Page 80 - Vector Analysis

76 CHAPTER 3. Multiple Integrals

Therefore,

U (f, P) ´ U (f, P1)

( sup A(R) inf A(R)) n mi ÿ

ÿ ÿ

ď f ´ f ν(∆k)

i=1 j=0 1ďkďN with

yj(i) P∆(ki)

( sup A(R) inf A(R)) n mi 2δrn´1

ÿ ÿ

ď f ´ f

i=1 j=0
( )
ď 2δrn´1(m1 + m2 + ¨¨¨ + mn + n) sup f A(R) ´ inf f R (A) ă ε ,
2

and the fact that U (f, P1) ´ L(f, P1) ă ε shows that
2

U (f, P) ´ I ď U (f, P) ´ I + U (f, P1) ´ U (f, P1)

ď U (f, P) ´ L(f, P1) + U (f, P1) ´ U (f, P1) ă ε .

Therefore, for any sample set tξ1, ¨ ¨ ¨ , ξN u for P,

N

ÿ f A(ξk)ν(∆k) ď U (f, P) ă I + ε .

k=1

Similar argument can be used to show that

N

ÿ f A(ξk)ν(∆k) ě L(f, P) ą I ´ ε

k=1

which concludes the Theorem. ˝

Definition 3.12. A bounded set A Ď Rn is said to have volume if the characteristic
function of A, denoted by 1A and given by

" 1 if x P A ,
0 otherwise ,
1A(x) =

ż

is Riemann integrable over A, and the number 1A(x) dx is called the volume of A and

A

is denoted by ν(A). If ν(A) = 0, then A is said to have volume zero.

Remark 3.13. Having defined the indicator function, then for a bounded function f : A Ñ
R with bounded domain A, any given partition P of A we have f A = f 1A; thus

U (f, P) = ÿ sup(f 1A)(x)ν(∆) and L(f, P) = ÿ inf (f 1A)(x)ν(∆) .
xP∆
∆PP xP∆ ∆PP