2023 Winter Real Analysis

Problem 1.

Consider a measure space and a sequence of measurable sets , such that

Show that almost every is an element of at most finitely many 's.

Proof.

Let be the characteristic function of . Then the condition is equivalent to

Thus the function is almost everywhere finite and hence the result is proved.

Problem 2.

Let be a -finite measure space and let be measurable. Let . Assign the Lebesgue measure on . Prove that is a measurable set on with respect to the product measure and that

Proof.

To see that is measurable, let . Then is measurable, being the composition of and the subtraction function. Since , it follows that is measurable. Now, by Tonelli's theorem, one has

where . So and the result follows.

Remark.

Is the above proof hard to understand? Think harder to understand the beauty and the insight there.

Problem 3.

Suppose that and are measure spaces and is a measurable map. Moreover, assume that for any measurable set , we have

Then for any measurable function , prove that if and only if , in which case,

Proof.

By considering positive and negative parts of the real and imaginary parts of , it suffices to consider the case that . If for a measurable set , then the result is clear by the assumption on . The case of simple functions follows. For a general , write as an increasing limit of simple functions and apply the Monotone convergence theorem.

Problem 4.

Let for (with respect to Lebesgue measure), and assume that .

(a). Show that a subsequence of tends to zero almost everywhere.

(b). Show by example that the sequence does not necessarily tend to zero almost everywhere.

Proof.

We shall be able to choose a subsequence such that

Then we are in the same situation as in Problem 2 of 2014 Spring Real Analysis Exam, and we can use the method there to complete the proof.

For part (b), use the usual typewriter example (characteristic functions of intervals of length wrapping around the interval).

Remark.

Some elaboration of the typewriter function. For any , define if and otherwise. Let

Then it is easy to verify that

On the other hand, we can verify that there are infinite many pairs of real numbers such that

for any . Therefore for any , is not convergent as .

Problem 5.

Let and .

(a). Show that the inclusions and are both false.

(b). Show that, for any , we have , and furthermore that for we have

where

Proof.

For part (a), functions and do the trick.

For part (b), without loss of generality, we assume that is a nonnegative function. We shall seek positive numbers such that

which is equivalent to

Let

Then are conjugate numbers, that is, . By Young's inequality,

Therefore we get

If we replace by for some positive number , we get

Taking integration on both sides, we get

where . In order to get the optimal , we take the derivative of the above right side with respect to to get

We can simplify the above to get

Thus for such a choice of , from (1) we get

The theorem is thus proved.

Remark.

A probably better way is to use the Hölder's inequality directly. We can write the inequality we want to prove as

Then since

the inequality follows.

However, our method actually proves the Hölder's inequality. Let be two nonnegative functions. By Young's inequality, we know that

it follows that

Replacing by and by , we get

The Hölder's inequality is obtained by minimizing the above right side with respect to .

Problem 6.

Let . Suppose that converges weakly to , that is, assume that

for all , where .

(a). Show that

(b). Give an example where the inequality in part (a) is strict.

Proof.

Let . Then

Using as a test function, we have

Using Holder inequality, we have

We obtain part (a) by combining the above two expressions.

Let and . Then is weakly convergent to by the Riemann-Lebesgue lemma, providing strict inequality in part (a).