at x = 0.
536. Let (S = \e n \ denote the rational points in the interval
21 = ( #, a). Let g(x) be continuous in 33 = ( 2 a, 2 #), and
= 0, at x = 0. Let 93* denote the interval $ after removing the
point x = 0. Let g have a derivative in 93*, such that
Then
A IP
is a continuous function in 21, and behaves at x = e m essentially
Ao;
as does at the origin.*
LJkX
To simplify matters, let us suppose that S does not contain the
origin. Having established this case, it is easy to dispose of the
general case. We begin by ordering the e n as in 233. Then
obviously if
e n = * , q > , p positive or negative,
we have ^
n > q.
Let
s
ife m =r,
s
>!>_!. (2
qs mn
* Cf. Dini, Theorie der Functioned etc., p. 192 sea. Leiozic. 1892.
NONINTUITIONAL CURVES 539
Let E(x) be the F series after deleting the m th term. Then
F (x) = a^(x ~ em ) + JE (x).
We show that E has a differential coefficient at x = e m , obtained
by differentiating E termwise. To this end we show that as h = 0,
(3
converges to ^ = 2^'CO , m*. (4
That is, we show
e>0 , rj > ,  D(A) (?<e , 0<A <rj. (5
Let us break up the sums 3), 4) which figure in 5), into three
parts r .5 oo
2 = 2 + 2 + 2. (6
1 1 r+l *+l
THUS J)fl t <7) r G t , + Z>,..fl [ r.. + A&. (7
< A + B + (7.
Since g'(e mn ) exists, the first term may be made as small as we
choose for an arbitrary but fixed r ; thus
A <\
Let us now turn to B. We have
s<\D n \+
provided g' (x) exists in the interval (e mn , e mn + A).
But by 2),
if 1
9<i. (8
2 W8
Thus by 1),
I #'Omn+ A')  ^ 2 a Mm a n a < JS/^r , jlfj a constant.
540 DEK1VATKS, EXTREMES, VARIATION
Hence a fortiori,
, ffl ^ , < ^ (9
Now the sum ,
converges if p > 0. Hence J5T P> 4 and 5,, may be made as small as
we choose, by taking p sufficiently large. Let us note that by 91,
ff P <~. (10
up*
Thus if p = Min (, /3),
for a sufficiently large r.
We consider finally O. We have
<  D. 
< Oi+ (7 2 4 <7 8 .
From 9) we see that
<7 3 <^5.<l,
for * sufficiently large. Since g(x) is continuous in 55,
\g(.x)\<N.
HeDC6 /* M ^ < 1 ^ ^ 1
a 
if ^.:JTI on using 10).
1*1
Taking * still larger if necessary, we can make
Thus
G v 0, < J.
NONINTUITIONAL CURVES 541
The reader now sees why we broke the sum 6) into three parts.
As h == 0, the middle term contains an increasing number of terms.
But whatever given value h has, 8 has a finite value.
Thus as A, B, O are each < e/3, the relation 5) is established.
Hence E has a differential coefficient at x = e m , and as
AJF __ A(0)
T" ~ am ~T~
our theorem is established.
537. Example 1. Let x
<7O)
Then for a; =*= 0, g' (x) =  L Here
3 Va;
ForarO, %/ (a;) = + ^ ? ^ (a;
Thus ,   
is a continuous function, and at the rational points e m in the in
terval 51,
RF (x) = 4 oo , J^F (a;) =  oo.
Hence the graph of F has a pantactic set of cuspidal tangents
in 21. The curve is not monotone in any interval of 9, however
small.
Example 2. Let ^
$r (x) = a; sin  , x ^=
a:
= , a: = 0.
Then 111
g f (x) as sin  cos  , a: s 0.
T iC ^
Here = 1. For x = 0,
+! , '* 1.
542 DERIVATES, EXTREMES, VARIATION
Then
is a continuous function in 31, and at the rational point e m ,
where E is the series obtained from F by deleting the m th term.
538. Pompeiu Curves.* Let us now show the existence of
curves which have a tangent at each point, and a paiitactic set of
vertical inflectional tangents.
We first prove the theorem (Borel):
Let B(x) = V ^ = V~ , a n > 0,
V ? n ^n
where (5 = \e n \ is an enumerable set in the interval SI, and
A = 2Va n 
in convergent. Then B converges absolutely and uniformly in a set
83 < 2(, and 3} is as near 21 as we choose.
The points 2) where adjoint B is divergent form a null set.
For let us enclose each point e n in an interval 8 n of length a "
k
with e n as center.
The sum of these intervals is
^ e,
"" ' 1C K
for k > sufficiently large. Let now k be fixed. A point x of 21
will not lie in any S n if
r n =  x  g n  > ~n.
Then at such a point,
k
Adjoint B < ^a n ~i= = k!,Vc^ = kA.
Va n
. Annalen, v. 68 (1907), p. 326.
NONINTUITIONAL CURVES 543
As & > 21 e, the points 3) where B does not converge ab
solutely form a null set.
539. 1. We now consider the function
f(x) = I a n (x  O* = 2/nO) (1
where @ = J^ n J is an enumerable pantactic set in an interval 21, and
ASa. (2
is a convergent positive term series.
Then F is a continuous function of x in ?(. For  x e n \ 3 is <
some M in ?{.
Let us note that each f n (x) is an increasing function and the
curve corresponding to it has a vertical inflectional tangent at the
point x = e n .
We next show that F (x) is an increasing function in 51. For let
x' < x 1 '. Then
/.(<> </(*")
JW*') <
Thus Ji(^) <
Hence ^(a/)
2. Let us now consider the convergence of
obtained by differentiating F term wise at the points of H (.
Let 3) denote the points in 31 where
diverges* We have seen 3) is a null set if
(5
544 DERIVATE8, EXTREMES, VARIATION
is convergent. Lei 21 = 3) f . Let x be a point of , i.e. a
point where 4) is convergent. We break 3) into two parts
such that in JDj, each n < 1. Then J> 2 is obviously convergent,
since each of its terms
a n ^ , c.
, :S'f n , where f n =
and the series 2) is convergent.
The series D l is also convergent. For as f n < 1, the term
t^t
and the series 4) converges by hypothesis, at a point x in &.
Hence 7>(.r) /s convergent at any point in (, r///ff G = 2( ?/'/^^ 5) is
<?0/MW#0?l.
3. Let C 1 ' denote the points in 21 where 3) converges. Let
2f= C+ A.
We next show that F\x) = D(^), for x in C. For taking x at
pleasure in C but fixed,
We now apply 156, 2, showing that Q is uniformly convergent
in (0*, 77). By direct multiplication we find that
Thus 6 ) gives
Q( h ) =
(x + h c n ) h Or h h  OO ~ O + (x 
Let us set
Then
NONINTUITIONAL (CURVES 545
for <  h <. ?;, 7; sufficiently small. As the series on the right is
independent <>f h, Q converges uniformly in (0*, rj). Thus
by 156, 2
F r = D , for any x in O.
4. Let now x be a point of A, not in S. At such a point we show
that
/"(*)= + 00, (8
and thus the curve F has a vertical inflectional tangent. For as
D is divergent at #, there exists for each AT an m, such that
But the middle term in 7) shows that for \h\< some ij f each
term in Q m is > * the corresponding term in D m . Thus
Since each term of ^ is > 0, as 7) shows,
Q(h) > M.
Hence 8) is established.
5. Let us finally consider the points x = e m . If 4> denotes the
series obtained from F by deleting the m th term, we have
, m .
Ax h i &x
As jPis increasing, the last term is >.0.
Hence !"()= + , in @.
vl a result we see the curve F ha* at each point a tangent. At an
enumerable pan tactic set F", it has points of inflection with vertical
tangents.
7. Let us now consider the inverse of the function F, which we
denote by
x=G(t^. (9
As x in 1) ranges over the interval 21, t =F(x) will range over
an interval S3, and by I, 381, the inverse function 9) is a one
valued continuous function of t in 83 which has a tangent at each
546 DERIVATES, EXTREMES, VARIATION
point of 33. If TFare the points in 33 which correspond to the
points V in 21, then the tangent is parallel to the axis at the
points W, or (?'() = 0, at these points. The points TFare pan
taetic in 33
Let Z denote the points of 33 at which GP '() =0. We show
that Z is of the 2 category, and therefore
CardZ=c.
For Cr ( () being of class <_! in 33> its points of discontinuity 8
form a set of the 1 category, by 486, 2. On the other hand, the
points of continuity of (?' form precisely the set Z, since the
points W are pantactic in 93 arid G 1 = in W. In passing let us
note that the points Z in 33 correspond 11 to a set of points $ at
which the series 3) diverges. For at these points the tangent to
F is vertical. But at any point of convergence of 3), we saw in
2 that the tangent is not vertical.
Finally we observe that 3) shows that
2 , n
3 <T p
Hence
ori
Summing up, we have this result :
8. Let the positive term series 2Va n converge. Let (, = \e n \ be
an enumerable pantactic set in the interval 21. The Pompeiu curves
defined by
F(x)=S.aJx e rf
have a tangent at each point in 31, whose, slope, is given by
when this series is convergent, i.e. for all x in 21 except a null set.
At a point set of the 2 category which embraces @, the tangents
are vertical. The ordinates of the curve F increase with x.
540. 1. Faber Curves.* Let F(x) be continuous in the interval
21 = (0, 1). Its graph we denote by F. For simplicity let
* Math. Annalen, v. 66 (1908), p. 81.
NONINTUITIONAL CURVES 547
_F(0) = 0, F(l) = 1 Q . We proceed to construct a sequence of
broken lines or polygons,
which converge to the curve F as follows :
As first line L Q we take the segment joining the end points of
F. Let us now divide 21 into n^ equal intervals
*
8 11' 8 12'" S l,n t (2
of length ., 1
o t = ,
1 i
and having
n, 12>*18 (3
as end points. As second line L v we take the broken line or
polygon joining the points on .F whose abscissae are the points 3).
We now divide each of the intervals 2) into w 2 equal intervals,
getting the n^ intervals
S 21 , S 22 , 8 23 ... (4
of length ., 1
On   1
and having
as end points. In this way we proceed on indefinitely. Let us
call the points
4=Kni
terminal points. The number of intervals in the r th division is
v r = n l  Wjj w r .
If L m (jx) denote the onevalued continuous function in SI whose
value is the ordinate of a point on L m , we have
, (6
since the vertices of L m lie on the curve F.
2. For each x in 21,
m (x) = F(x). (7
m=
For if # is a terminal point, 7) is true by 6).
548 DER1VATES, EXTREMES, VARIATION
If x is not a terminal point, it lies in a sequence of intervals
S 1 >^>
belonging to the 1, 2 division of 21.
Let r, __ , x
m C^m, ni a m, n+l)
Since F(aO is continuous, there' exists an s, such that
^(^)^(a m , n )<, m>* (8
for any x in S m . As L m (x) is monotone in S m ,
 Z/ m (z)  L m (a mn ) \ < \ L m (a mn )  L m (a m<n+l *) \
Thus I^C^^KJI^I. (9
Hence from 8), 9),
which is 7).
8. We can write 7) as a telescopic series. For
1= :i f (AA))
L^L^ ( 2  ij) = L, + (L,  X ) + (i 2  A)
etc. Hence
^(a) = lim i n (2) = i (^) + f ji^rc)  ^^(^l.
If we set
we have jP(ar)  t/ n (a?) , (11
o '
n 11 rl w
a ^(^) = 2/.(^) = i n ( : r). (12
The function / n (o;), as 10) shows, is the difference between the
ordinates of two successive polygons L n _ l , L n at the point x. It
may be positive or negative. In any case its graph is a polygon
NONINTUITIONAL CURVES 549
f n which has a vertex on the a?axis at the end point of each
interval S n ^. Let I n8 be the value of f n (x) at the point x = a M ,
that is, at a point corresponding to one of the vertices of f n . We
call l na the vertex differences of the polygon L n .
? n5 , n = Max ? .
s
Then l/nC*Ol<9n , in 21. (13
In the foregoing we have supposed F(x) given. Obviously if
the vertex differences were given, the polygons 1) could be con
structed successively.
We now show :
If 2 9n (14
is converge^
is uniformly convergent in 21, and is a continuous function in 21.
For by 13), 14), F converges uniformly in 21. As each f n (x)
is continuous, F is continuous in 21.
The functions so defined may be called Faber functions.
541. 1. We now investigate the derivatives of Faber $ functions,
and begin by proving the theorem :
If 2w r ..^ 8 =S^, (1
s
converge, the unilateral derivatives of F(x) exist in 21 = (0, 1 ) . More
over they are equal, except possibly at the terminal points A = \a mn \.
For let x be a point not in A. Let x r , x fr lie in V ' V*(x) ;
letx'x=h',x"x=h".
Let 0== F(x^F(xy
V h' h"
Then F r (x) exists at x, if
e>0 , 7;>0 , \Q\< , for any x r , x" in V. (2
550
DER1VATES, KXTKKMKS, VARIATION
Now
Q\<
F m (.X>~)  F m (x~)
F n (x")  F m (x)
+
W (v 1 ^
^m\ X j
>  ^mO)
h'
h"
h'
h"
But
Hence
Similarly
/.O') 
sufficiently large.
<?.<
Finally, if 77 is taken sufficiently small, x, x 1 ', r" will correspond
to the side of the polygon l/ m . Hence using 540, 12), we see
that Q l = 0. Thus 2) holds, and F'(x) exists at x.
If # is a terminal point a mn , and the two points x 1 ', rr" are taken
on the same side of a mn , the same reasoning shows that the uni
lateral derivatives exist at a mn . They may, however, be different.
2. Let Wj = w 2 = =2. For the differential coefficient F f (x) to
exist at the terminal point x, it is necessary that
Km 2 n p n = oo,
(3
(4
the points where the differential coefficient does not exist form a
pantactic set in 31.
Let us first prove 3). Let b < a< c be terminal points. Then
they belong to every division after a certain stage. We will
therefore suppose that 6, c are consecutive points in the n {h
division, and a is a point of the n 4 1 st division falling in the
interval 8 n = (6, c). If a differential coefficient is to exist at a,
a~c
must be numerically less than some M, as n = oo, and hence their
sum Q remains numerically < 2 M.
NONINTUITIONAL CURVES 551
Now
\ab\ = \ac\ = S n =  n+l .
Thus Q = 2+i j2 L n+l (a)  [i n (6) + L n <
or  @  = 4 2 n / n , ^ supposing a = a nt .
Hence 2 n y n < M,
which establishes 3).
Let us now consider 4). By hypothesis there exists a sequence
n l <n 2 <  = oc, such that
2 nm pn m > G , m = 1, 2 .,
(3 1 being large at pleasure. Hence at least one of the difference
quotients 5) belonging to this sequence of divisions is numerically
large at pleasure.
3< If X = 2C (1
i absolutely convergent, the functions F(x) have limited variation in
31.
Forf m (x) is monotone in each interval .. Hence in S^,
Var/ m = \l m>  L,. + i I <  l m ,\ +  L,. + i 
Hence in 21, Var/.,( a! )<22J B ..
Hence
VarJ 7 B (a;)<222Z m . = 2X , in .
m=l s
We apply now 531.
552
DERIVATES, EXTREMES, VARIATION
542. Faber Functions without Finite or Infinite Derivatives.
To simplify matters let us consider the following example.
The method employed admits easy generalization
and gives a class of functions of this type. We
use the notation of the preceding sections.
Let / { have as graph Fig. 1. We next
divide 21 = (0, 1) into 2 1! equal parts 8 n , S 12 and
take /!< as in Fig. 2. We now divide 31 into
2 2! equal parts S 21 , S 22 , S 23 , S 24 and take / 2 (#) as
in Fig. 3. The height of the peaks is Z 2 =
In the m th division SI falls into 2 m! equal parts
FIG. 1
FIG. 2
one of which may be denoted by
Its length may be denoted by the same letter,
thus i
In Fig. 4, S
division.
is an interval of the m 1 s
FIG. 8
AAA/
Fro. 4
1
The maximum ordinate of / m (V) is C= =  . ^ The
10 m z 10 W
part of the curve whose points have an ordinate < 2 l m have been
marked more heavily. The x of such points, form class 1. The
other x$> make up class 2. With each x in class 1, we associate
the points a m < /3 m corresponding to the peaks of f m adjacent to x.
Thus a m <x<fi m . If x is in class 2, the points m , y8 m are the
adjacent valley points, where f m = Q.
Let now # be a point of class 1. The numerators in
(1
have like signs, while their denominators are of opposite sign.
Thus the signs of the quotients 1) are different. Similarly if x
belongs to class 2, the signs of 1) are opposite. Hence for any a?,
NONINTUITIONAL CURVES
553
the signs of 1) are opposite. It will be convenient to let e m denote
either a^ or /3 m . We have
Hence
f m (x)f n (e n )
m!
410"
On the other hand, for any x^:r r in S m ,
2l m
x'x ~X"
Hence setting x 1 = e n , and letting n > w,
!/(.)/(*) I < Zm I .  * ! < ' ^
o o
wt w
1 2 ! 1 9nl!
10"^ ' 2^ 10 TO * 2 nl
J 1_
< 10 n " I0 m '
For if Iog 2 a l)e the logarithm of a with the base 2,
711
Iog 2 10 , for n sufficiently large.
Hence
Thus
n!
or
2 n ~ 1!
(8
2 n! 10 n '
and this establishes 4).
Let us now extend the definition of the f unctions f n (x) by giv
ing them the period 1. The corresponding Faber function F(x)
defined by 540, 12) will admit 1 as period. We have now
= 2 4
From 2) we have
77 > i /
^ i ^ 2 6 n
554 DERIVATES, EXTREMES, VARIATION
As to 2g, we have, using 4) and taking n sufficiently large,
.1 J_
' 9 ' 10"
m1
Similarly
2,<
/.(*) I < 2
Thus finally
As
8gn
S
Thus
e.x
18*.
3610" "'
As e n may be at pleasure B or /9 n , and as the signs of 1) are
opposite, we see that
awe? F(x) has neither a finite nor an infinite differential coefficient
at any point.
CHAPTER XVI
SUB AND INFRAUNIFORM CONVERGENCE
Continuity
543. In many places in the preceding pages we have seen how
important the notion of uniform convergence is when dealing
with iterated limits. We wish in this chapter to treat a kind of
uniform convergence first introduced by Arzeld, and which we
will call subuniform. By its aid we shall be able to give condi
tions for integrating and differentiating series termwise much
more general than those in Chapter V.
We refer the reader to Arzela's two papers, u Sulle Serie di
Funzioni," R. Accad. di Bologna, ser. V, vol. 8 (1899). Also
to a fundamental paper by Osgood, Am. Journ. of Math., vol. 19
(1897), and to another by ffobson, Proc. Lond. Math. Sac., ser. 2,
vol. 1 (1904).
544. 1. Let/^j ... x m , ^ n )=/(#, t) be a function of two
sets of variables. Let # = (2^ x m ) range over I in an raway
space, and ^ = (^... n ) range over X in an nway space. As a;
ranges over 3E and over !, the point (^ ... ^ )== (a;, ) will
range over a set 31 lying in a space 3J P , p = m + n.
Let T, finite or infinite, be a limiting point of X.
Let lim/(3? * t t ) = AO ... a? ) in I
*=T m
Let the point x range over 93<3E, while t remains fixed, then
the point (#, ) will range over a layer of ordinate t, which we
will denote by ? e . We say x belongs to or is associated with this
layer.
We say now that/= <, subuniformly in X when for each >0,
t/>0:
566
556 SUB AND INFRAUNIFORM CONVERGENCE
1 There exists a finite number of layers % t whose ordinates t
lie in Vf(r).
2 Each point # of is associated with one or more of these
layers. Moreover if x = a belongs to the layer 8 t , all the points
x in some V^(a) also belong to r
while (#, ) ranges over any one of the layers 8^. When w= 1,
that is when there is but a single variable x which ranges over an
interval, the layers reduce to segments. For this reason Arzela
calls the convergence uniform in segments.
2. In case that subuniform convergence is applied to the series
convergent in 21, we may state the definition as follows :
F converges subunif ormly in 21 when
1 For each e > 0, and for each v there exists a finite set of
layers of ordinates > v, call them
81, V" (2
such that each point x of 21 belongs to one or more of them, and if
x = a belongs to m , then all the points of 21 near a also belong
tog m .
2 ^.*"
as the point (#, ri) ranges over any one of the layers 2).
545. ^Example. Let
Here
The series converges uniformly in 21, except at x = 0. The
convergence is therefore not uniform in 21; it is, however, sub
uniform. For
n\x\
CONTINUITY 557
Hence taking m at pleasure and fixed,
\P m \ <e , s in 8 1 =(~S, 8),
sufficiently small. On the other hand,
Thus for w sufficiently large,
Hence we need only three segments 8 V $ 2 , $ 3 to get subuniform
convergence.
546. 1. Let /(a?!^, ^ n ) = <^(a? 1 # m ) in 3E, as = r,
finite or infinite. Let f(z* f) be continuous in H for each t near r.
For (f) to be continuous at the point x = a in X, it is necessary that
for each e > 0, there exists an tj> 0, and a d t for each t in F T) *(r)
such that
for each t in F^ and for any x in V dt (cC).
It is sufficient if there exists a single t=/3 in F^*(r) for which
the inequality 1) holds for any x in some F$(a).
It is necessary. For since < is continuous at x = a,
 <(#) </>(a)  <  , for any x in some F$(a).
o
Also since /= </>,
/(a, ~ ^( a ) I <  > for an J t in some ^i*( T )
o
Finally, since /is continuous in x for any near T,
/(a, 01 < ^ for an 7 ^ in some V^(a).
Adding these three inequalities we get 1), on taking
d t < 8, S .
558 SUB AND INFRAUNIFORM CONVERGENCE
It u sufficient. For by hypothesis
\f(x, /3)  <O)  > I , for any x in some F fi ,(a);
and hence in particular.
Also since /(#, /3) is continuous in #,
a?, /8, # < > for an * in some
Thus if S < ', S", these unequalities hold simultaneously. Add
ing them we get
<f>(x) <K) < i for any X in Ps(a),
and thus < is continuous at = a.
2. As a corollary we get :
Let V( x) = 2/ t ... ,(!  z m )
converge in 21, 0t'A /^r/>i 6mt^ continuous in 31.
tinuous at the point x a in 31, it is necessary that for each e > 0,
and for any cell 11^ > some /2 A i there exists a S M such that
\F^(x)\<e , /or awy x in ^(a).
It is sufficient if there exists an R K and a S >
a? in F s
547. 1. />^^ Urn /(^j ^ m , ^ t n ) = ^(^j x m } in J T finite
r=x
or infinite. Letf(.r, t) be continuous in Hi for each t near r.
1 //"/== <f> sub nn if or ml y in 3E, <j> is continuous in .
"2 If 3 is complete, and <f> is continuous in , / = </> subuniformly
n
7b prove 1. Let ^ = # be a point of 3. Let e > be taken at
pleasure and fixed. Then there is a layer 8^ to which the point
a belongs and such that
CONTINUITY 559
when (#, ) ranges over the points of S^. But then 1) holds for
t = ft and x in some V B (a). Thus the condition of 546, 1 is satis
fied.
To prove 2. Since <f> is continuous at x = a, the relation 1 )
holds by 546, 1, for each t in F^*(T) and for any x in V dt (a}.
With the point a let us associate a cube O a ^ lying in D dt (cC) and
having a as center. Then each point of 36 lies within a cube.
Hence by Borel's theorem there exists a finite number of these
cubes (7, such that each point of lies within one of them, say
0. A . ^, (2
But the cubes 2) determine a set of layers
8,, , V" (
such that 1) holds as (x, t) ranges over the points of 21 in each
layer of 8). Thus the convergence of /to < is sub uniform in J.
2. As a corollary we have the theorem :
Let F^x l .x m ) = ^,.. ln (x 1 ...x m )
converge in 26, each f, being continuous in 3. // JP converges sub
uniformly in , F is continuous in . // X z's complete and F is
continuous in 36, J5 7 converges sukuniformly in 3E.
548. 1. Let ^) = 2f.,., n (*,**.)
converge in 31.
j[/? fA^ convergence be uniform in 21 except possibly for the points
of a complete discrete set 93 = \ b \ . For each 6, let there exist a \
such that for any \ > \ ,
lim JF A <) = 0.
^ converges subuniformly in 21.
For let D be a cubical division of norm d of the space 9I TO in
which 21 lies. We may take d so small that $8 D is small at
pleasure. Let B D denote the cells of D containing points of 21
but none of S3. Then by hypothesis ^converges uniformly in JB D .
Thus there exists a /A O such that for any ft > /* ,
I ^0*0 1 < e ' * Qr an y x ^ ^ ^ n BD
560 SUB AND INFRAUNIFORM CONVERGENCE
At a point b of S3, there exists by hypothesis a Fs(5) and a X
such that for each X > X
 J\< j < , for any x in F 5 (6).
Let (7 6iA be a cube lying in J9 5 (5), having b as center. Since S3
is complete there exists a finite number of these cubes
C^ i C& jA8 (1
such that each point of S3 lies within one of them.
Moreover
for any x of 21 lying in the tc ih cube of 1).
As B D embraces but a finite number of cubes, and as the same
is true of 1), there is a finite set of layers such that
I < in each 2
The convergence is thus subunif orm, as X, /* are arbitrarily large.
2. The reasoning of the preceding section gives us also the
theorem :
in 36, r finite or infinite. Let the convergence be uniform in J except
possibly for the points of a complete discrete set (S \e\. For each
point e, let there exist an rj such that setting e(#, t) =/(., t) <(#),
lim e(#, f) = , for any t in F^*(r).
jce
Thenf= <f> subuniformly in 36.
3. As a special case of 1 we have the theorem :
Let *(*)=/,(*)+/,(*)+
converge in 21, and converge uniformly in 21, except at x = j, a:= <t .
.4^ re = t Ze^ there exist a i> t wcA that
, n t > v, , t = 1, 2 ... 5.
^=a t
F converges subuniformly in 21.
CONTINUITY 561
4. When
, N * x \
t) = $(x)
t=T
we will often set
/(*,*) = *(*) + o*o.
and call e the residual function.
549. Example 1.
f(x, n) = *" = <(X> = , for n ~ oo in 21 = (0 < a),
M.rP
a, , \ > , /* > 0.
The convergence is suburiiform in 31. For x = is the only
possible point of nonuniform convergence, and for any m,
I e(>, m) 1=^^=0 , asa:=0.
1 y e m^
/yi A/yd
.Example 2. /(#, n) =    = </> (a;) = , as n = <x>,
^ in 21 == (0 < a) , a, /3, X, p > , /t > X , <? > 0.
The convergence is uniform in 93 = (0 < a), where e > 0. For
 (*,n)<? A ^ , in 33
1 '""(? + w^e^
a a n A
^ n'*
< e , for n > some m.
Thus the convergence is uniform in 21, except possibly at x = 0.
The convergence is subuniform in 21. For obviously for a given n
lim/(#, n) = 0.
x=Q
550. 1. Let limf(x l x m t l ^ n ) = ^(^ a: m ) iw X, r finite
<=T
or infinite.
Let the convergence be uniform in H except at the points
562 SUB AND INFRAUNIFORM CONVERGENCE
For the convergence to be subuniform in , it is necessary that for
each b in SB, and for each > 0, there exists a t = near r, such that
}im\e(x, *)\>c. (1
x=b
For if the convergence is subuniform, there exists for each
and rj > a finite set of layers ?,, t in F' 7? *(r) such that
 e(#, )  < e , x in 8,.
Now the point # = b lies in one of these layers, say in 80 .
Then
 e(a?, /8)  < , for all # in some V*(ti).
But then 1) holds.
2. Example. Let ^ , ~ /i x
r ^v*0 = 2# n (l #).
o
This is the series considered in 140, Ex. 2.
F converges uniformly in 21 = ( 1, 1), except at x = 1.
we see that ,. TJ s \ i
hm F m (x) =  1.
Hence F is not subuniformly convergent in 31.
Integrdbility
551. 1. Infrauniform Convergence. It often happens that
f(x l " x m t^  n ) = (f>(x l  x m )
subuniformly in J except possibly at certain points (= \e\ form
ing a discrete set. To be more specific, let A be a cubical divi
sion of $ TO in which J lies, of norm 8. Let X denote those cells
containing points of J, but none of @. Since (5 is discrete,
Jf A = J. Suppose now/=< subuniformly in any JST A ; we shall
say the convergence is infrauniform in X. When there are no
exceptional points, infrauniform convergence goes over into sub
uniform convergence.
INTEGRABTLITY 563
This kind of convergence Arzela calls uniform convergence by
segments, in general.
2. We can make the above definition independent of the set @,
and this is desirable at times.
Let H = ( X, j) be an unmixed division of such that may be
taken small at pleasure. If f=<f> subuniformly in each X, we
say the convergence is infrauniform in 3E.
3. Then to each e, 77 >(), and a given Jf, there exists a set of
layers I x , t a , t \\\ F^*(T), such that the residual function e(o;, t)
is numerically < e for each of these layers. As the projections of
these layers { do not in general embrace all the points of J, we
call them deleted layers.
4. The points we shall call the residual points.
x 2
5. Example 1. __ V T~T3~*
This series was studied in 150. We saw that it converges uni
formly in Sl = (0, 1), except at #= 0.
As ,
and as this == 1 as x = for an arbitrary but fixed n, F does not
converge subuniformly in 21, by 550. The series converges infra
uniformly in 21, obviously.
6. Example 2. ^^ % x n (l  x}
o
This series was considered in 550, 2. Although it does not
converge subuniformly in an interval containing the point x = 1,
the convergence is obviously infrauniform.
552. 1. Let lim f (x l x m t n ) = </)(^ # TO ) be limited in ,
r finite or infinite. For each t near r, letf be limited and Rinteyrable
in H. For <f> to be Rintegrable in , it is sufficient thatf == <f> infra
uniformly in X. If Hi is complete, this condition is necessary.
564 SUB AND INFRAUNIFORM CONVERGENCE
It is sufficient. We show that for each e, o>> there exists a
division D of 9} m such that the cells in which
OSC <f> > Q) (1
have a volume < <r. For setting as usual
/=<+,
we have in any point set,
Osc<<0sc/+ Osce.
Using the notation of 551,
in the finite set of deleted layers I p ( 2 corresponding to
=}, 2 For each of these ordinates t L ,f(x, t ) is integrable
in 3. There exists, therefore, a rectangular division D of 9? OT ,
such that those cells in which
have a content < ^, whichever ordinate t t is used. Let E be a
A
division of 9J m such that the cells containing points of the residual
set y have a content < cr/2. Let F = D + E. Then those cells
of .F in which
O>^ or Osc ' e<>, *.)  >
J ^J
t = l, 2 have a content < o. Hence those cells in which 1)
holds have a content < a.
It is necessary, ifH is complete. For let
*i> 2 == T 
Since <fr and/(#, ^ n ) are integrable, the points of discontinuity of
<(V) and of f (x, t n ) are null sets by 462, 6. Hence if S, S< denote
the points of continuity of <(#) and /(a;, in X,
since 3 is measurable, as it is complete.
INTEGRABILITY 565
Let = Qdv{6/,
then =
by 410, 6.
Let SD = Dv((E, ),
then = , (1
as we proceed to show. For if 6r = I ,
But 6Ms a null set. Hence Meas Dv(, 6r) = 0, and thus
g = i = ), which is 1).
Let now be a point of 5), let it lie in S^, Sj, where ^, f 2
form a monotone sequence = r. Then since
there is an m such that
I (& O I < I fo r any n >m. (2
But ^ lying in ), it lies in S and S <n .
Thus
for any ^ in Fad). Hence
 <>, f n )  (f , O  < ^ , x in Fi(f ). (3
Now
e(x, O = e(a:, n )  e(f, U + c(f, * n ).
Hence from 2), 3),
 e(>, t n )\<e , for any x in F" 6 (:).
Thus associated with the point , there is a cube T lying in Dfi(f),
having  as center. As D = X 35 is a null set, each of its points
can be enclosed within cubes (7, such that the resulting enclosure
566 SUB AND INFRA UNIFORM CONVERGENCE
@ has a measure < <r, small at pleasure. Thus each point of J lies
within a cube. By Borel's theorem there exists a finite set of
these cubes
*1> ^2 '*' r ' 1' 2 '" ^'
such that each point of 3E lies within one of them. But corre
sponding to the F's, are layers
81, 8,,  8,
such that in each of them
Thus/ = </> subuniformly in X = (T l , T 2 ... T r ). Let y be the
residual set. Obviously $ < &. Thus the convergence is infra
uniform .
2. As a corollary we have :
Let F(x) = ^... in (x^^x m }
converge in 21. Let F be limited, and each f, be limited and R~in
tegrable in 21. For F to be Rintegrable in ?J, it in sufficient that F
converges infrauniformly in 21.
If 21 is complete* this condition is necessary.
553. Infinite Peaks. 1. Let lim/(a? t x m t l t n ) = <(V) in X,
t=T
T finite or infinite. Although f(x, t) is limited in I for each t
near r, and although <f>(x) is also limited in I, we cannot say that
/('#,  < some JMT (1
for any x in X and any near r, as is shown by the following
to
Example. Let /(a;, O = == 0(a?) = 0, as == oo for 2: in
= ( 00, GO).
It is easy to see that the peak of / becomes infinitely high {is
n = oo.
In fact, for x = , / = ^. Thus the peak is at least as high
V t e
as , which == oo .
e
INTEGRAB1LITY 567
The origin is thus a point in whose vicinity the peaks of the
family of curves f(x, t) are infinitely high. In general, if the
peaks of /vWiQ
in the vicinity F 6 of x become infinitely high as t = T, however
small 8 is taken, we say is a point with infinite peaks.
On the other hand, if the relation 1) holds for all x and t in
volved, we shall say /(#, ) is uniformly limited.
2. If lini /Ov.a^ ^ 1 ..^ n ) = </)(^ 1 ...^ m ), <wdf i/ /(, w
<=T
uniformly limited in 36, Am < / limited in H.
For a; being taken at pleasure in H and fixed, $(V) is a limit
point of the points /(#, t) us t = T. But all these points lie in
some interval ( 6r, 6r) independent of x. Hence <f> lies in this
interval.
3. If H is complete, the points $ in 3E with infinite peaks also form
a complete set. If these points $ are enumerable, they are discrete.
That $ is complete is obvious. But then $ = ^ = 0, as $ is
enumerable.
554. 1 . Let lim / ( x l x m t 1 n ) = ^(^ o; m ) m J, metric or
tT
complete. Let f (x, t) be uniformly limited in 3E, and Rintegrable
for each t near r. For the relation
lim (/(a:, 0= f<K*)
<=T /X /;
^o A0W, i^ Z8 sufficient that f '^ (j> infra uniformly in 3E. If Hi is
for each t complete, this condition is necessary.
For by 552, </> is J?integrable if /= </> infrauniformly, and when
X is complete, this condition is necessary. By 424, 4, each/ (x, )
is measurable. Thus we may apply 381, 2 and 413, 2.
2. As a corollary we have the theorem :
Let FC*)^ <>!*.)
converge in the complete or metric field 21 Let the partial sums F^ be,
uniformly limited in 21 Let each term/, be limited and Rintegrable
in 21. Then for the relation
f F=2 ft
J% *V l
568 SUB AND INFRAUNIFORM CONVERGENCE
to hold it is sufficient that F is infra uniformly convergent in 21.
21 is complete, this condition is necessary.
555. Example 1. Let us reconsider the example of 150,
We saw that we may integrate termwise in 21 = (0, 1), al
thongh jPdoes not converge uniformly in 21. The only point of
noniunl'orm convergence is x = 0. In 551, 5, we saw that it con
verges, however, infrauniformly in 21. As
I ^n(X) I < 1 * f r an y x in 21, and for every n,
all the conditions of 554 are satisfied and we can integrate the
series termwise, in accordance with the result already obtained
in 150.
Example. Let F(x) = V I ^  ^ ~ ^ x U 0.
F fM e nx * e (n ~ w
Then
We considered this series in 152, i. We saw there that this
series cannot be integrated termwise in 21 = (0 < a). It is, how
ever, subuniformly convergent in 21 as we saw in 549, Ex. 1. We
cannot apply 554, however, as F n is not uniformly limited. In
fact we saw in 152, l, that x = is a point with an infinite peak.
Example 3. F(x) = i# n (l  x).
o
We saw in 551, 6, that F converges infrauniformly in 21 = (0, 1).
Here F n (x)\ = \l ~ x\ < some M,
for any x in 21 = (0 < u), u <_ 1, and any n. Thus the F n are
uniformly limited in 21.
We may therefore integrate termwise by 554, 2. We may
verify this at once. For
x = 0.
INTEGRABILITY 569
Hence C u F(x)dx = u. (1
/o
On the other hand,
X u ?/ n+ l
F n dx = u   = u , as n = oo. (2
n f 1
n f 1
From 1), 2) we have
^n+1 U *M
 + 1 w 4 2 J '
556. 1. //' l /'^! " x m ^i O == $(#i x m) infrauniformly
in the metric or complete field , as t == T, T ^iii^ or infinite ;
2 /(a?, is uniformly limited in 36 flncif Rintegrable for each t
near r;
Then ^
uniformly with respect to the set of measurable fields ?l in I
If H is complete, condition 1 may be replaced by 3 (/>(#) is
Rintegrdble in %.
For by 552, 1, when 3 holds, 1 holds ; and when 1 holds, </>
is 7^integrable in X.
Now the points & t where
are such that ^
lim <g, = , by 412.
Let = (, + ,. Then
But
which establishes the theorem.
2. As a corollary we have :
If 1 .F(V)= 2/ tl ... ln (2i a; m ) converc/es wfrauniformlt/< and
each of its terms f^ is Rinteyrable in the metric or complete field 9f ;
570 SUB AND INFRAUNIFORM CONVERGENCE
2 F\(jx) is uniformly limited in SI/
Then
and the series on the right converges uniformly with respect to all
measurable 93 _<_ 21.
3. If 1 lim/(:r, ^ t n )=*<j>(x) is Rintegrable in the interval
t=T '
?l = (a < />), T finite or infinite ;
2 f(x y ix uniformly limited, and Rintegrable for each t near r;
Then lim f J f(x, t^dx = f*4>(x)dx = <S>(x),
f=T *<! *^
uniformly in ^ and <\>(x) is continuous in 21.
and also each termf t are Rintegrable in the interval 2l = (a< 6);
2 F^x) is uniformly limited in 21;
^^ a(rr)= 2 f'f^dx , m 21
*^a
i continuous.
For (? is a uniformly convergent series in 21, each of whose terms
is a continuous function of x.
Differentiability
557. 1 . If 1 lim /Or, ^ n ) == </>(^) m 21 = ( a < 6), r j^m'te or
tT
infinite ;
2fj(x, fy is Rintegrable for each t near T, and uniformly limited
m2l;
3 fj.(x, ^)= ^(^) infra uniformly in ?(, as t = r ;
Then at a point :r of continuity of ty in 21
4>'(X>=tO> (1
is the same
Or, ) = lini./OM) ( 2
DIFFERENTIABILITY 571
For by 554,
lira f /;(*, f)dx = PV(*)fe (8
tT ^d *J a
= liui [/(a:, )  /(, )] , by I, 538
t=*
= <K*)<Ka) , by 1.
Now by I, 537, at a point of continuity of >Jr,
From 3), 4), we have 1 ), or what is the same 2).
2. In the interval 31, if
1 F(x)= /' ti ... lri (#) converges ; (1
2 ~Eachfl(x) is limited and Rintegrable ;
3 .F( (#) is uniformly limited ;
4 (?(#)= 2// is infrauniformly convergent ;
TA^w dtf a point of continuity of Gr(j') in 31, wv ^//y differ mtiati*
the series 1) termwise, or F r (x) G(x).
3. /TI ^e interval 31, if
1 /(^, ^ O = <O) < === T, r finite or infinite :
2 f (x, f) is uniformly limited, and a continuous function of x ;
3 ty(x) = lim/^(a;, t) is continuous ;
tT
Then
or t^Aa^ f the same
^lim/O, 0=lim?/(rc, Q. (2
a^ /=T <=T ao;
For by 547, 1, condition 3 requires that /' = ^ subuniforml y
in SI. But then the conditions of 1 are satisfied and 1) and 2)
hold.
4. In the interval 31 let us suppose that
1 JF (/) = 2/4 . c n (*0 Converges ; (1
572 SUB AND INFRAUNIFORM CONVERGENCE
2 Each termf, is continuous;
3 F((x) is uniformly limited ;
4 6r(V)= 2/'(#) is continuous ;
Then we may differentiate 1) termwise, or F'(x) = Q(x).
558. Example 1. We saw in 555, Ex. 3 that
n+l
The series got by differentiating termwise is
#(aO=2z n (l20==l 0<a;<l
o C2
= , z = 0. v
Thus by 557, 4,
The relation 3) does not hold for x = 0.
Example 2.
 ^ gv ^  arctg ^
Vw f 1
Here JP(a;)=arctga;, for any x. (1
^ )=S ' ** (2
Hence Cr(x) is continuous in any interval 21, not containing
x = 0. Thus we should have by 557, 4,
JF(aO0(a!> * in 21. (3
This relation is verified by 1), 2). The relation 3) does not
hold for x = 0, since
J(0)=l, , 0(0) 0.
DIFFERENTIABILITY 573
Example 8.
= log (1 + a 2 ) , for any x.
In any interval 31, all the conditions of 557, 4, hold.
Hence F'(x)=G(x} , for any x in 21. (3
In case we did not know the value of the sums 1), 2) we could
still assert that 3) holds. For by 545, Gr is subuniformly con
vergent in 21, and hence is continuous.
Example 4*
l+nx
 J+O + i)* 1 = 1
(n + iy +* J e
ne n
Here
The series obtained by differentiating F termwise is
and hence
e x e nx
The peaks of the residual function
are of height = l/e. The convergence of Q is not uniform at
x = 0. The conditions of 557, 4, are satisfied and we can differ
entiate 1) termwise. This is verified by 2), 3).
574 SUB AND INFRAUNIFORM CONVERGENCE
559. 1. Ifl lim/(#, t 1 t n ) = <(V) is limited and Rinteyrable
tt
in the, interval 21 = ( a < b) ;
2 f(x, t) Is limited, and Hintegrable in $1, for each t near r;
8 T/r(a;)= lim (* f(x< f) = lim#O, t)
t=T J<1 * t=T
is a continuous function in ?I;
4 The point* & in 31 in whose vicinity the peaks of f(x, t) a*
t= r are infinitely high form an enumerable set ;
Then * x /v
0(x)= I 0<X) = liiii I /(^ t)dx = ^(x), (1
^/^ / T At
^ /v /**
lim I /(a;, t)ds = I lim/(^, O^,
/=T ^^ ' /tt f=T '
awrf ^/i6 set @ i complete and discrete.
For @ is discrete by 553, 3.
Let a be a point of A 31 (. Then in an interval a about ,
/(j?, t)  < some Tlf , x in a, any ^ near r. (^
Now by 556, 3, taking e>0 small at pleasure, there exists an
77 > such that
<e
for a^y x in a, and in F^ *(T). If we set x = a + h, we have
^ = o ? o i i^ = ir > ^ ^ + e'
Aa* h hJ h
Also by 556, 3, we have
for any x in a, and t in F^*(T). Thus
1 i* f < , 7 B(x) v( ) e" A$ 6^
I / ( X* 1 )uX      j~ . . ^r .  __
//V" ' h h A.r A
From M), 4 ) we have 4
e' A^ . e ;/
DIFFERENTIABILITY 575
Now e may he made small at pleasure, and that independent of
A. Thus the last relation mves
Ailr Atf P <
 , ror ;r in A.
A.r A#
As this holds however small h = A# is taken, we have
f/'V/r d0 f
Hence by 515, 3,
(x)=6(x)+ const , in 21.
For a? = a, ^(a ) 0(a) = ;
and thus
in .
2. As a corollary we have :
If] F(x) 2/' t1 ... t/t (#) is limited a)id R inteijrable in the inter
val 21 = ( a < J> ) ;
2 JF\(x) in limited and each term f\ i$ R httet/rable ;
3 G(x)= ^ I / t is continuous;
4 The points 6 m ?l i>^ wlioxe vicinity the peaks of F x (x) are in
finitely JiiifJt form an enumerable xet;
rt t y integrate the F series termwise.
560. 1 . IfT lim/( ar, t l  ^ n ) = </>(>) * ^ = ( a < fy r finite or
infinite ;
2 f' r (x^ f) is limited and Rinteyr able for each t near T;
3 The points (g o^ 1 ?l i ^Ao^ vicinity f^x, t) has infinite peaks
as t = r form an enumerable set:
4 <(#) *'* continuous at the points @;
5 ty(x) = limj^(r, O / limited and Rintegrable in ?l;
576 SUB AND INFRAUNIFORM CONVERGENCE
Then at a point of continuity of ^(#) in 31
or what is the same

ax t= r
For let 8 = (a < /3) be an interval in 31 containing no point of
(. Then for any x in
) , by 2.
^
Hence ~ x
lira I /;(*, f)dx = \im\f(x, f) /(, )}
/=T 'a /=T
== </><V)  <K) , byl. (2
By 556, 3, </>(^) is continuous in S. Thus <(#) i y continuous
at any point not in (g. Hence by 4 it is continuous in 31.
We may thus apply 559, l, replacing therein f(x, t) by/j(#, t).
We get
C x C x C x
t=T
Since 2) obviously holds when we replace a by a, this relation
with 3) gives
At a point of continuity, this gives 1) on differentiating.
2. I/* 1 jF(z) = 2/ ti ... tw (^) converges in the interval 21;
2 (?(a;) = /[(#) anc? eac?A q/ its terms are limited and R
integrable in 31;
8 The points of 31 m whose vicinity (? A (^) Aa infinite peaks as
\ == (X),form an enumerable set at which F(x) is continuous;
Then at a point of continuity of Gr(x) we have
or what is the same
DIFFERENTIABILITY 577
561. Example.
F
Hence
<*> , , , i\ ^ a
(x ^ = y (5^_ (*L+LX l _
w T 1 * nx ' * M ** J ~ **'
The series obtained by differentiating F termwise is
&(x )== \? [2nx _2n*a* __ 2(n
nx * nx2 ^ n
^(71+1)^ J
Here a r ^    I ?^  
Hence 9
is a continuous function of x.
The convergence of the Q series is not uniform at x = 0. For
set a n == I/ft,. Then
2
9
^,
To get the peaks of the residual function we consider the
points of extreme of
We find ., r o , o
' ^n 5 war + 2
"**
Thus y 1 = when
2 nV  5 wa ? + 1 = 0,
or when x = ~ or  , a, a constants.
Vw Vn
Putting these values in 3), we find that y has the form
y = c V/i.
Hence # = is the only point where the residual function has
an infinite peak. Thus the conditions of 560, 2, are satisfied, and
we should have F'(x) = &(x) for any x. This is indeed so, as 1 ),
2) show.
CHAPTER XVII
GEOMETRIC NOTIONS
Plane Curves
562. In this chapter we propose to examine the notions of
curve and surface together with other allied geometric concepts.
Like most of our notions, we shall see that they are vague and
uncertain as soon as we pass the confines of our daily experience.
In studying some of their complexities and even paradoxical
properties, the reader will see how impossible it is to rely on his
unschooled intuition. He will also learn that the demonstration
of a theorem in analysis which rests on the evidence of our
geometric intuition cannot be regarded as binding until the
geometric notions employed have been clarified and placed on a
sound basis.
Let us begin by investigating our ideas of a plane curve.
563. Without attempting to define a curve we would say on
looking over those curves most familiar to us that a plane curve
has the following properties :
1 It can be generated by the motion of a point.
2 It is formed by the intersection of two surfaces.
8 It is continuous.
4 It has a tangent at each point.
5 The arc between any two of its points has a length.
6 A curve is not superficial.
7 Its equations can be written in any one of the forms
y =/(*), a
(2
(3
and conversely such equations define curves.
578
PLANE CURVES 579
8 When closed it forms the complete boundary of a region.
9 This region has an area.
Of all these properties the first is the most conspicuous and
characteristic to the naive intuition. Indeed many employ this
as the definition of a curve. Let us therefore look at our ideas
of motion.
564. Motion. In this notion, two properties seem to be essen
tial. 1 motion is continuous, 2 it takes place at each instant in
a definite direction and with a definite speed. The direction of
motion, we agree, shall be given by dy/dx, its speed by ds/dt.
We see that the notion of motion involves properties 4, 5, and 7.
Waiving this point, let us notice a few peculiarities which may
arise.
Suppose the curve along which the motion takes place has an
angle point or a cusp as in I, 366. What is the direction of
motion at such a point? Evidently we must say that motion is
impossible along such a curve, or admit that the ordinary idea of
motion is imperfect and must be extended in accordance with the
notion of righthand and lefthand derivatives.
Similarly ds/dt may also give two speeds, a posterior and an
anterior speed, at a point where the two derivatives of s = </>()
are different.
Again we will admit that at any point of the path of motion,
motion may begin and take place in either direction. Consider
what happens for a path defined by the continuous function in
I, 367. This curve has no tangent at the origin. We ask how
does the point move as it passes this point, or to make the ques
tion still more embarassing, suppose the point at the origin. In
what direction does it start to move? We will admit that no
such motion is possible, or at least it is not the motion given us
by our intuition. Still more complicated paths of this nature are
given in I, 369, 371, and in Chapter XV of the present volume.
It thus appears that to define a curve as the path of a moving
point, is to define an unknown term by another unknown term,
equally if not more obscure.
565. 2 Property. Intersection of Two Surfaces. This property
has also been used as the definition of a curve. As the notion
580 GEOMETRIC NOTIONS
of a surface is vastly more complicated than that of a curve, it
hardly seems advisable to define a complicated notion by one still
more complicated and vague.
566. 3 Property, Continuity. Over this knotty concept philos
ophers have quarreled since the days of Democritus and Aristotle.
As far as our senses go, we say a magnitude is continuous when
it can pass from one state to another by imperceptible gradations.
The minute hand of a clock appears to move continuously, although
in reality it moves by little jerks corresponding to the beats of the
pendulum. Its velocity to our senses appears to be continuous.
We not only say that the magnitude shall pass from one state
to another by gradations imperceptible to our senses, but we also
demand that between any two states another state exists and so
without end. Is such a magnitude continuous ? No less a mathe
matician than Bolzano admitted this in his philosophical tract
Paradoxien des Unendlichen. No one admits it, however, today.
The different states of such a magnitude are pantactic, but their
ensemble is not a continuum.
But we are not so much interested in what constitutes a con
tinuum in the abstract, as in what constitutes a continuous curve
or even a continuous straight line or segment. The answer we
have adopted to these questions is given in the theory of irra
tional numbers created by Cantor and Dedekind [see Vol. I,
Chap. II], and in the notion of a continuous function due to
Cauchy and Weierstrass [see Vol. I, Chap. VII].
These definitions of continuity are analytical. With them we
can reason with the utmost precision and rigor. The consequences
we deduce from them are sufficiently in accord with our intuition
to justify their employment. We can show by purely analytic
methods that a continuous f unction /(#) does attain its extreme
values [I, 354], that if such a function takes on the value a at the
point P, and the value b at the point Q, then it takes on all inter
mediary values between a, J, as x ranges from P to Q [I, 357].
We can also show that a closed curve without double point does
form the boundary of a complete region [cf. 576 seq.].
567. 4 Property. Tangents. To begin with, what is a tangent ?
Euclid defines a tangent to a circle as a straight line which meets
PLANE CURVES 581
the circle and being produced does not cut it again. In com
menting on this definition Casey says, " In modern geometry a
curve is made up of an infinite number of points which are
placed in order along the curve, and then the secant through two
consecutive points is a tangent." If the points on a curve were
like beads on a string, we might speak of consecutive points. As,
however, there are always an infinite number of points between any
two points on a continuous curve, this definition is quite illusory.
The definition we have chosen is given in I, 365. That property
3 does not hold at each point of a continuous curve was brought
out in the discussion of property 1. Not only is it not necessary
that a curve has a tangent at each of its points, but a curve does
not need to have a tangent at a pantactic set of points, as we saw
in Chapter XV.
For a long time it was supposed that every curve has a tangent
at each point, or if not at each point, at least in general. Analytic
ally, this property would go over into the following : every con
tinuous function has a derivative. A celebrated attempt to prove
this was made by Ampere.
Mathematicians were greatly surprised when Weierstrass ex
hibited the function we have studied in 502 and which has no
derivative.
Weierstrass* himself remarks: "Bis auf die neueste Zeit hat
man allgemein angenommen, dass eine eindeutige und continuir
liche Function einer reellen Verlinderlichen auch stets eine erste
Ableitung habe, deren Werth nur an einzelnen Stellen unbestimmt
oder unendlich gross werden konne. Selbst in den Schriften von
Gauss, Cauchy, Dirichlet findet sich meines Wissens keine
Ausserung, aus der unzweifelhaft hervorginge, dass diese Mathe
matiker, welche in ihrer Wissenschaft die strengste Kritik iiberall
zu iiben gewohnt waren, anderer Ansicht gewesen seien."
568. Property 5. Length. We think of a curve as having
length. Indeed we read as the definition of a curve in Euclid's
Elements : a line is length without breadth. When we see two
simple curves we can often compare one with the other in regard
to length without consciously having established a way to measure
* Werke, vol. 2, p. 71.
582 GEOMETRIC NOTIONS
them. Perhaps we unconsciously suppose them described at a
uniform rate and estimate the time it takes. It may be that we
regard them as inextensible strings whose length is got by
straightening them out. A less obvious way to measure their
lengths would be to roll a straightedge over them and measure
the distance on the edge between the initial and linal points of
contact.
We ask how shall we formulate arithmetically our intuitional
ideas regarding the length of a curve ? The intuitionist says, a
curve or the arc of a curve has length. This length is expressed
by a number L which is obtained by taking a number of points
Pj, P 2 , P 3 " on the curve between the end points P, P', and
forming the sum
The limit of this sum as the points became pantactic is the
length L of the arc PP'.
Our point of view is different. We would say : Whatever
arithmetic formulation we choose we have no a priori assurance
that it adequately represents our intuitional ideas of length.
With the intuitionist we will, however, form the sum 1) and see if
it has a limit, however the points P t are chosen. If it has, we will
investigate this number used as a definition of length and see if it
leads to consequences which are in harmony with our intuition.
This we now proceed to do.
569. 1. Let = y =
be onevalued continuous functions of t in the interval 21 =
As t ranges over 21 the point a?, y will describe a curve or an arc
of a curve O. We might agree to call such curves analytic, in
distinction to those given by our intuition. The interval 21 is
the interval corresponding to C.
Let D be a finite division of 21 of norm c?, defined by
To these values of t will correspond points
P,P,,P,. (2
PLANE CURVES 583
on (7, which may be used to define a polygon P D whose vertices
are 2).
Let (m, m f 1) denote the side P m P m iii as well as its length.
If we denote the length of P D by the same letter, we have
P D = 2(m, m 4 1) = 2 VA^+~A^.
If lim P D (3
exists, it is called the length of the arc C, and is rectifiable.
2. (Jordan. ) For the arc PQ to be rectifiable^ it is necessary and
sufficient that the functions (/>, i/r in 1) have limited variation in 21.
j\ n .
Hence p > y I A I
But the sum on the right is the variation of < for the division 7).
If now <f> does not have limited variation in 21, the limit 3) does
not exist. The same holds for *\Jr. Hence limited variation is a
necessary condition.
The condition is sufficient. For
P D < 2 I A# 1+21 Ay I = Var <j> } Var \fr .
D D
As <, ^ have limited variation, this shows that
P = Max P D
D
is finite. We show now that
For there exists a division A such that
p _ 1 < p^ < p (5
Let A cause 21 to fall into v intervals, the smallest of which has
the length X. Let D be a division of 21 of norm d<d Q <\. .
Then no interval of D contains more than one point of A.
Let ED+ A.
~ ~ or P A .
Obviously P E >P
584 GEOMETRIC NOTIONS
Suppose that the point t K of A falls in the interval ( t , ^) of
D. Then the chord (/, i f 1) in P D is replaced by the two chords
(t, /c), (/e, 6 4 1) in P^. Hence
P E ^P D ^^O. K , * = 1,2 /*<*
where ^ = ^ ^ + ( ^, + X) _ ^ i + 1 ^
Obviously as <, ^ are continuous we may take d Q so small that
each
Gf^K<~ * for any d < d .
Hence p p e fi
J ^ /'/,< (^
From 5), 6) we have
P  Pj)< , for any d < rf ,
which gives 4).
3. If the arc PQ is rectifiable, any arc contained in PQ is also
rectifiable.
For 0, ^ having limited variation in interval 21, have a fortiori
limited variation in any segment of 31.
4. Let the rectifiable arc G fall into two arcs (7j, (7 2 . If s, s x , $ 2
are ^Ae lengths of C, <7j, (7 2 , ^/t^Ti
8=8^ S 3 . (7
For we saw that Cj, (7 2 are rectifiable since O is. Let Slj, 21 2
be the intervals in 21 corresponding to 6\, (7 2 . Let D v J9 2 be
divisions of 2lj , 2I 2 of norm d. Then
Sj = lim P A , * 2 = lim P D ^
But Dj, D 2 effect a division of 21, and since
s = lim P K (8
<r=0
with respect to the class of all divisions of 21, the limit 8) is the
same when E is restricted to range over divisions of the type of D.
Now
PD = PDI 4 Pj) t
Passing to the limit, we get 7).
PLANE CURVES 585
The preceding reasoning also shows that if C l , (7 2 are rectifiable
curves^ then is, and 7) holds again.
5. If 1) define a rectifiable curve, its length 8 is a continuous func
tion s() of t.
For </>, v/r having limited variation,
where the functions on the right are continuous monotone increas
ing functions of t in the interval 91 = (# < i).
For a division D of norm d of the interval A3! = (t, t + h) we
have
+ 2  Ay 
4
where S^ = (^(^h A) <(), and similarly for the other func
tions. As $j is continuous, 8^ == 0, etc., as A=0. We may
therefore take rj > so small that S^ , S(f> 2 , SI/TJ , S^r 2 < e/4, if A < rj.
Hence As = (^4 A) (0 < Max P^< e , if < h > y.
Thus s is continuous.
6. 2%e length s of the rectifiable arc C corresponding to the inter
val (a < t) is a monotone increasing function oft.
This follows from 4.
7. If x, y do not have simultaneous intervals of invariability, s(f)
is an increasing function of t. The inverse function is onevalued
and increasing and the coordinates x, y are one valued functions of s.
That the inverse function t (s) is onevalued follows from I, 214.
We can thus express t in terms of s, and so eliminate t in 1).
570. 1. If (f> f , yfr r are continuous in the interval 31,
For
lim 2VA02 + A^2. (2
586
GEOMETRIC NOTIONS
Now A0 K = <'(X)Mc A ^* = ^'(OMc (3
where t' K , tf" lie in the interval At
As <', vr' are continuous they are uniformly continuous. Hence
for any division D of norm < some d ,
where \ a K \ ,  &  < some ?;, small at pleasure, for any K. Thus
and we may take
ThuS
Hence
= lira
rf=0
(.)*
lira
8
< ,
which establishes 1).
For simplicity we have assumed $ ; , ^ to be continuous in 31.
This is not necessary, as the following shows.
2. Let aj, a n , Jj, 4 n >0 but not all = 0.
rrii
611  Vo?+ +o2 V4fT^ + 1  < S I a m  b m  ,
For
m = 1, 2  ra. (4
)(Vaf+
Hence
VofT^
But
f +  + VAJ +
This in 5) gives 4).
PLANE CURVES 587
Let us apply 4) to prove the following theorem, more general
than 1.
3. (2?atV0.) If $ , ^ are limited and Rintegrable, then
8 =
For by 4),
or 4> K  V K = 77.' Osc </>'(0 + i/l Osc i/r'<T) , in S K = A,
where 77*', 77^' are numerically <1. Thus
 28,*,  2S^  = 28,1k' Osc $' + 28^ Osc <p. (6
As (//, ^ f are integrable, the right side = 0, as d === 0. Now
lira 28^
rfo
Thus passing to the limit in 6), we have
lim 2A* K V^C^TT^C^) 5 " = f
^a
This with 2), 3) gives 1) at once.
571. Volterra's Curve. It is interesting to note that there are
rectifiable curves for which <'(0> ^'(0 are no ^ both Mintegr able.
Such a curve is Volterra's curve, discussed in 503. Let its equa
tion be y =/(#). Then f\x) behaves as
.1 1
z x sin  cos 
^ X
in the vicinity of a non null set in 21 = (0, 1). Hence f r (x) is
not jRintegrable in 21. But then it is easy to show that
does not exist. For suppose that
588 GEOMETRIC NOTIONS
were 72integrable. Then # 2 = 1 t/'(X) 2 ^ 8 jRintegrable, and
hence /'(#) 2 also. But the points of discontinuity of /' 2 in 21 do
not form a null set. Hence/' 2 is not Uintegrable.
On the other hand, Volterra's curve is rectifiable by 569, 2, and
528, 1.
572. Taking the definition of length given in 569, 1, we saw
that the coordinates
must have limited variation for the curve to be rectifiable. But we
have had many examples of functions not having limited variation
in an interval 21. Thus the curve defined by
y = x sin  , x =
x (4
does not have a length in 21 = ( 1, 1) ; while
1
y = x* sin  , x =
^ * ' (5
=0 , z=0
does.
It certainly astonishes the naive intuition to learn that the
curve 4) has no length in any interval B about the origin how
ever small, or if we like, that this length is infinite, however small
S is taken. For the same reason we see that
No arc of Weierstrass' curve has a length (or its length is infinite)
however near the end points are taken to each other, when ab>\.
573. 1. 6 Property. Spacefilling Curves. We wish now to
exhibit a curve which passes through every point of a square, i.e.
which completely fills a square. Having seen how to define one
such curve, it is easy to construct such curves in great variety, not
only for the plane but for space. The first to show how this may
be done was Peano in 1890. The curve we wish now to define is
due to Hilbert.
We start with a unit interval 21 = (0, 1) over which t ranges,
and a unit square 93 over which the point x, y ranges. We define
PLANE CURVES
589
as onevalued continuous functions of t in 21 so that xy ranges over
93 as t ranges over 21. The analytic curve C defined by 1) thus
completely fills the square $8.
We do this as follows. We effect a division of H into four
equal segments 8J, B'%, S' z , 84, and of 35 into equal squares rj{, rf^
y' B , ifv as in Fig. 1.
We call this the first division or D r The corre
spondence between 21 and 93 is given in first
approximation by saying that to each point P in
S[ shall correspond some point Q in rj[ .
We now effect a second division D 2 by dividing
each interval and square of D^ into four equal
parts.
We number them as in Fig. 2,
FIG. I.
Si'
Sn
16
Fio. 2.
As to the numbering of the rfs we observe the
following two principles : 1 we may pass over the
squares 1 to 16 continuously without passing the
same square twice, and 2 in doing this we pass
over the squares of Dj in the same order as in
Fig. 1. The correspondence between 21 and 93 is
given in second approximation by saying that to each point P in
8[' shall correspond some point Q in ij( f . In this way we continue
indefinitely.
To find the point Q in 93 corresponding to P in 21 we observe
that P lies in a sequence of intervals
8' >S" >"' >... =0, (2
to which correspond uniquely a sequence of squares
if >V /'"> =0. (3
The sequence 3) determines uniquely a point whose coordinates
are onevalued functions of t, viz. the functions given in 1).
The functions 1) are continuous in 21.
For let t' be a point near t ; it either lies in the same interval as
t in D n or in the adjacent interval. Thus the point Q r corre
590 GEOMETRIC NOTIONS
spending to t r either lies in the same square of D n as the point Q
corresponding to , or in an adjacent square. But the diagonal
of the squares = 0, as n = oo. Thus
Thus
both = 0, as t = t.
As t ranges over 21, the point x, y ranges over every point in the
square 33.
For let Q be a given point of 93. It lies in a sequence of
squares as 3). If Q lies on a side or at a vertex of one of the 77
squares, there is more than one such sequence. But having taken
such a sequence, the corresponding sequence 2) is uniquely de
termined. Thus to each Q corresponds at least one P. A more
careful analysis shows that to a given Q never more than four
points P can correspond.
2. The method we have used here may obviously be extended
to space. By passing median planes through a unit cube we
divide it into 2 3 equal cubes. Thus to get our correspondence
each division D n should divide each interval and cube of the pre
ceding division D n _ l into 2 s equal parts. The cubes of each divi
sion should be numbered according to the 1 and 2 principles of
enumeration mentioned in 1.
By this process we define
as onevalued continuous functions of t such that as t ranges over
the unit interval (0, 1), the point a?, y, z ranges over the unit
cube.
574. 1. Hitterf s Curve. We wish now to study in detail the
correspondence between the unit interval 21 and the unit square
93 afforded by Hilbert's curve defined in 573. A number of inter
esting facts will reward our labor. We begin by seeking the
points P in 21 which correspond to a given Q in 93
To this end let us note how P enters and leaves an rj square.
Let B be a square of D n . In the next division B falls into four
PLANE CURVES 591
squares B l J5 4 and in the wf 2 d division in 16 squares B^ 3 .
Of these last, four lie at the vertices of B ; we call them vertex
squares. The other 12 are median squares. A simple considera
tion shows that the rj squares of JD n +2 are so numbered that we
always enter a square B belonging to 2> n , and also leave it by a
vertex square.
Since this is true of every division, we see on passing to the
limit that the point Q enters and leaves any rj square at the ver
tices of 77. We call this the vertex law.
Let us now classify the points P, Q.
If P is an end point of some division D n > we call it a terminal
point, otherwise an inner point, because it lies within a sequence
of 8 intervals 8' > 8" > = 0.
The points Q we divide into four classes :
1 vertex points, when Q is a vertex of some division.
2 inner points, when Q lies within a sequence of squares
V>T?">  =0.
3 lateral points, when Q lies on a side of some rj square but
never at a vertex.
4 points lying on the edge of the original square 93. Points
of this class also lie in 1, 3.
We now seek the points P corresponding to a Q lying in one of
these four classes.
Class 1. Q a Vertex Point. Let D n be the first division such
that Q is at a vertex. Then Q lies in four squares rj L , 77,, rj K , ij t of
D n .
There are 5 cases :
) ij k I are consecutive.
/3) ij k are consecutive, but not I.
7) LJ are consecutive, but not k I.
8) ij, also k /, are consecutive,
e) no two are consecutive.
A simple analysis shows that a), #) are not permanent in the
following divisions ; 7), 8) may or may not be permanent ; e) is
permanent.
592 GEOMETRIC NOTIONS
Now, whenever a case is permanent, we can enclose? Q in a se
quence of j] squares whose sides = 0. To this sequence corre
sponds uniquely a sequence of 8 intervals of lengths = 0. Thus
to two consecutive squares will correspond two consecutive inter
vals which converge to a single point P in 21. If the squares are
not consecutive, the corresponding intervals converge to two dis
tinct points in 21. Thus we see that when 7) is permanent, to Q
correspond three points P. When S) is permanent, to Q corre
spond two points P. While when Q belongs to e), four points P
correspond to it.
Class "2. Q an Tuner Point. Obviously to each Q corresponds
one point P and only one.
Class 3. Q a Lateral Point. To fix the ideas let Q lie on a ver
tical side of one of the ?/\s. Let it lie between rj^ ijj of D n . There
are two cases :
) j = i + 1.
We see easily that a) is not permanent, while of course /S) is.
Thus to each Q in class 3, there correspond two points P.
Class 4. Q lies on the edge of 8. If Q is a vertex point, to it
may correspond one or two points P. If Q is not a vertex point,
only one point P corresponds to it.
To sum up we may say :
To each inner point Q corresponds one inner point P.
To each lateral point Q correspond two points P.
To each edge point Q correspond one or two points P.
To each vertex point Q, correspond two, three, or four points P.
2. As a result of the preceding investigation we show easily
that :
To the points on a line parallel to one of the sides of S3 correspond
in 21 an apantactic perfect set.
3. Let us now consider the tangents to Hilbert's curve which
we denote by H.
PLANE CURVES
593
Let Q be a vertex point. We saw there were three permanent
cases 7), 8), e).
In cases 7), 8) we saw that to two consecutive 8 intervals cor
respond permanently two contiguous ver
tical or horizontal squares.
Thus as t ranges over L  j ' ^ ' Q
8 t i 8 t+1 , the point #, y ranges
over these squares, and the secant line
joining Q and this variable point cr, y oscillates through 180.
There is thus no tangent at Q. In case e) we see similarly that
the secant line ranges through 90. Again there is no tangent
at Q.
In the same way we may treat the three other classes. We find
that the secant line never converges to a fixed position, and may
oscillate through 360, viz. when Q is an inner point. As a result
we see that Hubert's curve has at no point a tangent, nor even a
unilateral tangent.
4. Associated with Hilbert's curve IT are two other curves,
The functions <, i/r being onevalued and continuous in 31, these
curves are continuous and they do not have a multiple point. A
very simple consideration shows that they do not have even a
unilateral tangent at a pantactic set of points in 21.
575. Property 7. Equations of a Curve. As already remarked,
it is commonly thought that the equation of a curve may be
written in any one of the three forms
y =/(*), (i
4>(x,y)=0, (2
and if these functions are continuous, these equations define con
tinuous curves.
Let us look at the Hilbert curve H. We saw its equation
could be expressed in the form 3). JSTcuts an ordinate at every
point of it for which < y < 1. Thus if we tried to define H by
594 GEOMETRIC NOTIONS
an equation of the type 1), /(V) would have to take on every
value between and 1 for each value of x in 21 = (0, 1). No such
functions are considered in analysis.
Again, we saw that to any value x = a in 21 corresponds a perfect
apan tactic set of values \t a \ having the cardinal number c. Thus
the inverse function of x = </>() is a manyvalued function of x
whose different values form a set whose cardinal number is c.
Such functions have not yet been studied in analysis.
How is it possible in the light of such facts to say that we may
pass from 8) to 1) or 2) by eliminating t from 3). And if we
cannot, how can we say a curve can be represented equally well
by any of the above three equations, or if the curve is given by
one of these three equations, we may suppose it replaced by one
of the other two whenever convenient. Yet this is often done.
In this connection we may call attention to the loose way
elimination is treated. Suppose we have a set of equations
We often see it stated that one can eliminate ^ t n and obtain
a relation involving the #'s alone. Any reasoning based on such
a procedure must be regarded as highly unsatisfactory, in view of
what we have just seen, until this elimination process has been
established.
576. Property 8. Closed Curves. A circle, a rectangle, an
ellipse are examples of closed curves. Our intuition tells us that
it is impossible to pass from the inside to the outside without
crossing the curve itself. If we adopt the definition of a closed
curve without multiple point given in I, 362, we find it no easy
matter to establish this property which is so obvious for the simple
closed curves of our daily experience. The first to effect the
demonstration was Jordan in 1892. We give here * a proof due
to de la Valle~ Poussin.f
Let us call for brevity a continuous curve without double point
. * The reader is referred to a second proof due to Brouwer and given in 698 seq.
t Cours # Analyse, Paris, 1903, Vol. 1, p. 307.
PLANE CURVES 595
a Jordan curve. A continuous closed curve without double point
will then be a closed Jordan curve. Cf . I, 362.
577. Lei C be a closed Jordan curve. However small or> is
taken, there exists a polygonal ring R containing C and such that
1 Each point of R is at a distance < cr from C.
2 Each point of C is at a distance < a from the edges of R.
For let x = <<T) , y = i/r(0 C 1
be continuous onevalued functions of in T = (a < 6) defining C.
Let D = (a, a v a 2 6) be a division of T of norm d. Let
a, j, 2 be points of corresponding to a, a l If d! is suffi
ciently small, the distance between two points on the arc
(7 t = ( t _i, t ) is <e', small at pleasure. Let A be a quadrate
division of the a?, y plane of norm S. Let us shade all cells con
taining a point of (7 t . These form a connected domain since O t is
continuous. We can thus go around its outer edge without a
break.* If this shaded domain contains unshaded cells, let us
shade these too. We call the result a link A,. It has only one
edge E t , and the distance between any two points of E, is ob
viously < f + 2 V2 8. We can choose d, S so small that
c' + 2V2~8 < (7, arbitrarily small. (1
Then the distance between any two points of A. is < <r. Let e"
be the least distance between nonconsecutive arcs (7 4 , We take
B so small that we also have
' (2
Then two nonconsecutive links A^ Aj have no point in common.
For then their edges would have a common point P. As P lies
on _Z7 t its distance from O, is < V2 S. Its distance from G j is also
< V2 B. Thus there is a point P t on C t , and a point Pj on 0$ such
that
* Here and in the following, intuitional properties of polygons are assumed as
known.
59G GEOMETRIC NOTIONS
But by hypothesis e" < 77. Hence
e"<2V2S,
which contradicts 2).
Thus the union of these links form a ring R whose edges are
polygons without double point. One of the edges, say (? t , lies
within the other, which we call Gr e . The curve lies within R.
The inner polygon (? 4 must exist, since nonconsecutive links have
no point in common.
578. 1. Interior and Exterior Points. Let tr l > cr 2 > =0.
Let JSj, /2 2 be the corresponding rings, and let
be their inner and outer edges. A point P of the plane not on
which lies inside some 6r t we call an interior or inner point of O.
If P lies outside some 6r c , we call it an exterior or outer point of C.
Each point P not on O must belong to one of these two classes.
For let p = Dist (J, (7); then p is > some er n . It therefore lies
within 6j^ n) or without (r^ w) , and is thus an inner or an outer point.
Obviously this definition is independent of the sequence of rings
\JR n \ employed. The points of the curve (7 are interior to each
Gr^ and exterior to each Gf[ n) .
Inner points must exist, since the inner polygons exist as al
ready observed. Let us denote the inner points by 3 and the
outer points by O. Then the frontiers of $ and Q are the curve C.
2. We show now that
1 Two inner points can be joined by a broken line L, lying in 3.
2 Two outer points can be joined by a broken line L e lying in O.
3 Any continuous curve $ joining an inner point i and an outer
point e has a point in common with O.
To prove 3, let
be the equations of $, the variable t ranging over an interval
T=(a</9), t=a corresponding to i and t=zj3 to e. Let t r be
PLANE CURVES 697
such that a<t< t f gives inner points, while t = t f does not give an
inner point. Thus the point corresponding to t = t r is a frontier
point of 3 and hence a point of O.
To prove 1. If A, B are inner points, they lie within some Gr, .
We may join J., JS, 6? t by broken lines L a , L^ meeting (3\ at the
points .A', J3', say. Let Gr^ be the part of (? t lying between A \
B 1 . Then
La + Gab + Lb
is a broken line joining A to jB.
The proof of 2 is similar.
579. 1. Let P f , P" correspond to t = t f , t = t n , on the curve
defined by 577, 1). If '<", we say P f precedes P" and write
Any set of points on C corresponding to an increasing set of
values of t is called an increasing set.
As t ranges from a to 6, the point P ranges over C in a direct
sense.
We may thus consider a Jordan curve as an ordered set, in the
sense of 265.
2. (I)e la Valise Poussin.) On each arc (7 t of the curve <7, there
exists at least one point P t  such that
may be regarded as the vertices of a closed polygon without double
point and whose sides are all < e.
For in the first place we may take 8 > so small that no square
of A contains a point lying on nonconsecutive arcs O t of C. Let
us also take A so that the point a corresponding to t = a lies
within a square, call it A^, of A. As t increases from t = a, there
is a last point P l on where the curve leaves S r The point P l
lies in another square of A, call it $ 2 , containing other points of
C. Let P 2 be the last point of O in S%. In this way we may
continue, getting a sequence 1).
There exists at least one point of 1) on each arc 0, . For other
wise a square of A would contain points lying on nonconsecutive
arcs O K . The polygon determined by 1) cannot have a double
598 GEOMETRIC NOTIONS
point, since each side of it lies in one square. The sides are < e,
provided we take SV2 < e, since the diagonal is the longest line
we can draw in a square of side S.
580. Existence of Inner Points. To show that the links form a
ring with inner points, Schonfliess* has given a proof which may
be rendered as follows :
Let us take the number of links to be even, and call them L^
Z 2 ,  2n . T nen A> ^3' ^6'" l* e en ti re ly outside each other.
Since L^ L% overlap, let P be an inner common point. Simi
larly let Q be an inner common point of _L 2 , L 3 . Then P, Q
lying within J& 2 may be joined by a finite broken line b lying
within L 2 . Let 5 2 be that part of it lying between the last point
of leaving L and the following point of meeting Z 8 . In this
way the pairs of links
L^L Z ; L 3 L 5 ;
define finite broken lines
No two of these can have a common point, since they lie in
nonconsecutive links. The union of the points in the sets
L l > *2 L S ' *4 " Al ' b 2n
we call a ring, and denote it by 9t. The points of the plane not
in $R fall into two parts, separated by 9t. Let Z denote the part
which is limited, together with its frontier. We call Z the inte
rior of 3t. That Z has inner points is regarded as obvious since
it is defined by the links
which pairwise have no point in common, and by the broken lines
^2 ' J 4 ' *6 "
each of which latter lies entirely within a link.
Let S
* Die Entwickelung der Lehre von den Punktmannigfaltigkeiten. Leipzig, 1908,
Part 2, p. 170.
PLANE CURVES 599
Then these 8 have pairwise no point in common since the J a
have not.
Let Z = 8 2 4 8 4 + + ? 2n +
Then $ > 0. For let us adjoin L% to 3t, getting a ring 9? 2 whose
interior call J 2 . That ! 2 has inner points follows from the fact
that it contains 4 , 8 6 Let us continue adjoining the links
L^ L Q Finally we reach .L 2n , to which corresponds the
ring 9? 2n , whose interior, if it exists, is Z 2n  If 2 ^ oes no ^ ex *st,
5E 2w _ 2 contains only 2n . This is not so, for on the edge of L^
bounding , is a point P, such that some O p (P) contains points
of no L except L r In fact there is a point P on the edge of L^
not in either L 2 or l/ 2n , as otherwise these would have a point in
common. Now, if however small p > is taken, j?) p (P) contains
points of some L other than L^ , the point P must lie in L K which
is absurd, since L^ has only points in common with .Z/ 2 , L% n , and
P is not in either of these. Thus the adjunction of L 2 , L,
L 2n produces a ring 9? 2n whose interior 2n does not reduce to ;
it has inner points.
581. Property 9. Area. That a figure defined by a closed
curve without double point, i.e. the interior of a Jordan curve,
has an area, has long been an accepted fact in intuitional geometry.
Thus Lindemann, Vorlesun<jen ilber Greometrie, vol. 2, p. 557, says
" einer allseitig umgrenzten Figur kommt ein bestimmter Flachen
inhalt zu." The truth of such a statement rests of course on
the definition of the term area. In I, 487, 702 we have given a
definition of area for any limited plane point set 21 which reduces
to the ordinary definition when 21 becomes an ordinary plane figure.
In our language 21 has an area when its frontier points form a
discrete set. Let
define a Jordan curve 6, as t ranges over 2 T =(a<6). The
figure 21 defined by this curve has the curve as frontier. In I,
708, 710, we gave various cases in which ( is discrete. The
reasoning of I, 710, gives us also this important case :
If one of the continuous functions <, ^ defining the Jordan curve
S, has limited variation in T, then is discrete.
600
GEOMETRIC NOTIONS
l(i 17
It was not known whether would remain discrete if the con
dition of limited variation was removed from both coordinates,
until Osgood * exhibited a Jordan curve which is not discrete.
This we will now discuss.
582. 1. Osgood' 8 Curve. We start with a unit segment
T = (0, 1) on the t axis, and a unit square /S in the xy plane.
We divide Tin to 17 equal parts
/Tf /jj rn / \
"M'*2'"*17' V
and the square S into 9 equal
squares
/Sj, $g, 5 ' A3j 7 , (2
by drawing 4 bands B l which
are shaded in the figure. On
these bands we take 8 segments,
marked heavy in the figure.
Then as t is ranging from left
to right over the even or black
intervals T^ T^  T 16 marked heavy in the figure, the point a?, y
on Osgood's curve, call it ), shall range univariantly over the
segments 3).
While t is ranging over the odd or white intervals 7p T 3 T 17
the point xy on 5 shall range over the squares 2) as determined
below.
Eacli of the odd intervals 1) we will now divide into 17 equal
intervals T tj and in each of the squares 2) we will construct
horizontal and vertical bands J? 2 as we did in the original square
S. Thus each square 2) gives rise to 8 new segments on )
corresponding to the new black intervals in 7, and 9 new squares
& L J corresponding to the white intervals. In this way we may
continue indefinitely.
The points which finally get in a black interval call & the
others are limit points of the /3's and we call them X. The point
* Trans. Am. Math. Soc., vol. 4 (1903), p. 107.
PLANE CURVES 601
on O corresponding to a point has been defined. The point of
O corresponding to a point X is defined to be the point lying in
the sequence of squares, one inside the other, corresponding to the
sequence of white intervals, one inside the other, in which X falls,
in the successive divisions of T.
Thus to each t in T corresponds a single point #, y in S. The
aggregate of these points constitutes Osgood's curve. Obviously
the #, y of one of its points are onevalued functions of t in J 7 , say
* = <KO , y = t(0 ( 4
The curve ) has no double point. This is obvious for points of
O lying in black segments. Any other point falls in a sequence
of squares
Si>St>s VK ...
to which correspond intervals
in which the corresponding t's lie. But only one point t is thus
determined.
The functions 4) are continuous. This is obvious for points ft
lying within the black intervals of T. It is true for the points X.
For X lies within a sequence of white intervals, and while t ranges
over one of these, the point on ) ranges in a square. But these
squares shut down to a point as the intervals do. Thus </>, ty are
continuous at t = X. In a similar manner we show they are con
tinuous at the end points of the black intervals.
We note that to t = corresponds the upper lefthand corner
of /S^ and to t = 1, the diagonally opposite point.
2. Up to the present we have said nothing as to the width of
the shaded bands ^ 2>
&\ ^V"
introduced in the successive steps. Let
be a convergent positive term series whose sum A < 1. We
choose S l so that its area is a r J? 2 so t ' ia ^ ^ ts area ^ s a 2> e ^ c 
Then = , = 1A (5
602 GEOMETRIC NOTIONS
as we now show. For ) lias obviously only frontier points ;
hence O = 0. Since O is complete, it is measurable and
Let = S O, and B={B n \. Then (9 < B. For any point
which does not lie in some B n lies in a sequence of convergent
squares A^ > S^ > which converge to a point of ). Now
On the other hand, B contains a null set of points of ), viz. the
black segments. Thus
= S = A , and hence 6 = 1  A
and 5) is established.
Thus Osgood's curve is continuous, has no double point, and its
upper content is 1 A.
3. To get a continuous closed curve (7 without double point
we have merely to join the two end points a, /3 of Osgood's curve
by a broken line which docs not cut itself or have a point in com
mon with the square S except of course the end points a, /3.
Then (7 bounds a figure g whose frontier is not discrete, and $
does not have an area. Let us call such curves closed Osgood
curves.
Thus we see that there exist regions bounded by Jordan curves
which do not have area in the sense current since the Greek
geometers down to the present day.
Suppose, however, we discard this traditional definition, and
employ as definition of area its measure. Then wo can say :
A figure g formed of a closed Jordan curve J and its interior 3 ;
has an area, viz. Meas g.
For Front $ = J. Hence g is complete, and is therefore meas
ureable.
We note that a _ f a
We have seen there are Jordan curves such that
J>0.
DETACHED AND CONNECTED SETS 608
We now have a definition of area which is in accordance with the
promptings of our geometric intuition. It must be remembered,
however, that this definition has been only recently discovered,
and that the definition which for centuries has been accepted leads
to results which flatly contradict our intuition, which leads us to
say that a figure bounded by a continuous closed curve has an
area.
583. At this point we will break off our discussion of the
relation between our intuitional notion of a curve, and the con
figuration determined by the equations
where <, ^ are onevalued continuous functions of t in an interval
T. Let us look back at the list of properties of an intuitional
curve drawn up in 563. We have seen that the analytic curve
1) does not need to have tangents at a pantactic set of points on
it ; no arc on it needs have a finite length ; it may completely fill
the interior of a square ; its equations cannot always be brought
in the forms y =/(#) or J 7 (^)=0, if we restrict ourselves to
functions /or F employed in analysis up to the present; it does
not need to have an area as that term is ordinarily understood.
On the other hand, it is continuous, and when closed and with
out double point it forms the complete boundary of a region.
Enough in any case has been said to justify the thesis that
geometric reasoning in analysis must be used with the greatest
circumspection.
Detached and Connected Sets
584. In the foregoing sections we have studied in detail some
of the properties of curves defined by the equations
Now the notion of a curve, like many other geometric notions, is
independent of an analytic representation. We wish in the fol
lowing sections to consider some of these notions from this point
of view.
(504 GEOMETRIC NOTIONS
585. 1. Let 21, 33 be point sets in wway space 9t m . If
Dist(2l, 33)>0,
we say 21, 33 are detached. If 21 cannot be split up into two parts
S3, such that they are detached, we say 21 has no detached parts.
If 21 = 33 } and Dist (S3, )>0, we say S3, 6 are detached parts
of 21.
.Let the set of points, finite or infinite,
a, a v 2 , b (1
be such that the distance between two successive ones is < e. We
call 1) an esequence between a, 6 ; or a sequence with segments
(#i ^ a i+i) f length < . We suppose the segments ordered so
that we can pass continuously from a to b over the segments without
retracing. If 1) is a finite set, the sequence is finite, otherwise
infinite.
2. Let 21 have no detached parts. Let a, b be two of its points.
For each e > 0, there exists a finite ^sequence between a, b, and lying
in 21.
For about a describe a sphere of radius e. About each point of
21 in this sphere describe a sphere of radius e. About each point
of 21 in each of these spheres describe a sphere of radius e. Let
this process be repeated indefinitely. Let S3 denote the points of
21 made use of in this procedure. If S3 < 21, let = 21  33. Then
Dist (S3, )>e, and 21 has detached parts, which is contrary to
hypothesis. Thus there are sets of espheres in 21 joining a and b.
Among these sets there are finite ones. For let $ denote the
set of points in 21 which may be joined to a by finite sequences ; let
= 21  g. Then Dist (g, )>e. For if <e, there is a point/
in $, and a point g in @ whose distance is < e. Then a and g can
be joined by a finite esequence, which is contrary to hypothesis.
3. If 21 has no detached parts, it is dense.
For if not dense, it must have at least one isolated point a.
But then a, and 21 a are detached parts of 21, which contradicts
the hypothesis.
4. Let 21, S3, be complete and 21 = ($, ). If 21 has no de
tached parts, S3, have at least one common point.
IMAGES 605
For if 53, S have no common point, S = Dist (53, S) is > 0.
But S cannot > 0, since 53, S would then be detached parts of 21.
Since S = and since S3, are complete, they have a point in
common.
5. If 21 is such that any two of its points may be joined by an
esequence lying in 21, where e > is small at pleasure, 21 has n<r
detached parts.
For if 21 had 53, ( as detached parts, let Dist (33, <) = S. Then
8 > 0. Hence there is no sequence joining a point of 53 with a
point of S with segments < 8.
6. If 21 is complete and has no detached parts, it is said to be
connected. We also call 21 a connex.
As a special case, a point may be regarded as a connex.
1. If 21 is connected, it is perfect.
For by 3 it is dense, and by definition it is complete.
8. If 21 is a rectilinear connex, it has a first point a and a last
point yS, and contains every point in the interval (a, /:?).
For being limited and complete its minimum and maximum
lie in 21 and these are respectively a and /3. Let now
*i>* 2 > = 
There exists an e r sequence C between a, /3. Each segment has
an e 2 sequence (7 2 . Each segment of <7 2 has an e 3 sequence (7 3 ,
etc. Let be the union of all these sequences. It is pantactic
in (a, /3). As 21 is complete,
21 = (, 0).
Images
586. Let a^AOi'O *n=/n(*i'"O C 1
be onevalued functions of t in the point set . As t ranges over
Z, the point x = (x l # n ) will range over a set 21 in an wway
space 3t n . We have called 21 the image of . Cf. I, 238, 3.
If the functions / are not onevalued, to a point t may correspond
several images x', x f! finite or infinite in number. Conversely
606 GEOMETRIC NOTIONS
to the point x may correspond several values of t. If to each
point t correspond in general r values of #, and to each x in
general 8 values of , we say the correspondence between J, 21 is
r to s. If r = s = 1 the correspondence is 1 to 1 or unifold ; if
r > 1, it is manifold. If r = 1, 21 is a simple image of J, other
wise it is a multiple image. If the functions 1) are onevalued
and continuous in J, we say 21 is a continuous image of J.
587. Transformations of the Plane. Example 1. Let
u = x sin y , v = x cos y. (1
We have in the first place
U 2 f V* = iE 2 .
This shows that the image of a line x = a, a=(), parallel to the
*/axis is a circle whose center is the origin in the u, v plane, and
whose radius is a. To the yaxis in the x, y plane corresponds
the origin in the u, v plane.
From 1) we have, secondly,
u
 = tan y.
v
This shows that the image of a line y = 6, is a line through the
origin in the u, v plane.
From 1) we have finally that u, v are periodic in y, having the
period 2 TT. Thus as #, y ranges in the band J5, formed by the
two parallels y TT, or TT < y < TT, the point u, v ranges over
the entire u, v plane once and once only.
The correspondence between B and the w, v plane is unifold,
except, as is obvious, to the origin in the w, v plane corresponds
the points on the t yaxis.
Let us apply the theorem of I, 441, on implicit functions. The
determinant A is here
= x.
sin y, cos y
x cos y, x sin y
As this is = when x, y is not on the yaxis, we see that the
correspondence between the domain of any such point and its
image is 1 to 1. This accords with what we have found above.
IMAGES
607
It is, however, a much more restricted result than we have found ;
for we have seen that the correspondence between any limited
point set 21 in B which does not contain a point of the yaxis and
its image is unifold.
588. Example 2. Let
the radical having tlie })ositive sign.
first quadrant Q in the x, y plane.
From 1 ) we have at once
<L u> < 1 ,
Hence the image of Q is a band
From 1) we get secondly
2 , (1
Let us find the image of the
v > 0.
parallel to the vaxis.
Hence
y = uv
, # = y Vl
,2 2
f ^ = fl
(2
Thus the image of a circle in $ whose center is the origin and
whose radius is a is a segment of a right line v == a.
When x = y = 0, the equations 1) do not deline the correspond
ing point in the t&, v plane. If we use 2) to define the corre
spondence, we may say that to the line v = in B corresponds the
origin in the #, y plane. With this exception the correspondence
between Q and B is uniform, as 1), 2) show.
The determinant A of 1) is, setting
r = V# 2 f # 2 ,
x y *
d(u, v} _
7*
x
r
r 3
Z + y*
for any point #, y different from the origin.
589. Example 8. Reciprocal Radii. Let be the origin in the
#, y plane and fl the origin in the u, v plane. To any point
P = (#, ?/) in the #, y plane different from the origin shall cor
respond a point Q = (u, v) in the u, v plane such that flQ has
608 GEOMETRIC NOTIONS
the same direction as OP and such that OP flQ = 1. Analyti
cally we have
x = \y , u = \v , X > 0,
and
From these equations we get
u
and also
x 2 2 2
ir + v 2 ir f tr
The correspondence between the two planes is obviously unifold
except that no point in either plane corresponds to the origin in
the other plane. We find for any point a?, y different from the
origin that . , , ^
A = ( Ml v ) = __ 1
~~ '
Obviously from the definition, to a line through the origin in
the x, y plane corresponds a similar line in the w, v plane. As xy
moves toward the origin, u, v moves toward infinity.
Let x, y move on the line x = a = 0. Then 1) shows that u, v
moves along the circle
a(w a + v 2 ) u =
which passes through the origin. A similar remark holds when
x, y moves along the line y = b = 0.
590. Such relations between two point sets 81, 93 as defined in
586 may be formulated independently of the functions f. In fact
with each point a of 81 we may associate one or more points 6 X , J 2
of 93 according to some law. Then 93 may be regarded as the
image of 81. We may now define the terms simple, manifold, etc.,
as in 586. When b corresponds to a we may write b ~ a.
We shall call 93 a continuous image of 31 when the following con
ditions are satisfied. 1 To each a in 81 shall correspond but one
b in 93, that is, 93 is a simple image of 81. 2 Let b ~ a, let a t , a 2
be any sequence of points in 81 which = a. Let b n ~ a n . Then
b n must =s b however the sequence \a n \ is chosen.
IMAGES 609
When S3 is a simple image of 21, the law which determines
which b of 93 is associated with a point a of 21 determines obviously
n onevalued functions as in 586, 1), where ^ t m are the m co
ordinates of a, and x l x n are the n coordinates of 6. We call these
functions 1) the associated functions. Obviously when S3 is a
continuous image, the associated functions are continuous in 21.
591. 1. Let 93 be a simple continuous image of the limited complete
set 21. Then 1 93 is limited and complete. If 2 21 is perfect and
only a finite number of points of 21 correspond to any point of 93, then
93 is perfect. If 3 21 is a connex, so is 93.
To prove 1. The case that 93 is finite requires no proof. Let
b 1 , b 2 ... be points of 93 which = /3. We wish to show that ft lies
in 93. To each b n will correspond one or more points in 21; call
the union of all these points a. Since 93 is a simple image, a is an
infinite set. Let a r # 2 .. be a set of points in a which = a, a
limiting point of 21. As 21 is complete, lies in 21. Let b ~ a.
Let b. n ~ a n . As a n =^= a, b, n = /3. But 93 being continuous, b,
must = b. Thus ft lies in 93. That 93 is limited follows from the
fact that the associated functions are continuous in the limited
complete set 21. To prove 2. Suppose that 3} had an isolated
point b. Let b ~ a. Since 21 is perfect, let a l , 2 = a. Let
b n ~ a n . Then as $8 is continuous, b n = 6, and b is not an isolated
point. To prove 3. We have only to show that there exists
an esequence between any two points a, ft of 93, small at pleasure.
Let a ~ a, ft ~ b. Since 21 is connected there exists an ^sequence
between a, b. Also the associated functions are uniformly con
tinuous in 21, and hence y may be taken so small that each segment
of the corresponding sequence in 93 is > e.
2. Let /(j ... t m } be onevalued and continuous in the connex 21,
then the image of 21 is an interval including its end points.
This follows from the above and from 585, 8.
3. Let the correspondence between 21, 93 be unifold. If 93 is a
continuous image of 21, then 21 is a continuous image of 93.
For let \b n \ be a set of points in 93 which = b. Let a n ^ 6 n ,
a ~ b. We have only to show that a n = a. For suppose that it
does not, suppose in fact that there is a sequence # tl , # l$ which
610 GEOMETRIC NOTIONS
= a = a. Let /3 ~ . Then i tl , J tl  = /3. But any partial se
quence of \b n \ must = b. Thus b = /3, hence a = a, hence a n = a.
4. A Jordan curve J is a unifold continuous image of an interval
T. Conversely if J is a unifold continuous image of an interval T,
there exist two onevalued continuous functions
, y =
such that as t ranges over T, the point x, y ranges over J. In case
J is closed it may be regarded as the image of a circle F.
All but the last part of the theorem has been already established.
To prove the last sentence we have only to remark that if we set
x r cos t , y = r sin t
we have a unifold continuous correspondence between the points
of the interval (0, 2 ?r*) and the points of a circle.
5. The first part of 4 may be regarded as a geometrical definition
of a Jordan curve. The image of a segment of the interval T or
of the circle F, will be called an arc of J.
592. Side Lights on Jordan Curves. These curves have been
defined by means of the equations
y = *() (1
As t ranges over the interval T = (a < 5), the point P = (#, y)
ranges over the curve J. This curve is a certain point set in the
x, y plane. We may now propose this problem : We have given
a point set & in the #, y plane ; may it be regarded as a Jordan
curve ? That is, do there exist two continuous onevalued func
tions 1) such that as t ranges over some interval T 7 , the point P
ranges over the given set S without returning on itself, except
possibly for t = a, t = J, when the curve would be closed?
Let us look at a number of point sets from this point of view.
593. Example 1.
y = gin 1 ^ x in the i nterva i g _ (_ i ? i) 9 but *
x
= , for x = 0.
IMAGES 611
Is this point set S a Jordan curve ? The answer is, No. For a
Jordan curve is a continuous image of an interval 21. By 591, 1,
it is complete. But S is not complete, as all the points on the
y axis, 1 < y < 1 are limiting points of 6, and only one of them
belongs to , viz. the origin.
2. Let us modify S by adjoining to it all these missing limiting
points, and call the resulting point set C. Is G a Jordan curve ?
The answer is again, No. For if it were, we can divide the inter
val T into intervals 8 so small that the oscillation of <, fy in any
one of them is < o>. To the intervals 8, will correspond arcs O t on
the curve, and two nonconsecutive arcs C t are distant from each
other by an amount > some e, small at pleasure. This shows that
one of these arcs, say C K , must contain the segment on the ^/axis
1 < y < ! Hut then Osc ^ = 2 as t ranges over the correspond
ing S K interval. Thus the oscillation of ^r cannot be made < e,
however small S K is taken.
3. Let us return to the set & defined in 1. Let A, E be the
two end points corresponding to x = 1, x = 1. Let us join them
by an ordinary curve, a polygon if we please, which does not cut
itself or &. The resulting point set $ divides all the other points
of the plane into two parts which cannot be joined by a contin
uous curve without crossing $. For this point of view $ must be
regarded as a closed configuration. Yet $ is obviously not complete.
On the other hand, let us look at the curve formed by removing
the points on a circle between two given points a, b on it. The
remaining arc 8 including the end points a, b is a complete set, but
as it does not divide the other points of the plane into two sepa
rated parts, we cannot say 8 is a closed configuration.
We mention this circumstance because many English writers
use the term closed set where we have used the term complete.
Cantor, who first introduced this notion, called such sets abge
schlossen, which is quite different from geschlossen = closed.
_i
594. Example 2. Let p = e ', for in the interval 21 = (0, 1)
except 9 = 0, where p = 0. These polar coordinates may easily be
replaced by Cartesian coordinates
i * 1
*cos0 = e*sin0 > in a >
612 GEOMETRIC NOTIONS
except 6 = 0, when x, y both = 0. The curve thus defined is a
Jordan curve.
Let us take a second Jordan curve
J),
with p = for 6 = 0, If we join the two end points on these
curves corresponding to 6 = 1 by a straight line, we get a closed
Jordan curve <7, which has an interior $, and an exterior ).
The peculiarity of this curve J is the fact that one point of it,
viz. the origin x y = 0, cannot be joined to an arbitrary point
f 3 hy a finite broken line lying entirely in Q ; nor can it be
joined to an arbitrary point in O by such a line lying in )
595. 1. It will be convenient to introduce the following terms.
Let 21 be a limited or unlimited point set in the plane. A set
of distinct points in 51
a l , a 2 , a 3 . (1
determine a broken line. In case 1) is an infinite sequence, let a n
converge to a fixed point. If this line has no double point, we call
it a chain, and the segments of the line links. In case not only the
points 1) but also the links lie in 31, we call the chain a path. If
the chain or path has but a finite number of links, it is called
finite.
Let us call a precinct a region, i.e. a set all of whose points are
inner points, limited or unlimited, such than any two of its points
may be joined by a finite path.
2. Using the results of 578, we may say that,
A closed Jordan curve J divides the other points of the plane into
two precincts, an inner Q and an outer ) Moreover, they have a
common frontier which is /.
3. The closed Jordan curve considered in 594 shows that not
every point of such a closed Jordan curve can always be joined to
an arbitrary point of 3 r O by a finite path.
Obviously it can ly an infinite path. For about this point, call
it.P, we can describe a sequence of circles of radii r = 0. Between
any two of these circles there lie points of $ and of ), if r is suf
IMAGES 613
ficiently small. In this way we may get a sequence of points in 3,
viz. /j, I 2 == P. Any two of these I m , I m+l may be joined by a
path which does not cut the path joining /j to I m . For if a loop
were formed, it could be omitted.
4. Any arc $ of a closed Jordan curve J can be joined by a path
to an arbitrary point of the interior or exterior, which call 21.
For let J= $ + . Let k be a point of $ not an end point.
Let S = Dist(&, ), let a be a point of 21 such that Dist(#, &)
< A S. Then T , . OA 1 ,,
2 77= Dist(>, 8) > 8.
Hence the link = (a, A) has no point in common with 8. Let
b be the first point of I in common with $. Then the link
m = (a, 6) lies in 21. If now a is any point of 21, it may be joined
to a by a path p. Then p + m is a path in 21 joining the arbi
trary point a to a point b on the arc $.
596. Example 3. For in 21 = (0*, 1) let
p = a(l + O,
and ,, , ( + ik
p = a(l + e v e/ )
These equations in polar coordinates define two nonintersecting
spirals S^, S 2 which coil about p = a as an asymptotic circle F.
Let us join the end points of the spirals corresponding to 6 = 1
by a straight line L. Let & denote the figure formed by the
spirals S^ /S^, the segment L and the asymptotic circle F. Is
a closed Jordan curve ? The answer is, No. This may be seen
in many ways. For example, 6 does not divide the other points
into two precincts, but into three, one of which is formed of points
within F.
Another way is to employ the reasoning of 593, 2. Here the
circle F takes the place of the segment on the ^axis which figures
there.
Still another way is to observe that no point on F can be joined
to a point within S by a path.
597. Example 4. Let S be formed of the edge @ of a unit
square, together with the ordinates o erected at the points
GEOMETRIC NOTIONS
x ~, of length , r&= 1, 2 Although 6 divides the other
points of the plane into two precincts $ and ), we can say that
S is not a closed Jordan curve.
For if it were, 3 and O would have to have S as a common
frontier. But the frontier of ) is (, while that of 3 is S.
That & is not a Jordan curve is seen in other ways. For
example, let 7 be an inner segment of one of the ordinates o.
Obviously it cannot be reached by a path in D.
Brouwer's Proof of Jordan's TJieorem
598. We have already given one proof of this theorem in 577
seq., based on the fact that the coordinates of the closed curve are
expressed as onevalued continuous functions
Brouwer's proof * is entirely geometrical in nature and rests
on the definition of a closed Jordan curve as the unifold continu
ous image of a circle, cf. 591, 5.
If 21, 33, are point sets in the plane, it will be convenient to
denote their frontiers by g^, g^  so that
5^= Front 21 , etc.
We admit that any closed polygon $ having a finite number of
sides, without double point, divides the other points of the plane
into an inner and an outer precinct ty t , $ e respectively. In the
following sections we shall call such a polygon simple, and usu
ally denote it by ty.
We shall denote the whole plane by (.
Let 21 be complete. The complementary set A is formed, as
we saw in 328, of an enumerable set of precincts, say A = \A n \.
* Math. Annalen, vol. 69 (1910), p. 169.
BROUWER'S PROOF OF JORDAN'S THEOREM 615
599. 1. If a precinct 31 and its complement* A each contain a
point of the connex S, then $% contains a point of S.
For in the contrary case c = JDv(8l, S) is complete. In fact
33 = ?l f gf^ is complete. As & is complete, Dv(58, S) is com
plete. But if ggi does not contain a point of , c = Dv($8, S).
Thus on this hypothesis, c is complete. Now c = Dv(A, (5) is
complete in any case. Thus S = c + c, which contradicts 585, 4.
2. If ^J t , *Pe, the interior and exterior of a simple polygon $ each
contain a point of a connex @, then ^ contains a point of S.
3. Let $ be complete and not connected. There exists a simple
polygon $ such that no point of $ lies on ^3, while a part of $ lies in
$,, and another part in ty e .
For let itj, $ 2 be two nonconnected parts of & whose distance
from each other is p > 0. Let A be a quadrate division of the
plane of norm S, so small that no cell contains a point of ^ 1 and
$ 2 . Let A! denote the cells of A containing points of $ r It is
complete, and the complementary set A 2 = @ AJ is formed of one
or more precincts. No point of St 1 lies in A 2 or on its frontier.
Let Pj, P 2 be points in $j, $ 2 respectively. Let D be that
precinct containing P 2 . Then $ D embraces a simple polygon $
which separates P l and P 2 .
4. Let $j, $ 2 be two detached connexes. There exists a sim.ple
polygon $ which separates them.. One of them is in *i)3 the other in
^ and no point of either connex lies on $.
For the previous theorem shows that there is a simple polygon
$ which separates a point P l in ^ l from a point P a in $ 2 and no
point of ff x or $ 2 lies on ty. Call this fact F.
Let now P l lie in *ip t . Then every point of $j lies in $ t . For
otherwise ^ and ^ each contain a point of the connex $j . Then
2 shows that a point of l lies on $, which contradicts J 7 .
5. Let $8 be a precinct determined by the connex (. Then
6 = Front 53 is a connex.
* Since the initial sets are all limited, their complements may be taken with ref
erence to a sufficiently large square ; and when dealing with frontier points, points
on the edge of jQ may be neglected.
616 GEOMETRIC NOTIONS
For suppose b is not a connex. Then by 3, there exists a simple
polygon $ which contains a part of b in ^ and another in ty e ,
while no point of b lies on $. Hence a point /3' of b lies in $ t ,
and another point ft" in $ e . As 35 is a precinct, let us join /3',
ft" by a path t; in 33. Thus $ contains at least one point of v,
that is, a point of 33 lies on *ij3. As b and ^ have no point in
common, and as one point of ty lies in 33i all the points of ty lie
in S3. Hence Dv($, 6) = 0. (1
As b is a part of ( and hence some of the points of 6 are in ^ e
and some in ty L , it follows from 2 that a part of ^ lies in g. This
contradicts 1).
6. Let $j, $ 2 be two connexes without double point. By 8
there exists a simple polygon ty which separates them and has
one connex inside, the other outside $.
Now $ = $} 4 $2 i s complete and defines one or more precincts.
One of these precincts contains ty.
For say ty lay in two of these precincts as 21 and 33 Then the
precinct 21 and its complement (in which 33 lies) each contain a
point of the connex $. Thus $% contains a point of $. But $$
is a part of , and no point of $ lies on ty.
That precinct in Comp $ which contains ty we call the inter
mediate precinct determined by $^ U 2 , or more shortly the pre
cinct between $ x , $ 2 and denote it by Inter ($ r $ 2 ).
7. Let $j, $ 2 t> e t wo detached connexes, and let I = Inter ($ r $ 2 ).
Then $ 19 $ 2 can be joined by a path lying in f, except its end points
which lie on the frontiers of ^ $ 2 respectively.
For by hypothesis p = Dist(^ ) 1 , $ 2 )>0. Let P l be a point of
5^ such that some domain b of P contains only points of $j and
of f. Let Q l be a point of f in b. Join P^ Q 1 by a right line, let
it cut 5% fi 1 ' 8 ^ a t the point P 1 . In a similar way we may reason
on $ 2 , obtaining the points P", Q 2 . Then P' Q^^P" is the path
in question. If we denote it by v> we may let v* denote this
path after removing its two end points.
8. Let $j, $ 2 be two detached connexes. A path v joining $j,
$ 2 and lying in f = Inter ($j, $ 2 )> en d points excepted, determines
one and only one precinct in I .
BKOUWEirS PROOF OF JORDAN'S THEOREM 617
For from an arbitrary point P in f, let us draw all possible
paths to v. Those paths ending on the same side (left or right)
of v certainly lie in one and the same precinct f r or fy in f. Then
since one end point of v is inside, the other end point outside $,
there must be a part of $ which is not met by v and which joins
the right and left sides of v. We take this as an evident property
of finite broken lines and polygons without double points.
Thus ti and ! r are not detached ; they are parts of one precinct.
9. Two paths jjj, # 2 without common point, lying in f and joining
$j, $' 2 , split t into two precincts.
Let i = f v l ; this we have just seen is a precinct. From any
point of it let us draw paths to # 2 . Those paths ending on the
same side of v% determine precincts t$, i r which may be identical.
Suppose they are. Then the two sides of v 2 can be joined by a
path tying in f, which does not touch v 2 (end points excepted),
has no point in common with Vj, and together with a segment of
v 2 forms a simple polygon ^ which has one end point of v^ in $ t ,
the other end point in ty e . Thus by 2, ty contains a point of the
connex v l . This is contrary to hypothesis.
Similar reasoning shows that
10. The n paths v l v n pairwise without common point, lying in
f, and joining the connexes $j, $ 2 split I into n precincts.
Let us finally note that the reasoning of 595, 4, being independ
ent of an analytic representation of a Jordan curve, enables us to
use the geometric definition of 591, 6, and we have therefore the
theorem
11. Let 21 he a precinct whose frontier is a Jordan curve. Then
there exists a path in 21 joining an arbitrary point of 21 with any arc
Having established these preliminary theorems, we may now
take up the body of the proof.
600. 1. Let 21 be a precinct determined by a closed Jordan curve
J. Then g = Front 21 is identical with J.
If J determines but one precinct 21 which is pantactic in (, we
have obviously g = J.
618 GEOMETRIC NOTIONS
Suppose then that SI is a precinct, not pantactic in (. Let S3
be a precinct =31 determined by ft. Let 6 = Front S3. Then
b <_ <L </. Suppose now b < J. As Jis a connex by 591, l, g is a
connex by 599, 6. Similarly since 8? is a connex, b is a connex.
Since b < <7i let 5 ~ b on the circle F whose image is J. We
divide b into three arcs J x , 6 2 , ^3 to which ~ b x , b 2 , b 8 in b.
Let /9 = Inter (b x , b 3 ).
Then by 599, 11, we can join b p b 3 by a path v 1 in SI, and by a
path v 2 in S3. By 599, 9, these paths split /3 into two precincts
$i> /3 2 . We can join v^ v% by a path u^ lying in ySj, and by a
path u 2 lying in /3 2 .
Now the precinct S3 and its complement each contain a point of
the connex u^. Hence by 599, 1, b contains a point of u v Simi
larly b contains a point of u%. Thus u^ u 2 cut b, and as they
do not cut bj, b 3 by hypothesis, they cut b 2 . Thus at least one
point of fii and one point of /3 2 He in b 2 .
Let p be a point of /3 l lying in b 2 , let p ~p on the circle. Let
b 1 be an arc of 6 2 containing p. Let b' ~b f . As the connex b ;
has no point in common with Front /3 X , b' must lie entirely in /3 X
by 599, 1. This is independent of the choice of b', hence the
connex b 2 , except its end points, lies in /3 V Thus /3 2 can contain
no point of b 2 , which contradicts the result in italics above.
Thus the supposition that b < J is impossible. Hence b = J,
and therefore g = *?
As a corollary we have :
2. A Jordan curve is apantactic in (.
3. A closed Jordan curve J cannot determine more than two
precincts.
For suppose there were more than two precincts
!, ^, a,  (i
Let us divide the circle F into four arcs whose images call <7 X , J%,
J v J t
Then by 1, the frontier of each of the precincts 1) is J. Thus
by 599, 9, there is a path in each of the precincts 2^, Slg join
ing J l and 7 8 . These paths split
DIMENSIONAL INVARIANCE 619
I = Inter (J^, Jg)
into precincts fj, f a
Now as in 1, we show on the one hand that each f t must contain
a point of J" 2 or J" 4 , and on the other hand neither 7 2 nor J 4 can
lie in more than one f t .
4. A closed Jordan curve J must determine at least two precincts.
Suppose that J determines but a single precinct 21. From a
point a of 51 we may draw two nonintersecting paths u^ u% to
points ij, J 2 of J.
Since the point a may be regarded as a connex, a and e/are two
detached connexes. Hence by 599, o, the paths w x , u z split ?l into
two precincts Sip 21 2 . Let / = (^, M 2 , <7). The points 6 X , 6 2
divide <7into two arcs Jj, J2, and
are closed Jordan curves. Regarding a and Jj as two detached
connexes, we see/! determines two precincts, a^ 0%. By 599, 1, a
path which joins a point a l of ^ with a point a 2 of 2 must cut j\
and hence y. It cannot thus lie altogether in Slj or in 21 2 Thus
both j, a 2 do not lie in 2lj, nor both in 21 2 . Let us therefore
say for example that 2^ lies in c^, and SL^ in 2 . Hence by 2,
2lj is pantactic in c^, and 2T 2 in 2 . By 1, each point of j\ is com
mon to the frontiers of ^ and of 03, and hence of ?Ij and of 21 2 ,
as these are pantactic.
Let P be a point of J^ . It lies either in a t or 0%. Suppose it
lies in a l . Then it lies neither in 2 nor on Front 2 , and hence
neither in S1 2 nor on Front 21 2 . But every point of / 2 and also
every point of j\ lies on Front 1 2 . We are thus brought to a
contradiction. Hence the supposition that J determines but a
single precinct is untenable.
Dimensional Invariance
601. 1. In 247 we have seen that the points of a unit interval
/, and of a unit square S may be put in one to one correspondence.
This fact, due to Cantor, caused great astonishment in the mathe
matical world, as it seemed to contradict our intuitional views
620 GEOMETRIC NOTIONS
regarding the number of dimensions necessary to define a figure.
Thus it was thought that a curve required one variable to define
it, a surface two, and a solid three.
The correspondence set up by Cantor is not continuous. On
the other hand the curves invented by Peano, Hilbert, and others
(cf. 573) establish a continuous correspondence between /and S,
but this correspondence is not one to one. Various mathemati
cians have attempted to prove that a continuous one to one corre
spondence between spaces of m and n dimensions cannot exist.
We give a very simple proof due to Lebesc/ue.*
It rests on the following theorem :
2. Let 21 be a point set in 9? m . Let Q < 31 be a standard cube
0<# t # t <2cr , i=l, 2" m.
Let Sj, E 2 "* be a finite number of complete sets so small that each
lies in a standard cube of edge or. If each point of 21 lies in one of
the S's, there is a point of 21 which lies in at least m f 1 of them.
Suppose first that each 6 t is the union of a finite number of
standard cubes. Let (Sj denote those GTs containing a point of
the face f x of Q lying in the plane x 1 = a r The frontier JJi of @i
is formed of a part of the faces of the CTs. Let F l denote that
part of ^ which is parallel to fj. Let O 1 = ^(Q, FI). Any
point of it lies in at least two ('s.
Let @ 2 denote those of the S's not lying altogether in @j and
containing a point of the face f 2 of Q determined by x 2 = a 2 . Let
.F 2 denote that part of Front ( 2 which is parallel to f 2 . Let
Q 2 = jDt^Qj, .F 2 ). Any point of it lies in at least three of the &'s.
In this way we may continue, arriving finally at Q m , any point
of which lies in at least m f 1 of the S's.
Let us consider now the general case. We effect a cubical
division of space of norm d<&. Let 0, denote those cells of D
which contain a point of & t . Then by the foregoing, there is a
point of 21 which lies in at least m + 1 of the (7's. As this is true,
however small d is taken, and as the (Ts are complete, there is at
least one point of 21 which lies in m 4 1 of the S's.
* Math Annalen, vol. 70 (1911), p. 166.
DIMENSIONAL INVAKIANCE (521
3. We now note that the space 9? m may be divided into congruent
cells so that no point is In more than m 4 1 cells.
For m = 1 it is obvious. For m = 2 we may
use a hexagonal pattern. We may also use
a quadrate division of norm 8 of the plane.
These squares may be grouped in horizontal
bands. Let every other band be slid a distance
^ 8 to the right. Then no point lies in more
than 3 squares. For m = 3 we may use a
cubical division of space, etc.
In each case no point of space is in more than m f 1 cells.
Let us call such a division a reticulation of 9f m .
4. Let 21 be a point set in 3J m having an inner point a. There is
no continuous unifold image 33 of 31 in 9i n , w=w, such that l)~a is
an inner point of $8.
For let n > m. Let us effect a reticulation H of 9t m of norm p.
If S > is taken sufficiently small A = J9 2 (a) lies in 21. Let
E '= Da(fl) ; if p is taken sufficiently small, the cells
0^0, 0. (1
of R which contain points of E, lie in A. Let the image of E be
@, and that of the cells 1) be
Si, 6a <.. (2
These are complete. Each point of ( lies in one of the sets 2).
Hence by 2, they contain a point /? which lies in n + 1 of them.
Then a~/3 lies in n f 1 of the cells 1). But these, being part of
the reticulation R, arc such that no point lies in more than m + 1
of them. Hence the contradiction.
602. 1. Sehonflies* Theorem. Let
u =/(, #) , v = g(x, #) (1
be one valued and continuous in a unit square A whose center is
the origin. These equations define a transformation T. If T is
regular, we have seen in I, 742, that the domain ^0 P (P) of a point
P = (#, y) within A goes over into a set E such that if Q~P
then D a (Q) lies in E, if cr >0 is sufficiently small.
622 GEOMETRIC NOTIONS
These conditions on /, g which make T regular are sufficient,
but they are much more than necessary as the following theorem
due to Schonfliess * shows.
2. Let A B+cbea unit square in the x, y plane, whose center
is the origin and whose frontier is c.
u =/(#, y} , v = g(x, y)
be onevalued continuous functions in A. As (x, y) ranges over A,
let (u, v) range over 21 = 93 f c where c ~ c. Let the correspondence
between A and 21 be uniform. Then c is a closed Jordan curve and
the interior c t of c is identical with 93.
That C is a closed Jordan curve follows from 576 seq., or 598
seq. Obviously if one point of 93 lies in C all do. For if /3 t , /8 e
are points of 93, one within c and the other without, let J t ~/3 t ,
J e ~&. Then J t , b e lying in B can be joined by a path in B
which has no point in common with c. The image of this path is
a continuous curve which has no point in common with c, which
contradicts 578, 2.
Let
be the equation of c in polar coordinates.
If < /*, < 1, the equation
P = rt(P)
defines a square, call it <?^, concentric with c and whose sides are
in the ratio fi : 1 with those of c. The equations of C M ~ <v are
W =/ 50 cos ,
These C M curves have now the following property :
If a point (p, q) is exterior {interior) to c Mo , it is exterior (in
terior) to c^ifor all /z such that
I A 6 "~ MO I ^ some e > 0.
For let PH be the distance of (/?, q) from a point (w, v) on c^.
Then , _  _
*Goettingen Nachrichten, 1899. The demonstration here given is due to Osgood,
Goett. Nachr., 1900.
AREA OF CURVED SURFACES 623
is a continuous function of 0, //. which does not vanish for /x = /A O ,
when < 6 < 2 TT. But being continuous, it is uniformly con
tinuous. It therefore does not vanish in the rectangle
' < < 2 7T.
We can now show that if 33<C it is identical with c t . To this
end we need only to show that any point /3 of c t lies on some c^.
In fact, as /x = 0, c^ contracts to a point. Thus ft is an outer point
of some c^, and an inner point of others. Let /LC O be the maximum
of the values of /JL such that /3 is exterior to all c u , if /A</LC O .
Then /3 lies on c Mo . For if not, is exterior to (V + e , by what we
have just shown, and /^ is not the maximum of p.
Let us suppose that 33 lay without c. We show this leads to a
contradiction. For let us invert with respect to a circle f, lying
in C . Then c goes over into a curve f, and 31 goes over into
3) = S + f . Then @ lies inside f . Let , ?? be coordinates of a
point of 3). Obviously they are continuous functions of z, y in
^' and A 3) , c~f, uniformly.
By what we have just proved, ( must fill all the interior of f.
This is impossible unless 21 is unlimited.
3. We may obviously extend the theorem 2 to the case
. u m =
and A is a cube in raway space 5K m , provided we assume that c, the
image of the boundary of A, divides space into two precincts
whose frontier is c.
Area of Curved Surfaces
603. 1. The Inner Definition. It is natural to define the area of a
curved surface in a manner analogous to that employed to define
the length of a plane curve, viz. by inscribing and circumscrib
ing the surface with a system of polyhedra, the area of whose
faces converges to 0. It is natural to expect that the limits of
the area of these two systems will be identical, and this common
limit would then forthwith serve as the definition of the area of
the surface. The consideration of the inner and the outer sys
624
GEOMETRIC NOTIONS
terns of polyhedra afford thus two types of definitions, which
may be styled the inner and the outer definitions. Let us look
first at the inner definition.
Let the equations of the surface 8 under consideration be
x =
y =
z =
v),
(1
the parameters ranging over a complete metric set 21, and a?, ?/, z
being one valued and continuous in 21.
Let us effect a rectangular division D of norm d of the u, v
plane. The rectangles fall into triangles t K on drawing the
diagonals. Such a division of the plane we call quasi rectangular.
J*Q=(u Q iV Q ) , P 1= O + 8, v) , jP 2 = (V v o + 7 ?)
be the vertices of ^. To these points in the u, v plane corre
spond three points ^ t = (# t , y t , 2 t ), 4=1, 2, 3, of $ which form the
vertices of one of the triangular faces r K of the inscribed polyhe
dron n^ corresponding to the division D. Here, as in the follow
ing sections, we consider only triangles lying in 21. We may do
this since 21 is metric.
Let X K , Y K , Z K be. the projections of T K on the coordinate planes.
Then, as is shown in analytic geometry,
where
2 X
*J *X<C
y*
y\ ~
i'y , A'
i"y , A"z
and similar expressions for Y K , Z K .
Thus the area of 11^ is
the summation extending over all the triangles t K lying in the
set SI.
Let x, y, z have continuous first derivatives in 31. Then
OV
AREA OF CURVED SURFACES
625
with similar expressions for the other increments. Let
dy dz
du du
dy dz
dv dv
D ==
dx dz
dx dy
du du
dx dz
dv dv
, c=
du du
dx dy
dv dt;
+~~" O \ + *7 f /~f i .
JK.)V K , J K == ( \S K "f f
(2
Then
*,= (
where a K j3 K <y K are uniformly evanescent with d in 21 Thus if
A, B, do not simultaneously vanish at any point of 2k we liave
as area of the surface 8
lim '/)= I V^h^H O^dudv. (3
*=o J*
2. An objection which at once arises to this definition lies in
the fact that we have taken the faces of our inscribed polyhedra
in a Very restricted manner. We cannot help asking, Would we
get the same area for $ if we had chosen a different system of
polyhedra ?
To lessen the force of this objection we observe that by replac
ing the parameters u, v by two new parameters it', v' we may
replace the above quasi rectangular divisions which correspond to
the family of right lines u = constant, v = constant by the infinitely
richer system of divisions corresponding to the family of curves
u f = constant, v r = constant. In fact, by subjecting u f , v r to cer
tain very general conditions, we may transform the integral 3)
to the new variables u f , v r without altering its value.
But even this does not exhaust all possible ways of dividing 21
into a system of triangles with evanescent sides. Let us there
fore take at pleasure a system of points in the u, v plane having
no limiting points, and join them in such a way as to cover the
plane without overlapping with a set of triangles t K . If each
triangle lies in a square of side c?, we may call this a triangular
division of norm d. We may now inquire if /$/> still converges
to the limit 3), as d = 0, for this more general system of divisions.
It was generally believed that such was the case, and standard
treatises even contained demonstrations to this effect. These
leinonst rations are wrong ; for Schwarz * has shown that by
* Werke, vol. 2, p. 309.
626
GEOMETRIC NOTIONS
properly choosing the triangular divisions D, it is possible to
make S D converge to a value large at pleasure, for an extensive
class of simple surfaces.
604. 1. Schwarzs Example. Let C be a right circular cylin
der of radius 1 and height 1. A set of planes parallel to the base
at a distance  apart cuts out a system of circles F x , F 2 Let
71
us divide each of these circles into m equal
arcs, in such a way that the end points of
the arcs on Fj, F 3 , F 5 lie on the same
vertical generators, while the end points of
F 2 , F 4 , F 6  lie on generators halfway
between those of the first set. We now
inscribe a polyhedron so that the base of
one of the triangular facets lies on one
circle while the vertex lies on the next circle above or below, as
in the figure.
The area t of one of these facets is
m
1
1 COS
m/
Thus
m * ri* 2m
There are 2 m such triangles in each layer, and there are n
layers. Hence the area of the polyhedron corresponding to this
triangular division D is
Sj> = 2k = 2 mn sin v + 4 sin * ~
m * n* 2m
Since the integers 7, n are independent of each other, let us
consider various relations which may be placed on them.
Case 1. Let n = \m. Then
m

2 m
= 2
. 7T /
7T
4
sin /
sm
7T ml. i 7T*
2m
m TT I X 2 /^ 2 2 4 ?ft 4
1
7T
m V
2m
as m = oo.
AREA OP CURVED SURFACES
627
Case 2. Let n = \m*. Then
m
Bin 51
m
1 *
44 7r 
m n
2 m
X m 2m
7T
2m
= 27T\[
l + X 2 , as m = oo.
4
Case 3. Let n = Xm 3 . Then
sin
7T
m
sin ~~
w 2 X 2
= +oo
as m == QO.
2. Thus only in the first case does S D converge to 2 TT, which
is the area of the cylinder C as universally understood. In the
2 and 3 cases the ratio h/b = 0. As equations of C we may
take
x = cos u , y = sin u , 2 = v.
Then to a triangular facet of the inscribed polyhedron will cor
respond a triangle in the u, v plane. In cases 2 and 3 this tri
angle has an angle which converges to TT as m = GO. This is not
so in case 1. Triangular divisions of this latter type are of great
importance. Let us call then a triangular division of the u, v
plane such that no angle of any of its triangles is greater than
TT e, where e > is small at pleasure but fixed, positive triangu
lar divisions. We employ this term since the sine of one of the
angles is > some fixed positive number.
605. The Outer Definition. Having seen one of the serious diffi
culties which arise from the inner definition, let us consider briefly
the outer definition. We begin with the simplest case in which
the equation of the surface S is
z =/(#, y), (1
/ being onevalued and having continuous first derivatives. Let
us effect a metric division A of the x, y plane of norm S, and on
628 GEOMETRIC NOTIONS
each cell d K as base, we erect a right cylinder (7, which cuts out an
element of surface 8*. from S. Let ^ be an arbitrary point of 8 K
and Z K the tangent plane at this point. The cylinder O cuts out
of Z K an clement A/S^ . Let V K be the angle that the normal to Z K
makes with the 2axis. Then
1
COS V K =
^v
\dy)*
and A S r  ^"
cos V K
The area of S is now defined to be
Urn 2A& (2
when this limit exists. Tlie derivatives being continuous, we have
at once that this limit is
)
Air
which agrees with the result obtained by the inner definition in
IJ03, 3).
The advantages of this form of definition are obvious. In the
first place, the nature of the divisions A is quite arbitrary ; however
they are chosen, one and the same limit exists. Secondly, the most
general type of division is as easy to treat as the most narrow, viz.
when the cells d K are squares.
Let us look at its disadvantages. In the first place, the elements
AiS K do not form a circumscribing polyhedron of S. On the con
trary, they are little patches attached to S at the points ty K , and
having in general no contact with one another. Secondly, let us
suppose that S has tangent planes parallel to the 2axis. The de
rivatives which enter the integral 603, 3) are no longer continuous,
and the reasoning employed to establish the existence of the limit
2) breaks down. Thirdly, we have the case that z is not one
valued, or that the tangent planes to S do not turn continuously,
or do not even exist at certain points.
AREA OF CURVED SURFACES 629
To get rid of these disadvantages various other forms of outer
definitions have been proposed. One of these is given by Ooursat
in his Cours d* Analyse. Instead of projecting an arbitrary
element of surface on a fixed plane, the xy plane, it is projected on
one of the tangent planes belonging to that element. Hereby the
more general type of surfaces defined by 603, 1) instead of those
defined by 1) above is considered. The restriction is, however,
made that the normals to the tangent planes cut the elements of
surface but once, also the first derivatives of the coordinates are
assumed to be continuous in 21. Under these conditions we get
the same value for the area as that given in 603, 3).
When the first derivatives of a?, /, z are not continuous or do
not exist, this definition breaks down. To obviate this difficulty
de la ValleePomsin has proposed a third form of definition in his
Cours d" Analyse, vol. 2, p. 30 seq. Instead of projecting the
element of surface on a tangent plane, let us project it on a plane
for which the projection is a maximum. In case that S has a con
tinuously turning tangent plane nowhere parallel to the zaxis, de
la ValleePoussin shows that this definition leads to the same
value of the area of S as before. He does not consider other cases
in detail.
Before leaving this section let us note that Jordan in his Cours
employs the form of outer definition first noted, using the paramet
ric form of the equations of S. In the preface to this treatise the
author avows that the notion of area is still somewhat obscure, and
that he has not been able u a definir d'une mani^re satisfaisante
1'aire d'une surface gauche que dans le cas ou la surface a un plan
tangent variant suivant une loi continue."
606. 1. Regular Surfaces. Let us return to the inner definition
considered in 603. We have seen in 604 that not every system of
triangular divisions can be employed. Let us see, however, if w r e
cannot employ divisions much more general than the quasi rec
tangular. We suppose the given surface is defined by
the functions $, i/r, ^ being onevalued, totally differentiable func
tions of the parameters u, v which latter range over the complete
680
GEOMETRIC NOTIONS
metric set 31. Surfaces characterized by these conditions we
shall call regular. Let
be the vertices of one of the triangles t K , of a triangular division
D of norm d of 21. As before let $ , ^ $ 2 be the corresponding
points on the surface S. Then
r
and similar expressions hold for the other increments. Also
dy
du
4 2 Jf '
I ' ***
where JT denotes the sum of several determinants, involving the
infinitesimals
a' a" 8' 8"
^y t tt y i Pz > Pz *
Similar expressions hold for Y K , Z*. We get thus
where A^ B, are the determinants 2) in 603. Then the area of
the inscribed polyhedron corresponding to this division D is
Let us suppose that
as it, v ranges over 1. Also let us assume that
Vt Vf *7f
A* * K 6*
(2
(3
AREA OF CURVED SURFACES 631
remain numerically < e for any division D of norm d< d , small
at pleasure, except in the vicinity of a discrete set of points, that
is, let 3) be in general uniformly evanescent in 8, as d = 0. Then
where in general
n .or
1 ' Cont 31
If now J., J5, (7 are limited and .Bintegrable in 81, we have at
once
lim S D =fdudv^A* + B* + <7 2
as in 603.
2. We ask now under what conditions are the expressions 3)
in general uniformly evanescent in 21 ? The answer is pretty evi
dent from the example given by Schwarz. In fact the equation
of the tangent plane Z at ^Q is
On the other hand the equation of the plane T=
is
x y
o,
r
or finally
Thus for 3) to converge in general uniformly to zero, it is nec
essary and sufficient that the secant planes T converge in general
uniformly to tangent planes. Let us call divisions such that the
faces of the corresponding inscribed polyhedra converge in general
uniformly to tangent planes uniform triangular divisions. For
such divisions the expressions 3) are in general uniformly evanes
cent, as d s= 0. We have therefore the following theorem :
3. Let W be a limited complete metric set. Let the coordinates
X) y, z be onevalued totally differentiate functions of the parame
632 GEOMETRIC NOTIONS
ters u, v in 21, such that A 2 + B* + C 2 is greater than some positive
constant^ and is limited and Rintegrable in ?l. Then
# = lim S D =
rfO
D denoting the class of uniform triangular divisions of norms d.
This limit we shall call the area of S. From this definition we
have at once a number of its properties. We mention only the
following :
4. Let ?(j, 2l m be unmixed metric sets whose union is 21. Let
$1, S m be the pieces of S corresponding to them. Then each S K
has an area and their sum is S.
5. Let 21 A be a metric part of 21, depending on a parameter \ = 0,
such that 21 A = 21. Then
limS^S.
A =
(). The area of S remains unaltered when S is subjected to a dis
placement or a transformation of the parameters as in I, 744 seq.
607. 1. Irregular Surfaces. We consider now surfaces which
do not have tangent planes at every point, that is, surfaces for
which one or more of the first derivatives of the coordinates rr, #, z
do not exist, and which may be styled irregular surfaces. We
prove now the theorem :
Let the coordinates x, y, z be onevalued functions of u, v having
limited total difference quotients in the metric set 21. Let D be a
positive triangular division of norm d<d Q . Then
Max S D
is finite and evanescent with 21.
For let the difference quotients remain < JJL. We have
But
cosec
=
AREA OF CURVED SURFACES 633
where K is the angle made by the sides P^P^ P^P^ As D is a
positive division, one of the angles of t K is such that cosec K is
numerically less than some positive number M. Thus
where /A, M are independent of K and d. Similar relations hold
for \Y K \, \Z K \. Hence
Sj> < 2 6 fji *M . t K = 6
where 77 > is small at pleasure, for d Q sufficiently small.
2. Let 21 awe? a;, y, z be as in 606, 3, except at certain points form
ing a discrete set a, the first partial derivatives do not exist. Let
their total difference quotients be limited in 21. Then
lim^= f
rf=0 J
where D denotes a positive trianyular division of norm d.
Let us first show that the limit on the left exists. We may
choose a metric part S3 of 21 such that S = 21 S3 is complete and
exterior to 21 and such that 93 is as small as we please. Let S
denote the area of the surface corresponding to S. The triangles
t K fall into two groups : Q^ containing points of 93 ; 6? 2 containing
only points of S. Then
S D = 2 VXJ + 17 + 2 = 2 + 2.
6', G,
But 93 may be chosen so small that the first sum is < e/4 for
any d < d Q . Moreover by taking d still smaller if necessary, we
have
2
03
Hence
Similarly for any other division D f of norm
<S/>.fl<s<e/2 , d f <d Q
decreasing d Q still farther if necessary. Thus
03i GEOMETRIC NOTIONS
Hence lira S D exists, call it S. Since S exists we may take d Q
so small that
\SS0\<e/2 , d<d .
This with 1) gives
tf
that is,
k = lim f
./g
by I, 724.
608. 1. The preceding theorem takes care of a large class of
irregular surfaces whose total difference quotients are limited.
In case they are not limited we may treat certain cases as follows:
Let us effect a quadrate division of the u, v plane of norm d,
and take the triangles t K so that for any triangular division D
associated with d, no square contains more than n triangles, and
no triangle lies in more than v squares; w, v being arbitrarily
large constants independent of d. Such a division we call a
quasi quadrate division of norm d. If we replace the quadrate by
a rectangular division, we get a quasi rectangular division.
We shall also need to introduce a new classification of functions
according to their variation in 31, or along lines parallel to the
u, v axes. Let D be a quadrate division of the w, v plane of norm
d<d . Let
W K = Osc/(w, v) , in the cell d*.
Then Max Sc^c?
is the variation of / in 81. If this is not only finite, but evanes
cent with 81, we say/ has limited fluctuation in SI. Obviously this
may be extended to any limited point set in wway space.
Let us now restrict ourselves to the plane. Let a denote the
points of 21 on a line parallel to the waxis. Let us effect a divi
sion D f of norm d 1 . Let a>' K = Osc/(w, v) in one of the intervals
of D'. Then
rj a = Max 2o>
is the variation of /in a.
AREA OF CURVED SURFACES 635
Let us now consider all the sets a lying on lines parallel to the
is, and let
If now there exists a constant Or independent of a such that
that is, if ?; a is uniformly evanescent with tr, we say that/(w, v)
has limited fluctuation in 31 with respect to u.
With the aid of these notions we may state the theorems :
2. Let the coordinates x, y, z be onevalued limited functions in
the limited complete set 31. Let x, y have limited total difference
quotients, while z has limited variation in 31. Let D denote a quasi
quadratic division of norm d<d Q . Then
Max S D
. t D
is finite.
For, as before,
2jr ic <At.Ai' + Aj,'.A;.
But p denoting a sufficiently large constant,
Ai!, A^ are </**.
Let o) t = Osc z in the square s t . If the triangle t K lies in the
squares * 4 , s v
Thus, n denoting a sufficiently large constant,
2 JST.
the summation extending over those squares containing a triangle
of D. But z having limited variation,
2ft> t d < some M.
Hence 2 ^ ^ 2 ( ^ ( ^
Finally, as in 607,
2 \Z \ <some M r .
The theorem is thus established.
636 GEOMETRIC NOTIONS
3. The coordinates #, y, 2, being as in 2, except that z has limited
fluctuation in 21, and D denoting a quasi quadrate division of
norm d < c? ,
Max S D
_ D
is finite and evanescent with 21.
The reasoning is the same as in 2 except that now M, M f are
evanescent with 21.
4. Let the coordinates x, y, z have limited total difference quo
tients in 21, while the variation of z along any line parallel to the u
or v axis is < M. Let 21 lie in a square of side s = 0. Then
Max S D <sGr,
D
where G is some constant independent of s, and D is a quasi rectan
gular division of norm d < d Q ,
For here
22  X K  < 2  A'y 1  A"*  + 2  A"y   A'* 
where M 1 denotes a sufficiently large constant ; d u , d v denote the
length of the sides of one of the triangles t K parallel respectively
to the u, v axes, and <W M , co v the oscillation of z along these sides.
Since the variation is < M in both directions,
Ms.
V
Similarly
2a) v d u < M,.
The rest of the proof follows as before.
5. The symbols having the same meaning as before, except that z
has limited fluctuation with respect to u, v,
The demonstration is similar to the foregoing. Following the
line of proof used in establishing 607, 2 and employing the
theorems just given, we readily prove the following theorems :
AREA OF CURVED SURFACES 637
6. Let 21 be a metric set containing the discrete set a. Let b be
a metric part of 21, containing a such that 33 = 21 b is exterior to a,
and b == 0. Let the coordinates #, y, z be onevalued totally differ
entiable functions in 33, such that A 2 4 B* h O 2 never sinks below a
positive constant in any 33, is properly Rintegrable in any 93, and
improperly integrable in 21. Let x, y have limited total difference
quotients, and z limited fluctuation in b. Then
lim So = f V2 2
d=o 'a
lim S = V2 2 "+ & + C*dudv
where A, B, are the determinants in 603, 2), and D is any quasi
quadrate division of norm d.
7. Let the symbols have the same meaning as in 6, except that
1 a reduces to a finite set.
2 z has limited variation along any line parallel to the u, v axes.
3 D denotes a uniform quasi rectangular division. Then
= fVZ 2 + & + CPdudv.
^2l
8. :7% symbols having the same meaning as in 6, except that
1 z has limited fluctuation with respect to u, v in b.
2 D denotes a uniform quasi rectangular division. Then
lim S D = I V^l 2 + * + C^ttdto.
d=0
0. If we call the limits in theorems 6, 7, 8, area, the theorems
606, 3, 4, 5 still hold.
INDEX
(Numbers refer to pages)
Abel's identity, 87
series, 87
Absolutely convergent integrals, 31
series, 79
products, 247
Addition of cardinals, 292
ordinals, 312
series, 128
Adherence, 340
Adjoint product, 247
series, 77, 139
set of intervals, 337
Aggregates, cardinal number, 278
definition, 276
distribution, 295
enumerable, 280
equivalence, 276
eutactic, 304
exponents, 294
ordered, 302
power or potency, 278
sections, 307
similar, 303
transfinite, 278
uniform or 11 correspondence, 276
Alternate series, 83
Analytical curve, 582
Apantactic, 325
Area of curve, 599, 602
surface, 623
Arzela, 365, 555
Associated simple series, 144
products, 247
multiple series, 145
normal series, 245
logarithmic series, 243
inner sets, 365
Associated, outer sets, 365
nonnegative functions, 41
Baire, 326, 452, 482, 587
Bernouillian numbers, 265
Bertram's test, 104
Bessel functions, 238
Beta functions, 267
Binomial series, 110
Bocher, 165
Bonnet's test, 121
Borel, 324, 542
Brouwer, 614
Cahen's test, 340
Cantor's 1 and 2 principle, 316
theorem, 450
Category of a set, 326
Cauchy's function, 214
integral test, 99
radical test, 98
theorem, 90
Cell of convergence, 144
standard rectangular, 359
Chain, 612
Class of a function, 468, 469
Conjugate functions, 238
series, 147
products, 249
Connex, 605
Connected sets, 605
Contiguous functions, 231
Continuity, 452
infra, 487
semi, 487
supra, 487
Continuous image, 608
689
640
INDEX
Contraction, 287
Convergence, infrauniform, 562
monotone, 176
uniform, 156
at a point, 157
in segments, 556
subuniform, 555
Coproduct, 212
Curves, analytical, 582
area, 599, 602
Faber, 546
Jordan, 595, 610
Hilbert, 590
length, 579
nonintuitional, 537
Osgood, 600
Pompeiu, 542
rectifiable, 583
spacefilling, 588
D'Alembert, 96
Deleted series, 139
Derivates, 494
Derivative of a set, 330
order of, 331
Detached sets, 604
Dilation, 287
Dini, 176, 185, 438, 538
series, 86
Discontinuity, 452
at a point, 454
of 1 kind, 416
of 2 kind, 455
pointwise, 457
total, 457
Displacement, 286
Distribution, 295
Divergence of a series, 440
Division, complete, 30
separated, 366, 371
unmited, 2
of series, 196
of products, 253
Divisor of a set, 23
quasi, 390
Divisor, semi, 390
Du Bois Reymond, 103
< c , (S^osets, 473
Elimination, 594
Enclosures, complementary c, 355
deleted, 452
distinct, 344
divisor of, 344
e, 355
measurable, 356
nonoverlapping, 344
null, 366
outer, 343
standard, 359
Enumerable, 280
Equivalent, 276
Essentially positive series, 78
negative series, 78
Euler's constant, 260
Eutactic, 301
Exponent*, 29 1
Exponential series, 96
Extremal sequence, 374
Faber curves, 516
Fluctuation, 63 1, 635
Fourier's coefficient, 416
constants, 416
series, 416
Function, associated nonnegative func
tions, 41
Basel's, 238
Beta, 267
Cauchy's, 214
class of, 468, 469
conjugate, 233
contiguous, 231
continuous, 452
infra, 487
semi, 487
supra, 487
discontinuous, 452
of 1 kind, 410
of 2 kind, 455
INDEX
641
Function, Gamma, 267
Gauss' II(a;), 238
hyperbolic, 228
hypergeometric, 228
lineooscillating, 528
maximal, 488
measurable, 338
minimal, 488
monotone, 137
null, 385
oscillatory, 488
pointwise discontinuous, 457
residual, 561
Riemann's, 459
totally discontinuous, 457
truncated, 27
uniformly limited, 160, 567
Volterra's, 501, 583
Weierstrass', 498, 523, 581, 588
Gamma function, 267
Gauss' function U(x), 238
test, 109
Geometric series, 81, 139
Harnack, divergence of series, 440
sets, 354
Hermite, 300
Hubert's curves, 590
Hobson, 389, 412, 555
Hyperbolic functions, 228
Hypercotnplete sets, 472
Hypergeometric functions, 229
series, 112
Images , simple, multiple, 606
unifold, manifold, 606
continuous, 606, 608
Integrals, absolutely convergent, 31
L or Lebesgue, proper, 372
improper, 403, 405
improper, author's, 32
classical, 20
de la ValleePoussin, 27
inner, 20
Integrals, R or Riemannian, 372
Integrand set, 385
Intervals, of convergence, 90
adjoint set of, 337
set of, belonging to, 337
Inversion, geometric, 287
of a series, 204
Iterable sets, 14
Iterated products, 251
series, 149
Jordan curves, 595, 610
variation, 430
theorem, 436
Kdnig, 527
Rummer's test, 106, 124
Lattice points, 137
system, 137
Law of Mean, generalized, 505
Layers, 555
deleted, 563
Lebesque or L integrals, 372
theorems, 413, 424, 426, 452, 475 S
520, 619
Leibnitz's formula, 226
Length of curve, 579
Lindermann, 300, 599
Lineooscillating functions, 528
Link, 612
Liouuille numbers, 301
Lipschitz, 438
Logarithmic series, 97
Luroth, 448
Maclaurin's series, 206
Maximal, minimal functions, 488
Maximum, minimum, 521
at a point, 485
Measure, 348
lower, 348
upper, 343
Mertens, 130
Metric sets, 1
642
INDEX
Monotone convergence, 176
functions, 137
MooreOsgood theorem, 170
Motion, 579
Multiplication of series, 129
cardinals, 293
ordinals, 314
infinite products, 253
Normal' form of infinite product, 245
Null functions, 385
sets, 348
Numbers, Bernouillian, 265
cardinal, 278
class of ordinal numbers, 318
limitary, 314.
Liouville, 301
ordinal, 310
rank of limitary numbers, 331
Ordered sets, 302
Order of derivative of a set, 331
Oscillation at a point, 464
Oscillatory function, 488
Osfjfood curves, 600
Moore theorem, 170
theorems, etc., 178, 555, 622
Pantactic, 325
Path, 612
Peaks, 179
infinite, 566
Poly ant, 153
Point sets, adherence, 340
adjoint set of intervals, 337
apantactic, 325
associated inner set, 365
outer set, 365
Baire sets, 326
category 1 and 2, 326
coherence, 340
conjugate, 51
connected, 605
convex, 605
detached, 604
Point sets, divisor, 23
<B., <E = s e ^, 473
Harnack sets, 354
hypercomplete, 472
images, 605, 606
integrand sets, 385
iterable, 14
measurable, 343, 348
metric, 1
negative component, 37
null, 348
pautactic, 325
positive component, 37
potency or power. 278
projection, 10
quasi divisor, 390
reducible, 336
reticulation, 621
semidivisor, 390
separated intervals, 337
sum, 22
transfinite derivatives, 330
union, 27
wellordered, 304
Pointwise discontinuity, 457
Pompeiu, curves, 542
Potency or power of a set, 278
Power series, 89, 144, 187, 191
Precinct, 612
Pringsheim, theory of convergence, 113
theorems, etc., 141, 215, 216, 217,
220, 273
Projection, 10
Products, absolute convergence, 247
adjoint, 247
associate simple, 247
conjugate, 249
coproduct, 242
iterated, 251
normal form, 245
Quasidivisor, 390
Raabe's test, 107
Rank of limitary numbers, 331
INDEX
643
Rate of convergence or divergence, 102
Ratio test, 96
Reducible sets, 335
Remainder series, 77
of Taylor's series, 209, 210
Rectifiable curves, 583
Regular points, 428
Residual function, 561
Reticulation, 621
Richardson, 32
Riemann's function, 459
theorem, 444
R or Riemann integrals, 372
Rotation, 286
Scheefer, theorem, 516
Schoitfliess, theorems, 598, 621
Schtrarz, theorem, etc., 448, 626
Section of an aggregate, 307
Segment, constant, or of invariability,
521
Semidivisor, 390
Separated divisions, 366, 371
functions, 403
sets, 366
of intervals, 337
Sequence, extremal, 374
w tuple, 137
Series, Abel's, 87
absolute convergent, 79
adjoint, 77, 139
alternate, 83
associate logarithmic, 243
normal, 245
simple, 144
multiple, 144
Bessels, 238
binomial, 110
cell of convergence, 144
conjugate, 147
deleted, 139
Dini's, 86
divergence of, 440
essentially positive or negative, 78
exponential, 96
Series, Fourier's, 416
geometric, 81, 139
harmonic, 82
general of exponent p,, 82
hypergeometric, 112
interval of convergence, 90
inverse, 204
iterated, 149
logarithmic, 97
Maclaurin's, 206
power, 89, 144, 187
rate of convergence or divergence,
102
remainder, 77
simple convergence, 80
Taylor's, 206
tests of convergence, see Tests
telescopic, 85
trigonometric, 88
twoway, 133
Similar sets, 303
Similitude, 287
Simple convergence of series, 80
Singular points, 26
Spacefilling curves, 588
Steady convergence, 176
Submeasurable, 405
Sum of sets, 22
Surface, area, 623
irregular, 632
regular, 629
Taylor's series, 206
Telescopic series, 85
Tests of convergence, Bertram, 104
Bonnet, 121
Cahen, 108
Cauchy, 98, 99
d'Alembert, 96
Gauss, 109
Rummer, 106, 124
Pringsheim, 123
Raabe, 107
radical, 98
ratio, 96
644
INDEX
Tests of convergence, tests of 1 and 2 Uniformly limited function, 160, 567
kind, 120
Weierstrass, 120
Theta functions, 135, 184, 256
Total discontinuity. 457
Transfinite cardinals, 278
derivatives, 330
Translation, 286
Trigonometric series, 88
Truncated function, 27
Twoway series, 133
Undetermined coefficients, 197
Unifold image, 606
Uniform convergence, 156
at a point, 157
correspondence, 276
Union of sets, 22
ValleePoussin (dela), 27, 594
Van Vleck sets, 361
Variation, limited or finite, 429, 530
positive and negative, 430
Volterra curves, 501, 587
Wall is formula, 260
Weierstrass' function, 498, 523, 588
test, 120
Wellordered sets, 304
Wilson, W. A., vii, 395, 401
Young, W. H., theorems, 360, 363
Zeros of power series, 191
SYMBOLS EMPLOYED IN VOLUME II
(Numbers refer to pages)
Frontal. F^, 614
f*,20
J*
ft, i
U,{ },22
Dv, 22
Adj J, 31
/A,M,31
SU.0,32. S*/, a , 0,34
fy, *_., 34
A H ,A H ,AdjA,77. A n , p , 78
^ = ^...^,138; 1, = ^...^ 139
R v  ,..., 139
31  23, 276 ; % & 8, 303
Card a, 278
e = ,280; c, 287
91,, 290
So, 307
Ord 31, 311
w ,311; O, 318
KpKj, 318,323
Z r Z 2 ..,318
a<> = 21", 330; a<> = 5l a , 331
8 = Mea^ , 343 ; g = Meas , 348
8 = Meas $, 348
T, /, /', 372, 403, 405
Sdv, Qdv, 300
V* 429; Var/= V,, 429
Osc / = oscillation in a given set.
Osc/, 454
Disc/ 454
xa=a
e, e.io, 473
/>),/(*), 488
/'(*),/'(*), 493
/', .ff/', /', X/', 7/', Uf, 5(o),
.R(a), 494
A(a, )8), 494
INDEX
645
The following symbols are defined in Volume I and are repeated here for
the convenience of the reader.
l)ist(a, x) is the distance between
a and x
D 6 (), called the domain of the point
a of norm 8 is the set of points or,
such that Dist (a, x) < 8
F (5 (</), called the vicinity of the point
a of norm 8, refers to some set $,
and is the set of points in />$()
which lie in $1
7)5* (a) 5 TV (a) are the same as the
above sets, omitting . They are
called deleted domains, deleted vi
cinities
a n == means a n converges to a
f(x) = , means /(x) converges to a
A line of symbols as :
< 0, w, I  M I < c, n > m
is of constant occurrence, and is to
be read : for each c > 0, there exists
an index /, such that  a a n  < e,
for every n > m
Similarly a line of symbols as :
>0, 8>0, /(*) !<,* iii IV()
is to be read : for each c > 0, there
exists a 8 > 0, such that
!/(*) !<,
for every x in Fg* (a)