MATH501_HW3 - Serge Ballif MATH 501 Homework 3 In problems below L stands for the-algebra of Lebesgue measurable subsets of d m for Lebesgue measure on

# MATH501_HW3 - Serge Ballif MATH 501 Homework 3 In problems...

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Serge Ballif MATH 501 Homework 3 October 10, 2007 In problems below L stands for the σ -algebra of Lebesgue measurable subsets of d , m for Lebesgue measure on L , ( X, M , μ ) stands for the measure space, and L 1 = L 1 ( X, M , μ ). 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 > 0 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 Lusin’s 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 : d be a Lebesgue measurable function. Show there exists a Borel measurable function g : d such that f = g a.e.  #### You've reached the end of your free preview.

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• Fall '08
• WYSOCKI,KRZYSZTOF
• Math, Algebra, Sets, measure, FK, Lebesgue measure, Lebesgue integration
• • • 