2013 Spring Real Analysis

Problem 1.

For each of (a) and (b) either prove or give a counterexample. Suppose that for every , is a -measurable function, and

Then

(a) in measure;

(b) for -almost all , .

Proof.

I think we shall assume that , otherwise take whose average is zero would provide counterexample.

Assume that . Then (a). is correct, because for any ,

(b). is not correct. See Problem 4 of 2023 Winter Real Analysis Exam.

Problem 2.

Let be Borel measurable for every . Define to be the set of points in such that exists and is finite. Show that is a Borel measurable set.

Proof.

We can write

so it is Borel.

Problem 3.

Does there exist a nowhere dense subset of

(a) of Lebesgue measure greater than 9/10?

(b) of Lebesgue measure 1?

If yes, construct such a set, if no, prove why not.

Proof.

(a). Let be a nowhere dense subset of with measure . Then would give such an example. There are multiple ways to construct . For example, let be a sequence of positive numbers such that

Then can be inductively construct by removing open intervals in the middle of the intervals from the -step.

(b). is not possible. Let be such a nowhere dense set with full measure. Then , where is the closure of . By definition, contains no interior point. In particular . But since is a closed set, it must miss an open ball. Thus the measure of can not be equal to .

Problem 4.

Let , and . Show that the integral

exists a.e. and that .

Proof.

We have

This completes the proof.

Problem 5.

Show that for ,

as .

Proof.

By Hölder's inequality, . Then the result follows from the Riemann–Lebesgue lemma. (See Problem 1 of 2008 Fall Real Analysis Exam.)

Problem 6.

Assume that . Let

Show that for any bounded interval . (Hint: First prove for intervals with length .)

Proof.

We shall assume that is absolutely continuous.

We start with any open interval of length . Then since there is not overlapping, we have

By Intermediate value theorem, there is a such that . By the fundamental theorem of calculus, we have

for all .

For any interval of longer length, we can chop them small so that each small piece is of length .

Remark.

I am not sure if the condition of absolute continuity can be dropped. If we take a singular function whose derivative is a.e. , then the condition of becomes vacuous. In this case, exists and is a.e. finite. But I don't know how to prove that it is bounded on bounded intervals.