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# With the claim established the rest is trivial let u

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With the claim established, the rest is trivial. Let U be open in R 2 . then U = [ n N ( I n × J n ) for intervals I n , J n , hence h - 1 ( U ) = [ n N h - 1 ( I n × J n ) is measurable being a countable union of measurable sets. (b) The easiest way to prove h is measurable is to use the characterization by open sets: h is measurable if and only h - 1 ( U ) is measurable for every open subset U of R . Now h - 1 ( U ) = { x R : φ ( f ( x ) , g ( x )) U } = { x R : ( f ( x ) , g ( x )) φ - 1 ( U ) } . Here we may have to think how to apply part (a) and I admit that the fact that the h of this part has nothing to do with the h of part (a) can be confusing. Let us define H ( x ) = ( f ( x ) , g ( x )) so H : R R 2 and h = φ H . If I, J are open intervals, then H - 1 ( I × J ) = { x R : ( f ( x ) , g ( x )) I × J } = { x R : f ( x ) I } ∩ { x R : g ( x ) J } = f - 1 ( I ) g - 1 ( J ) so that H - 1 ( I × J ) is measurable for all open rectangles I × J . By part (a), H - 1 ( W ) is measurable for all open sets W R 2 . Returning now to our open subset U of R , since φ is continuous, φ - 1 ( U ) is open in R 2 , hence h - 1 ( U ) = ( φ H ) - 1 ( U ) = H - 1 ( φ - 1 ( U ) ) is measurable. 4

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10. Let { f n } n =1 be a sequence of measurable functions on a measurable subset E of R , converging pointwise to a function f . Assume: Z E | f n | ≤ 1 for all n N . Prove: Z E | f | ≤ 1 . A dash of Fatou could help. Solution. Because each f n is measurable, | f n | is non-negative measurable and Fatou’s lemma applies: (1) Z E lim inf n →∞ | f n | ≤ lim inf n →∞ Z | f n | . Since f n f , it follows that lim inf n →∞ | f n | = lim n →∞ | f n | = | f | . Since R | f n | ≤ 1 for all n , we also have lim inf n →∞ R | f n | ≤ 1. Thus (1) can be expanded to Z E | f | = Z E lim inf n →∞ | f n | ≤ lim inf n →∞ Z | f n | ≤ 1 . We are done. 11. Let f n : R R be measurable for all n N . Prove: If X n =1 Z R | f n | < , then X n =1 f n converges a.e. In other words, you have to prove that the set of points x R for which the series of real terms X n =1 f n ( x ) does not converge is a null set. Solution. By Lebesgue’s Monotone convergence Theorem (series version), since | f n | is non-negative, measurable for all n , we have Z R X n =1 | f n | = X n =1 Z | f n | < . The function x 7→ n =1 | f n ( x ) | is thus integrable, hence n =1 | f n ( x ) | < a.e. Since absolute convergence implies convergence, we are done. (I believe I once mentioned this in class.) 12. Let f : R R be integrable. Assume Z E f = 0 for all measurable subsets E of R . Prove f = 0 a.e. Solution. We proved in class: If f 0 is measurable and R E f = 0, then f = 0 a.e. on E . Let A = { x R : f ( x ) 0 } , B = { x R : f ( x ) < 0 } . Then since A is measurable, we have R A f = 0; by the mentioned class result, f = 0 a.e. on A since f 0 on A . Similarly R b f = 0, thus also R B ( - f ) = 0; since - f > 0 on B we conclude - f = 0 a.e. on B, hence f = 0 a.e. on B . Since R = A B and the union of two null sets is a null set, f = 0 a.e. on R .
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• Spring '11
• Speinklo
• Empty set, Open set, Topological space, Dominated convergence theorem, Lebesgue integration, Non-measurable set

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