Unformatted text preview: 5 # E . This proves the claim.
Since the claim holds, for every ( > 0 we have found a plurirectangle P1
contained in E and and a plurirectangle P2 is containing E such that
0 % meas P2 ' meas P1 % (.
It follows by the previous theorem that E is Peano—Jordan measurable. 79 Friday, April 01, 2011
The next theorem shows that the integral of a nonnegative function is given
by the volume of the subgraph.
Theorem 141 Let R ) RN be a rectangle and let f : R * [0, /) be a bounded
function. Then f is Riemann integrable over R if and only if the set
Sf := {(x, y ) ! R 1 [0, /) : 0 % y % f (x)}
is Peano—Jordan measurable in RN +1 and in this case
A
measN +1 Sf =
f (x) dx.
R Proof. Step 1: Assume that f is Riemann integrable over R. Given ( > 0, by
Theorem 126 there exists a partition P ( of R such that
0 % U (f, P ( ) ' L (f, P ( ) % (.
Write P ( = {R1 , . . . , Rn } and set
I
J
Ti := Ri 1 0, inf f ,
Ri and
P1 := n
E Ti , I
J
Ui := Ri 1 0, sup f
Ri P2 := i=1...
View
Full Document
 Spring '14
 Trigraph, Ri, measure, Lebesgue integration, Meas

Click to edit the document details