MATH501_HW4 - Serge Ballif MATH 501 Homework 4 In problems...

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Serge Ballif MATH 501 Homework 4 October 22, 2007 In problems 1 and 5, ( X, M , μ ) stands for the measure space and L 1 = L 1 ( X, M , μ ). 1. Let ( X, M , μ ) be a measure space such that μ ( X ) < . Assume that ( f n ) is a sequence of integrable functions f n : X 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 n →∞ f n ( x ) = 0 a.e. In particular, on the set (0 , ), the sequence is bounded by the L 1 function 1 x 2 . Thus lim n →∞ R [ a, ) f n dm = R [ a, ) 0 dm = 0. Case 2: Suppose now that a = 0. The sequence is no longer bounded by an L 1 function, so we cannot invoke the dominated convergence theorem. The substitu- tion u = nx and du = n dx gives us for each n , R 0 n 1+ n 6 x 2
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