UMD Analysis Qualifying Exam/Jan07 Real

Problem 1Edit

Suppose that   is measurable and that  . Prove that


Solution 1Edit

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.

Problem 3Edit

Let  , where  . Prove that  , and  .

Solution 3Edit

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 5Edit

Suppose   are measurable sets, and there is an integrable function   such that  . Prove that there is a measureable set   such that  .

Solution 5Edit

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.