2019 Fall Real Analysis

Problem 1.

Let be a compact set. Suppose we are taking the Lebesgue measure. Solve 3 of the following problems.

(a). Prove or disprove that necessarily ;

(b). Prove or disprove that necessarily ;

(c). Prove or disprove that necessarily ;

(d). Prove or disprove that necessarily .

Proof.

We assume that is a bounded domain that contains an interior point .

(a). is true. Let and assume that

Then by Hölder's inequality, we have

(b). is not true. We take

where .

Since is an interior point of , there exists an such that . We shall prove that but . First, we have

where is the area of the unit sphere in . On the other hand,

The second term in the above right is finite because

For the first term, we have

Thus .

Both (c). (d). are not true. See Problem 5 of 2023 Winter Real Analysis for the proof.

Problem 2.

Let be a sequence of measurable set in and

(a) Assume as . Prove or disprove that necessarily .

(b) Suppose there exists an infinite subsequence such that . Prove or disprove that necessarily . Discuss separately the cases of finite and infinite .

Proof.

We can write

We prove (b). first. Assume that

Then

almost everywhere. Moreover, is a monotonicity increasing sequence of functions. Assume that is finite, then by the Monotone Convergence Theorem, we must have

Since for infinitely many 's, we have

If is an infinite measure, then either or positive can happen. For example, if . Then .

Now we treat (a)., which is very similar to Problem 3 of 2022 Spring Real Analysis Exam. Assume that

Then

for any . Thus .

However, if

Then if we can find a subsequence of such that

and if we assume that is finite, then by (b)., it is possible that but otherwise could have been of zero measure. So it is inconclusive.

Problem 3.

Let be a finite measure on and define

Prove that is finite -a.e. for . Here denotes the Lebesgue measure on .

Proof.

By Fubini-Tonelli theorem, for any , we know that

For any , we have . Thus

is independent to . Therefore we have

Hence is Lebesgue integrable on any interval , and therefore the function is almost everywhere finite.

Remark.

In the Fubini's Theorem, we need to assume that the double integrability. In the Tonelli's Theorem, the nonnegativity of the function is used to replace the double integrability. Nevertheless, we have the following result.

Theorem. Let be a nonnegative function on . Assume that the iterated integral is finite

Then so is the double integral

We can use the Levi's Theorem to prove it. For any , let be the characteristic function of the interval over . We consider the functions

for any . Since the functions are bounded with compact support for any , we can use Fubini's Theorem to conclude that

Since the above inequality is true for any , by the Levi's Theorem, we conclude the double integrability of the function.

Problem 4.

Let be an absolutely continuous function on such that . Suppose that

Prove that .

Proof.

Let be a sequence of real numbers. Then by the Fatou's lemma, we have

Thus is a constant and since , it must be identically zero.

Problem 5.

Let be a measure space and suppose that is a sequence of measurable functions such that

and exists and is finite for any .

(a). Prove that there exists such that

(b). Prove that, for the in part (a), we have

Proof.

Define a functional on such that

By assumption, is a linear functional. Thus by the Riesz representation theorem, there is a function such that

for all . This proves (a)..

Taking , we shall get

This proves (b)..

Problem 6.

(a) Suppose . Show that

(b) Give an example of such that

Proof.

Without loss of generality, we assume that is nonnegative.

(a). We consider

Using Cauchy's inequality, we have

Therefore we have

(b). Let be a sequence of nonnegative real numbers. Define to be , where is the integer part of . Let

Then

On the other hand

It is thus easy to verify that .

Let . Then by (1), we know that