UMD Analysis Qualifying Exam/Jan07 Real

Problem 1 Suppose that is measurable and that . Prove that Solution 1 Since then a.e. on [0,1]. This implies that a.e. on [0,1] since on (0,1). Then by Lebesgue Dominated Convergence, we have Now to handle the interval : Case 1: For every we have and increases monotonically to 1. So by the same argument as above, Lebesgue Dominated convergence gives us and we're done. Case 1: Notice that for every we have , the right-hand-side must necessarily integrate to infinity. So by monotonicity of the integral we have that for all . This gives as desired. Problem 3 Let , where . Prove that , and . Solution 3 Dividing by , we can assume without loss of generality that (similarly for with their appropriate norms). Thus we want to show that . The proof hinges on Young's Inequality which tells us that Problem 5 Suppose are measurable sets, and there is an integrable function such that . Prove that there is a measureable set such that . Solution 5 We claim that can only take on the values 0 and 1. To see this, suppose the contrary, suppose differs from 0 or 1 on a set . We can exclude the case because otherwise we can modify on a null set to equal an indicator function without affecting the integral. Then On , is a strictly positive function. Then for any sufficiently small there exists some with such that on for some positive constant . Then . Thus we have shown that we can obtain a positive lower bound for completely independent of the choice of . This contradicts . Hence can only assume values 0 and 1 almost everywhere. Since , then it is certainly measurable. Hence is measurable. And we're done.