UMD Analysis Qualifying Exam/Aug12 Real

Problem 1 Compute the following limit. Justify your answer. Solution 1 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 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 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 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 (i) Fix epsilon greater than zero. Then, consider the sets Sn={x in [0,1] : n-1<=f(x)<n}. The partial sums of the integrals of f over Sn 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 Ac 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.