UMD Analysis Qualifying Exam/Aug12 Real

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Problem 1[edit | edit source]

Compute the following limit. Justify your answer.

Solution 1[edit | edit source]

We will use the dominated convergence theorem. First, note that for and ,


and this function is in , with


Therefore, by the LDCT,

Problem 3[edit | edit source]

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 | edit source]

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 ,.



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 | edit source]

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 | edit source]

(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.