Page 86 - Vector Analysis

82 CHAPTER 3. Multiple Integrals

Proof. By (a), (d) and (e) of Proposition 3.14,

żżż

0 = 1B(x) dx = 1B(x) dx = 1B(x) dx

AAB

and ˝
żżż

0 = 1B(x) dx = 1B(x) dx = 1B(x) dx .

AAB

ż

Therefore, 1B(x) dx = 0 which implies that B has volume zero.

B

k

Lemma 3.22. Let A1, ¨ ¨ ¨ , Ak Ď Rn be bounded sets of volume zero. Then Ť Aj has volume

zero. j=1

Proof. It suffices to prove the case for k = 2. Suppose that A1 and A2 are bounded sets
of volume zero, and A = A1 Y A2. By Lemma 3.21, A1 X A2 has volume zero; thus (e) of
Proposition 3.14 and Corollary 3.16 imply that

żż żż

1A(x) dx = 1A(x) dx ě 1A(x) dx + 1A(x) dx = 0

A A1YA2 A1 A2

and

żż żż

1A(x) dx = 1A(x) dx ď 1A(x) dx + 1A(x) dx = 0 .

A A1YA2 A1 A2

ż ˝

Therefore, 1A(x) dx = 0 which implies that A has volume zero.

A

Theorem 3.23. Let A Ď Rn be a bounded set such that B A has volume zero, and f : A Ñ R

be a bounded function. If f is continuous except perhaps on a set of volume zero, then f is

Riemann integrable over A.

Proof. Let R be a closed cube such that A Ď R and B A X B R = H. We show that f A = f 1A
is Riemann integrable over R and by (a) of Proposition 3.14, we then obtain that

żż ż ż ż

f (x) dx = (f 1A)(x) dx = (f 1A)(x) dx = (f 1A)(x) dx = f (x) dx

AR R R A

which implies that f is Riemann integrable over A.

Let ε ą 0 be given. Suppose that the collection of discontinuities of f is D, and

B = B A Y D. Since B A and D has volume zero, Lemma 3.22 implies that B has volume

zero; thus (a) of Proposition 3.14 then implies (with B Ď R in mind) that

żż żż

1B(x) dx = 1B(x) dx = 0 and 1B(x) dx = 1B(x) dx = 0 .

RB RB