2018 Spring Real Analysis

Problem 1.

Let , and let , be sequences of real numbers such that for all and . Show that for almost all .

Proof.

We consider

Since , we have . So the function is almost everywhere finite, and the conclusion follows.

Problem 2.

Let and assume that in as . Let be a bounded sequence in , and let , . Prove that the sequence of function has a subsequence that converges in , almost everywhere, and in measure.

Proof.

The sequence has a convergent subsequence. For the sake of simplicity, we assume that the subsequence is the sequence itself, and the limit is . Then we can prove that is convergent to . Thus subsequence of is convergent to some in almost everywhere.

In the above proof, we need to use the fact that

Problem 3.

Let for some . Assume that for every ,

where is the Lebesgue measure. Show that

Proof.

We have

By Hölder's inequality, we have

The theorem follows.

Problem 4.

Let be such that

for all Lebesgue measurable sets with . Prove or disprove that for almost all .

Proof.

Yes, a.e..

Let be a positive number. Let . Consider , where is chosen so that the two intervals don't intersect. Then . From

by taking derivative with respect to , we get . Since both and are arbitrary, we conclude that is a constant function. But since , we conclude that almost everywhere.

Problem 5.

Let be a positive finite Borel measure on and let be continuous. Assume that for every Borel set with . Show that there is a Borel measurable function such that

for all continuous .

Proof.

Let be the measure such that . Then by assumption, . By Radon–Nikodym theorem, there is a function such that

The above equation is equivalent to that (1) is valid for all characteristic function . By taking limit, the equation must be true for all continuous functions.

Problem 6.

Let for some . Compute

Proof.

The limit is equal to .

First, we observe that by using the Hölder's inequality, we have

Thus

Next, let be a Riemann integrable function. Then we have

Summing over and taking limit, we get

Finally, for any , we let be Riemann integrable so that . Thw we have

A method of approximation can be used here to complete the proof.