2013 Fall Real Analysis

Problem 1.

Let for every . Show that

Proof.

First, we have

Thus

On the other hand, let and let . Then . We thus have

Thus

As a result

Since is arbitrary, we prove the result.

Problem 2.

Assume that is a subset of and the distance between any two points in is a rational number. Show that is a countable set. (Note that this is actually true in any dimension.)

Proof.

Let be two distinct points. Then any n must be in the intersection of two circles centered at respectively with rational radii. Since one can only have countably many circles with rational radius centred at a fixed point, and since two circles can intersect at most two points. must be countable.

Problem 3.

Let be a non-negative measurable function on . Prove that if is integrable, then a.e.

Proof.

By the Monotone convergence theorem, the series

is convergent. In particular

as . This can only happens when is zero almost everywhere.

Problem 4.

For each , we define as follows:

(a) Show that

(b) For , what is

Justify your answer.

Proof.

(a). Let . Then we have . Thus

As a result, we have

It is not hard to prove that if , then .

(b). We claim that

For any , for , we shall have if . Consider the Riemann sum

and

Their difference is no more than . On the other hand, the Riemann sum tends to the definite integral as . This completes the proof of the claim.

(c). If we only assume that is bounded measurable, the above claim is not true anymore. For example if we write , and define as

Then it is not hard ro verify that

while

Problem 5.

Let . Find where is the Lebesgue measure in . Fully justify your steps.

Proof.

We claim that is of measure zero. To see this, we observe that

Let be the sequence of rational numbers. Then

and apparently these are all of measure zero sets.

Finally, we have

Problem 6.

Suppose that with and -a.e. on . Show that for every ,

as .

Hint: First consider the case when is of finite measure.

Proof.

For any , there is a subset with finite measure such that

On , we use Egorov's theorem. There is a set with sufficiently small that outside , the convergence of is uniform. By the Dominated convergence theorem, we may assume that

Thus by Hölder's inequality, we have

Since and since the convergence of to is uniform on , the conclusion follows.