2017 Spring Real Analysis

Problem 1.

There are five separate statements listed below. For each, say whether it is true or false. For true statement cite an appropriate theorem or give a justification. For false statements provide counterexamples.

If is a sequence of measurable sets in a measure space such that , and is measurable, is it true that you can conclude that if

(a) ;

(b) are nested and decreasing;

(d) and is locally integrable;

(e) .

Proof.

(a). is not true. Let . Then ;

(b). is not true. Let ; , and let ;

(c). is not true, see the above counterexample;

(d). is not true. Let . Let be a sequence of . Let for some appropriate ;

(e). is true. It follows from Dominated convergence theorem.

Problem 2.

Consider Lebesgue measure on the real line.

Let be a measurable subset, and consider a sequence of real measurable function . Suppose that a.e. in . If , is it true that

(a) ;

(b) .

Prove or give a counterexample.

Proof.

(a). is true. For any , there is a measurable set with and uniformly on . By Hölder's inequality, we have

Then

Letting , we get

completing the proof.

(b). This is not true. Let . Then almost everywhere, and . But .

Problem 3.

Let be a measurable function on . Assume that for any , . Show that .

Proof.

We first observe that

On the other hand, by Fubini's theorem, we have

The theorem is proved by combining the above two inequalities.

Problem 4.

Let be a -finite measure space with .

(a) Prove that there exist , disjoint, such that for all and .

(b) Prove that there exists an -measurable real-valued funtion on such that but for all .

Proof.

Write

such that . Replacing by is needed, we may assume that all are disjoint.

Let . Then . Therefore there is a subsequence of such that

We take

Then satisfies the requirement.

(b). Define a function such that if . Then

is not integrable. But for any ,

Problem 5.

Let be a measure space. Show that

(a) ;

(b) It is not always true that .

Proof.

(a). We just need to prove that, if a function is in both and , then it is in . But an function is (essentially) bounded. Thus

This completes the proof.

(b). is not true. Let with respect to the Lebesgue measure. Let for , and for . Then for all , but is neither in nor in .

Problem 6.

Let be open in . Show that is equal to the union of pairwise disjoint open balls and a set of Lebesgue measure zero.

Hint: First show there are disjoint open balls with .

Proof.

Let . Without loss of generality, we assume that . By Vitali covering lemma (see the infinite version), there are disjoint open balls with . Let . Using the Vitali covering lemma on , there are disjoint open balls with . Inductively, we can define .

By construction, . Thus

This completes the proof.