21-356 Lecture 30

# This proves the claim since the claim holds for every

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Unformatted text preview: 5 # E . This proves the claim. Since the claim holds, for every ( > 0 we have found a pluri-rectangle P1 contained in E and and a pluri-rectangle 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...
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