Page 83 - Vector Analysis

§3.2 Properties Of The Integrals 79

Since ε ą 0 is given arbitrarily, we conclude that

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

AAA

ż żż

Similarly, we have (f + g)(x) dx ď f (x) dx + g(x) dx; thus (b) is established.

A AA

(c) It suffices to show the case c = ´1. Let ε ą 0 be given. Then there exist partitions
P1 and P2 of A such that

żż
´f (x) dx ´ ε ă L(´f, P1) and U (f, P2) ă f (x) dx + ε .

AA

Let P be the common refinement of P1 and P2. Then

żż
´f (x) dx ´ ε ă L(´f, P1) ď L(´f, P) ď ´f (x) dx

AA

and
żż
f (x) dx ď U (f, P) ď U (f, P2) ă f (x) dx + ε .

AA

By the fact that

L(´f, P) = ÿ inf A ÿ sup f A(x)ν(∆) = ´U (f, P) ,

∆PP xP∆ (´f ) (x)ν(∆) = ´

∆PP xP∆

we find that

żż
´f (x) dx ´ ε ă L(´f, P) = ´U (f, P) ď ´ f (x) dx

AA

and
żż
´f (x) dx ě L(´f, P) = ´U (f, P) ą ´ f (x) dx ´ ε .

AA

Therefore,
ż żż

´f (x) dx ´ ε ă ´ f (x) dx ă ´f (x) dx + ε .

A AA

Since ε ą 0 is given arbitrarily, we conclude (c).

(e) Since f is bounded on A, there exist M ą 0 such that ´M ď f (x) ď M for all x P A.

Therefore, ´1A ď f ď 1A on A; thus (c) and (d) imply that
M

ż ż ě ż f (x) = 1 ż
dx M
0 = 1A(x) dx = 1A(x) dx A f (x) dx
M
A A A