2018 Fall Real Analysis

Problem 1.

Let , be a sequence of Lebesgue measurable functions. Prove that the set of points for which the sequence fails to converge as is Lebesgue measurable.

Proof.

The set of divergence can be represented by

So it is measurable.

Problem 2.

Let be absolutely continuous. Suppose that for some . Show that for any , , we have

Proof.

We write

Then we have

We use Hölder Inequality to get

Thus we have

The theorem is proved.

Problem 3.

Let be a sequence of functions in , for some , such that for . Suppose that almost everywhere on as . Show that

(a). ,

(b). weakly as , i.e., for any , , we have

Proof.

Since , then . We use Fatou's Theorem to get

This proves part (a).

For part (b), we prove as follows. For any , we can choose a compact set such that

We can also find a set such that is sufficiently small so that

and moreover, uniformally on . Using the triangle inequality, we have

For the second term of the above right, we have the following estimate

For the first term of Equation (1), since is of finite measure, the fact that that unifromly implies in the -norm. Thus for sufficiently large , we can make

Putting the above two inequalities together completes the proof of part (b).

Problem 4.

Let be such that

for all with the property that

Prove that there are such that for almost all .

Proof.

For any continuous function , Define numbers such that

Define numbers such that

and

Then satisfies (2). By assumption,

On the other hand, we have

By definition of , we have

The (3) is for any function . This completes the proof.

Problem 5.

Let . Show that the series

converges absolutely for almost all .

Proof.

For any , we consider

where is the characteristic function of . It turns out that

Thus

As a result, for almost all , is absolute convergent.

Problem 6.

Let be a closed set with , where is the Lebesgue measure and . Set

where is the Euclidean distance from to . Show that for almost all and for all .

Proof.

If , then there is a such that . Thus

If , we note that

Therefore we have

Since

we have

Thus for almost all , is finite.