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Unformatted text preview: Serge Ballif MATH 501 Homework 3 October 10, 2007 In problems below L stands for the algebra of Lebesgue measurable subsets of R d , m for Lebesgue measure on L , ( X, M , ) stands for the measure space, and L 1 = L 1 ( X, M , ). 1. Let E L be such that m ( E ) < and let ( f k ) be a sequence of measurable functions defined on E . Assume also that for every x E there exists a constant M x such that  f k ( x )  M x for all k 1 . Prove that given > there exists a closed set F E and a constant M such that m ( E \ F ) < and  f k ( x )  M for all x F and all k 1 . Each function f k satisfies the criteria for Lusins Theorem: 1. f k is measurable, 2. f k is finite valued on E , and 3. E has finite measure. Hence, for every > 0, there exists a closed set F , with F E and m ( E F ) / 2 and such that f  F is continuous. The fact that m ( E ) < also tells us that there exists a compact set K such that K E and m ( E K ) / 2. The set F K is compact as an intersection of a closed set with a compact set. We also have F K E and m ( E ( F K )) = m (( E F ) ( E K )) m ( E F ) + m ( E K ) = . Let F = F K . Then f k is continuous on F since it is continuous on the larger set F . Hence, the function  f k  is continuous on F as a composition of continuous functions   f k . Therefore, as a continuous function on a compact set,  f k  always attains a maximum value M k . We need an M that bounds all M k . hmmmmmmmmmmmm 2. Let f : R d R be a Lebesgue measurable function. Show there exists a Borel measurable function g : R d R such that f = g a.e....
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 Fall '08
 WYSOCKI,KRZYSZTOF
 Math, Algebra, Sets

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