UMD Analysis Qualifying Exam/Aug06 Real
Contents
Problem 1a[edit]
Prove the following version of the RiemannLebesque Lemma: Let . Prove in detail that

Solution 1a[edit]
Note that .
Hence we can equivalently show
as
Claim[edit]
Let be a step function.
as
Proof[edit]
Step functions approximate L^1 functions well[edit]
Since , then
Hence, given , there exists such that
Problem 1b[edit]
Let be an increasing sequence of positive integers. Show that has measure 0.

Solution 1b[edit]
For the sake of brevity, let S be the set of such x. If S has positive measure, then it contains a subset of positive measure on which liminf[sin(n_{k}x)] is bounded below by some positive constant; ie, the integral of liminf[sin(n_{k}x)] over S will be positive. If S has zero measure, then the same integral will be zero. Thus, we must only compute an appropriate integral to show that m(S)=0.
Since we do not know that S has finite measure, take a sequence of functions f_{k}(x)=2^{x}*sin(n_{k}x), each of which is clearly integrable. By Fatou's lemma, the integral of liminf[f_{k}(x)] over S <= the liminf of the integrals of f_{k}(x) over S. However, each f_{k} is the L^{1} function 2^{x}*sin(n_{k}x). By the RiemannLebesgue Lemma, this goes to 0 as n_{k} goes to infinity.
Hence our original integral of a strictly positive function over S is bounded above by 0, so m(S)=0.
Problem 3[edit]
Suppose , where . Show that . 
Solution 3[edit]
Let then we can write
Hence .
Problem 5[edit]
Let ,
(a) Show that is differentiable a.e. and find . (b) Is absolutely continuous on closed bounded intervals ? 
Solution 5[edit]
Look at the difference quotient:
We can justify bringing the limit inside the integral. This is because for every , . Hence, our integrand is bounded by and hence is for all . Then by Lebesgue Dominated Convergence, we can take the pointwise limit of the integrand. to get
It is easy to show that is bounded (specifically by ) which implies that is Lipschitz continuous which implies that it is absolutely continuous.