2015 Fall Real Analysis

Problem 1.

Let be a measurable subset of . Assume that is 1-periodic, i.e. . Compute

Justify your answer.

Proof.

We shall prove that

Let

Then is also a periodic function. Let be an anti-derivative of , say,

Then

Thus anti-derivative of is also periodic. Since

we only need to prove that

For any , let be a smooth function such that

Then we have

Problem 2.

Suppose and assume that there exists such that for all measurable subsets , we have

Show that for . Show that the statement fails for by giving a counterexample.

Proof.

For any , let

Then by assumption, we have

Thus

Let . Then

On the other hand, the theorem is not true for . Here is the example. Let . Then . On the other hand, let be any measurable set. Let . Then

Problem 3.

Show that a function is measurable if and only if is a measurable subset of .

Proof.

See Problem 2 of 2023 Winter Real Analysis Exam.

Problem 4.

Let and set

Show that and in as .

Proof.

Without loss of generality, we assume that . Then using integration by parts, we have

Since , at infinity it is zero. For the second term of the above right, we have

Thus we have . By Lebesgue differentiation theorem, almost everywhere. Thus by Problem 1 of 2022 Fall Real Analysis Exam, we get in .

Problem 5.

Let be a measure space and let be a sequence of measurable functions satisfying the following:

and

Prove that for all ,

Proof.

Let

for all . Then

Thus

since . Thus is almost everywhere finite, and as a result, a.e..

Problem 6.

Let be Lebesgue measurable sets and assume that for every , there exists a null set such that

Here . Show that if is not a null set, then the complement of in is a null set.

Proof.

By the Lebesgue's density theorem, if , then there is a point which is dense. That is, for any , there is a such that . Thus by assumption, for any , we have . Thus the complement of doesn't have any dense set. Then by the Lebesgue's density theorem again, the complement of is a null set.