MATH501_HW4 - Serge Ballif MATH 501 Homework 4 October 22,...

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Serge Ballif MATH 501 Homework 4 October 22, 2007 In problems 1 and 5, ( X, M ) stands for the measure space and L 1 = L 1 ( X, M ). 1. Let ( X, M ) be a measure space such that μ ( X ) < . Assume that ( f n ) is a sequence of integrable functions f n : X C converging uniformly to the function f on X . Prove that R f n R f . The sequence ( f n ) is uniformly convergent with limit f if for every ± > 0, there exists a natural number N such that for all x in X and all n N , we have | f n ( x ) - f ( x ) | < ± . Claim 1. The function f is in L 1 . Proof. We need to show that R | f | < . Using the triangle inequality, together with the facts that μ ( X ) < and R | f n | < we can calculate Z | f | = Z | f - f n + f n | Z | f - f n | + | f n | = Z | f - f n | + Z | f n | Z ± + Z | f n | = ±μ ( X ) + Z | f n | < . Thus, f is in L 1 . To see that R f k R f we must show that for any ± > 0, there is an N > 0 such that for n N , | R f n - R f | < ± . Using the uniform convergence of the sequence ( f n ) we can choose N such that for n > N we have | f n - f | < ± μ ( X ) . Now, ± ± ± ± Z f n - Z f ± ± ± ± = ± ± ± ± Z f n - f ± ± ± ± Z | f n - f | Z ± μ ( X ) = ±. Hence, R f k R f . 2. Let f n ( x ) = n 1+ n 6 x 2 . Compute lim n →∞ R [ a, ) f n dm when a < 0 , a = 0 , and a > 0 . Page 1 of 5
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Serge Ballif MATH 501 Homework 4 October 22, 2007 Case 1: Suppose a > 0. Then | f n ( x ) | = | n 1+ n 6 x 2 | < 1 x 2 for all x . Since the positive function 1 x 2 is Riemann integrable for a > 0, it is Lebesgue integrable. Therefore, by the dominated convergence theorem, lim n →∞ R [ a, ) f n dm = R [ a, ) lim n →∞ f n dm . We calculate the limit lim n →∞ f n ( x ) = 0 for all x except x = 0, where lim n →∞ f n (0) = . Thus, lim
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This note was uploaded on 04/01/2008 for the course MATH 501 taught by Professor Wysocki,krzysztof during the Fall '08 term at Pennsylvania State University, University Park.

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MATH501_HW4 - Serge Ballif MATH 501 Homework 4 October 22,...

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