2016 Fall Real Analysis

Problem 1.

Let be a -finite measure space, and let be a sub--algebra of . Given a nonnegative -measurable function on , show that there is a nonnegative -measurable function on , such that

for every nonnegative -measurable function . In what sense is unique?

Proof.

Let be a measure on defined by

for any . Observe that is -finite and . So, by the Radon-Nikodym theorem, there exists an -measurable function such that

for any .

When is -measurable, we can find an increasing sequence of -simple functions converges to pointwise. So we have monotonically converging to and monotonically converging to . Then, by the Monotone convergence theorem, we have

as desired.

By the Radon-Nikodym construction, is unique up to .

Problem 2.

Let be a measure space. Suppose there exists an extended real valued -measurable function on such that -a.e. and is -integrable. Prove that is a -finite measure space.

Proof.

See Problem 3 of 2021 Spring Real Analysis Exam.

Problem 3.

Let . Set and for .

(i) Show that the series converges in if and only if

(ii) Suppose that for and that is bounded for for some . Show that converges uniformly for .

Proof.

To show that the series converges in , we need to show that . Observing that

we have

Thus is convergent if and only if (1) is valid.

Suppose that for . Let . Since for , we have

We claim that

for all nonnegative integer. If , it follows from assumption. If the above inequality if true for , then

Since

is convergent for any , the conclusion of the problem follows.

Problem 4.

Construct a nonnegative measurable function on such that

(i) for any ;

(ii) for any interval .

Proof.

Let be a sequence of rational numbers. Define

for all . Let

Then define ; if , then define -- if belows to several of these 's, define to be the maximal of such 's.

We make a remark of the function . Let

Then for any , we have

as . As a result, we have . Since means that belongs to infinitely many 's, we conclude that for almost all , belongs to at most finitely many intervals . Thus is almost everywhere finite.

Thus for almost all , it belong to a unique such that is maximized. Let

Then

Since on , we have

is integrable for all .

On the other hand, for any interval , then there is a subsequence such that . Since , we have .

Problem 5.

Assume that is a nonnegative measurable function on and

Show that a.e..

Proof.

We have

By our assumption, the above right is equal to 0. So we have a.e., which implies that a.e. on .

Problem 6.

Let be a sequence of measurable functions on such that for all and all and assume that

Show that uniformly on each compact set in , if .

Proof.

Let be compact. Let be any positive number. Then there exists a set such that

and on , uniformly. Without loss of generality, we assume that

for some small depending on , and as . Then for any , we have

We have

and we have

The theorem follows by the above two inequalities.