2020 Fall Real Analysis

Problem 1.

For , denote

as the set of rational numbers in . Denote

Prove that there exists a sequence of positive numbers such that

Proof.

This problem is very similar to

We estimate

Taking , we get

Thus the function is almost everywhere finite.

Problem 2.

is a sequence of measurable functions on . Prove that

if and only if converges to 0 in measure.

Proof.

I don't think the problem is correct. For example, if for and for , then doesn't converge to zero in measure, but (1) is valid.

We now assume that . For any , let

If as , then

This proves (1). On the other hand, if (1) is valid, then for any , we have

which implies that is convergent to in measure.

Problem 3.

Let be a measure space. Let be a real-valued -measurable function on . Suppose is -integrable on for every -integrable real-valued -measurable function on . Show that .

Proof.

Without loss of generality, we assume that and . For any , let

Assume that . Then for any , we have . Let be a subsequence of such that

Define a sequence of real numbers by

Define by if . Then

On the other hand, using the same method, we get

This is a contradiction.

Problem 4.

Let be a measure space. Consider an -measurable function on such that for some . Show that

Proof.

By the inequality

From the above inequality, we know that as . By the Dominated convergence theorem, we have

This completes the proof.

Problem 5.

A metric on a space is called an ultrametric if the triangle inequality is replaced by the following stronger property. For all :

Prove:

(1). If is an open ball in , then any point in is a center of . (This means that for all , for some .)

(2). Every open ball in is both open and closed.

Proof.

(1). Let , we shall prove that

By symmetry, we only need to prove the above left is contained in the above right. Let . Then we have . This completes the proof.

(2). Let be a sequence in . Let . Then for any . We choose large enough so that . Then , and this proves the open ball is closed.

Problem 6.

Compute

Be sure to justify your answer. (You may assume elementary facts about calculus, but state them.)

Proof.

Using integration by parts, we have

Thus we have

for a constant . Thus