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124 XIII INTEGRATION

(13.5.11) and the fact that all finite sets are negligible (for Lebesgue meas-
ure) (13.6).

If l = [a, b] is a bounded interval, the "Dirichlet function" (3.11)
cp_l — cp_l <p_Q (which is 0 at the rational points of I and on the complement of
I, and 1 at the irrational points of I) is integrable, and its integral is b — a,
since the set Q is negligible (13.6).

(13.7.9) A subset A ofX. is integrable if and only if, given any B > 0, there
exists a compact set K and an open set G such that KczAcG and
H*(G - K) g 6.

The condition is sufficient by virtue of the definition of integrable functions
and (12.7.4). Suppose conversely that A is integrable, and let e be such that
0 < s < 1. By hypothesis, there exists h e«/ such that (p_A^h and

(/z — cp_A) d/j,^ e.

Let G be the set of points x e X such that h(x) > 1 — e; then G is open
(12.7.2) and contains A. Clearly h ^> (1 — fi)<po» hence

1 f i

KG)£-----\hdii£-----(/z(A) -f e).

1 -ej 1 -e

Also there exists g e £P such that g ^ <p_A and f (<p_A — g) d^ g e, and by the

definition of 5^ the set of points x e X such that g(x) > 0 is relatively compact.
Choose 5 > 0 so that §n(A) ^ e, and let K be the set of all x e X such that
g(x) ^ <5; then K is closed in X (12.7.2) and therefore compact (3.17.3), since
it is contained in a relatively compact set. Clearly we have K c: A and
9 =* <PK + ^B > where B = A — K; hence

g dp £ /i(K) + ^(B) ^ XK) + ^(A) ^ ^(K) + e,
and finally //(A) g jg dp 4- e ^ A*(K) + 2e. Q.E.D.

There exist bounded subsets of R which are not integrable with respect to
Lebesgue measure (Section 13.21, Problem 6).

(13.7.10) Let n : X-> X' be a homeomorphism. It follows directly from the
definitions that if/' is any function on X' with values in R, we have

f V • *) dp, f/'dOrGO) = f (/' o n) dfi.



	124 XIII INTEGRATION (13.5.11) and the fact that all finite sets are negligible (for Lebesgue meas- ure) (13.6). If l = [a, b] is a bounded interval, the "Dirichlet function" (3.11) cp_l — cp_l <p_Q (which is 0 at the rational points of I and on the complement of I, and 1 at the irrational points of I) is integrable, and its integral is b — a, since the set Q is negligible (13.6). (13.7.9) A subset A ofX. is integrable if and only if, given any B > 0, there exists a compact set K and an open set G such that KczAcG and H(G - K) g 6. The condition is sufficient by virtue of the definition of integrable functions and (12.7.4). Suppose conversely that A is integrable, and let e be such that 0 < s <* 1. By hypothesis, there exists h e«/ such that (p_A^h and (/z — cp_A) d/j,^ e. Let G be the set of points x e X such that h(x) > 1 — e; then G is open (12.7.2) and contains A. Clearly h ^> (1 — fi)<po» hence 1 f i KG)£-----\hdii£-----(/z(A) -f e). 1 -ej 1 -e Also there exists g e £P such that g ^ <p_A and f (<p_A — g) d^ g e, and by the definition of 5^ the set of points x e X such that g(x) > 0 is relatively compact. Choose 5 > 0 so that §n(A) ^ e, and let K be the set of all x e X such that g(x) ^ <5; then K is closed in X (12.7.2) and therefore compact (3.17.3), since it is contained in a relatively compact set. Clearly we have K c: A and 9 =* <PK + ^B > where B = A — K; hence

	g dp £ /i(K) + ^(B) ^ XK) + ^(A) ^ ^(K) + e, and finally //(A) g jg dp 4- e ^ A*(K) + 2e. Q.E.D.

	There exist bounded subsets of R which are not integrable with respect to Lebesgue measure (Section 13.21, Problem 6). (13.7.10) Let n : X-> X' be a homeomorphism. It follows directly from the definitions that if/' is any function on X' with values in R, we have

	f V • ) dp,* f/'dOrGO) = f (/' o n) dfi.