UMD Analysis Qualifying Exam/Jan07 Real
Suppose that is measurable and that . Prove that
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: Edit
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: Edit
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.
Let , where . Prove that , and .
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
Suppose are measurable sets, and there is an integrable function such that . Prove that there is a measureable set such that .
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.
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.