Page 84 - Vector Analysis

80 CHAPTER 3. Multiple Integrals

żż

which implies that f (x) dx ď 0. Similarly, ´f (x) dx ď 0 which further implies

A A

ż

that f (x) dx ě 0. Therefore, by Corollary 3.8 we conclude that

A

żż

0 ď f (x) dx ď f (x) dx ď 0

AA

ż ˝

which implies that f is Riemann integrable over A and f (x) dx = 0.

A

Remark 3.15. Let A Ď Rn be bounded and f, g : A Ñ R be bounded. Then (b) of

Proposition 3.14 also implies that

ż żż żżż

(f ´ g)(x) dx ď f (x) dx ´ g(x) dx and f (x) dx ´ g(x) dx ď (f ´ g)(x) dx .

A AA AAA

Corollary 3.16. Let A, B Ď Rn be bounded such that A X B has volume zero, and f :
A Y B Ñ R be bounded. Then

żż ż ż żż

f (x) dx + f (x) dx ď f (x) dx ď f (x) dx ď f (x) dx + f (x) dx .

AB AYB AYB AB

Proof. Note that f 1A + f 1B = f 1AYB + f 1AXB on A Y B. Therefore, (a), (b) of Proposition
3.14 and Remark 3.15 implies that

żż ż ż ż

f (x) dx + f (x) dx = (f 1A)(x) dx + (f 1B)(x) dx ď (f 1A + f 1B)(x) dx

AB AYB AYB AYB

ż( )
= f 1AYB ´ (´f 1AXB) (x) dx

AYB

żż

ď f 1AYB(x) dx ´ (´f 1AXB)(x) dx

AYB AYB

żż

= f (x) dx ´ (´f )(x) dx

AYB AXB

which, with the help of Proposition 3.14 (e), further implies that

żż ż

f (x) dx + f (x) dx ď f (x) dx .

AB AYB

The case of the upper integral can be proved in a similar fashion. ˝

Having established Proposition 3.14, it is easy to see the following theorem (except (c)).
The proof is left as an exercise.