2022 Spring Real Analysis

Problem 1.

In a topological space , a subset is called a set if it can be written as the countable intersection of open sets. Consider a map between metric spaces and . Prove that

is a subset of .

Proof.

Let , we define the jumping of at by

where is the oscillation of on the given interval, given by

is continous at if and only if .

For any , the set

must be an open set. Thus

is a subset of .

Problem 2.

Let be a Lebesgue measurable set of finite Lebesgue measure. Suppose further that is a Lebesgue measurable function satisfying the following condition:

Prove that .

Proof.

We just need to prove that, for any , there exist finitely many disjoint subsets of such that . We prove this by induction. By continuity of the Lebesgue measure, there exists an such that . Thus we can write as disjoint union

with .

Inductively, for any , we can find such that .

Now we choose small enough such that and for that ,

whenever . Since , we must have

for all . Thus we have

is integrable.

Problem 3.

Suppose that is a finite measure space. Define a pseudometric on by setting

where denotes the symmetric difference of and , that is,

The relation defines an equivalence relation on . We denote the equivalence class of by . The pseudometric induces a metric on the set of equivalence classes by setting . Prove that the resulting metric space, that is, the set of -equivalence classes with the aforementioned metric, is complete.

Proof.

We define

Then it is not hard to prove that

Observe that we can write

Let be a Cauchy sequence. Then for any , we have

if . Let be a subsequence of such that

then taking limit with respect to the sequence, from (1), we shall get

This proves that the sequence is convergent to , proving the completeness of the metric space.

Remark.

Similar method would lead being convergent to . In fact, we shall have by taking in (1).

Problem 4.

Does there exist a Lebesgue measurable subset such that for all ? Justify your answer.

Proof.

Let

Then by assumption, we have

for any . It follows that for any open interval , we have

Thus over any open set , the integration of is zero. Hence over any measurable set the integration of is zero. Therefore is identically zero, a contradiction.

Problem 5.

Suppose that is a sequence of measurable functions such that for all . Further suppose that a.e.. Prove that:

(i). ;

(ii). in .

Proof.

Part (i) is the lower semi-continuity of weak convergence. One of the ways to prove it is to use the Egorov's theorem. For any , there exists a measurable set with such that on , the convergence is uniform. It follows that

by assumption. Since is arbitrary, by Levi's Monotone Convergence Theorems, we get .

To prove (ii), we let be a small number. Let be the set defined above, then since is uniformly convergent outside , there exists an such that

On the other hand, we have, by Cauchy inequality, that

Combining the above two estimates, we obtain

Since is arbitrary, we complete the proof.

Problem 6.

Let be such that . Show that .

Proof.

Without loss of generality, we may assume that . Thus the problem becomes: if , then .

By the AM-GM inequality , we know that, for any nonnegative number , we have

Replacing by for any positive number to be determined later, we have

Thus

Let , and integrate. We have

Taking , we get the desired estimate.

Remark.

Alternatively, we can use Hölder's inequality to get

The conclusion follows.