UMD Analysis Qualifying Exam/Aug12 Real
Problem 1[edit]
Compute the following limit. Justify your answer.

Solution 1[edit]
We will use the dominated convergence theorem. First, note that for and ,
Therefore,
and this function is in , with
.
Therefore, by the LDCT,
Problem 3[edit]
Assume is absolutely continuous on an interval and there is a continuous function such that a.e. Show that is differentiable at every and that everywhere on . 
Solution 3[edit]
If exists, then by definition, . So we need to show that this limit both exists and is equal to .
Then by the absolute continuity of , .
Since, is continuous, then for any there exists some such that for ,.
Therefore,
.
The same argument gives a lower bound, giving us altogether
. Therefore, the limit exists (i.e. is differentiable) and the difference quotient goes to .
Problem 5[edit]
Let be a nonnegative Lebesgue integrable function on . Denote by the Lebesgue measure on . (i) Prove that, for each , there exists a such that
(ii) Prove that, for each , there is a such that for each measurable subset : if , then 
Solution 5[edit]
(i) Fix epsilon greater than zero. Then, consider the sets S_{n}={x in [0,1] : n1<=f(x)<n}. The partial sums of the integrals of f over S_{n} comprise a monotonically increasing sequence of real numbers, bounded by the finite integral of f over [0,1].
Hence, this sequence converges, and the tail of the sequence, which is the integral of f over the set {x in [0,1] : f(x)>= n}, must eventually be less than epsilon for some n.
(ii) Fix epsilon greater than zero. By part (i), there exists some constant c such that, given the set A={x in [0,1] : f(x)>=c}, the integral of f over A is less than epsilon/2. On the complement of A (in [0,1]), f is bounded above by c, and so any set of measure less than epsilon/2c will produce an integral whose value less than epsilon/2.
If m(A) is nonzero, take delta to be the minimum of m(A) and epsilon/2c. Given any measurable E in [0,1] with m(E) less than delta, the portion in A and the portion in A^{c} will each have integrals with values at most epsilon/2, so the integral of f over E has a value of at most epsilon.
If m(A)=0, then f is bounded almost everywhere on [0,1], and we simply take delta to be epsilon/c.