2024 Spring

2024 Spring Real Analysis

Problem 1.

Assume that . Let . Show that

Proof.

Let be any positive number. Then for large enough

Since , for any , there exists an such that

Thus

Since is fixed, we conclude that

for any . This completes the proof.

Problem 2.

Suppose that and are nonnegative integrable functions on . Assume further that

Prove that for any measureable set ,

Proof.

By Fatou's Lemma, we have

Thus for any subsequence of we have

On the other hand, using the Fatou's Lemma again, we have

Thus we have

The above righ side is equal to

Since equality must be valid for all the above inequalities, we have

for any subsequence. This completes the proof.

Problem 3.

Let be non-decreasing. Recalling that this assumption implies that exists for almost all (with respect to Lebesgue measure ). Prove that

Proof.

Define

for any (if , then we define ). Then are nonnegarive and

almost everywhere.

We also have

Thus the problem follows from the Fatou's Lemma

Problem 4.

Suppose that is a measure space and and are integrable. For each , set

and

Part a: Assume that is -finite. Prove that

Here, for subsets and of a set ,

denotes their symmetric difference.

Part b: Show that the conclusion of part (a) holds even if is not -finite.

Proof.

Let

Then

Let . If we restrict the measure to , then since are integrable, is -finite. This answers Part b of the problem.

Let be the characterisic function with respect to a set . Then on

Then by Tonelli's theorem (which does need -finiteness), we have

Similarly, we have

Combininig the above equalities we get the conclusion.

Problem 5.

For and , let

and

Part a: Let be a continuous function on that vanishes outside of a compact set. Prove that the functions

converge uniformly to on as .

Part b: Let for some . Prove that the functions

converge to in as .

Proof.

We first observe that, for any ,

Thus we have the following identity

To prove Part a, we note that since is continuous with compact support, then is uniformly continuous. As a result, for any , there is a such that whenever . For such a , we choose large enough so that if , then . Thus

completing the proof.

To prove Part b, we take any . Then we can find a smooth function with compact support so that

Let . Then is a finite number that may depend on . We estimate

for some constant independent to . For large enough, this implies that

Finally, by Young's convolution inequality, we have

Thus by the triangle inequality, we have

and the conclusion follows.

Remark.

The specific expression of function is not important. The result holds true for any nonnegative continuous function with compact support such that

Problem 6.

Suppose is a (Lebesgue) measurable subset. For all , define

Prove that for a.e. ,

Proof.

We assume that . Let be a fixed point. We assume that without loss of generaly. Let . Then

As a result, we have

Thus

By the Lebesgue's density theorem, for almost all ,

The result follows.