2026 Spring Real Analysis

Problem 1.

Let . Prove that

Here denotes Lebesgue measure.

Proof.

For each , we have

If , this sum is . If , then

Therefore, for every ,

By Tonelli's theorem

and hence

Using the pointwise bound above, we get

Since ,

Therefore,

Problem 2.

Let be a semi-finite measure space, let , and suppose that

Show that for any , there exists such that

Proof.

Recall that is semi-finite means that if and , then there exists such that

We prove the claim by contradiction. Suppose there exists some such that every measurable subset with finite measure satisfies

Define

By assumption,

By the definition of supremum, for each , choose with such that

Let

Then and , so by the definition of ,

Also,

Hence

Now set

Since is increasing, continuity from below gives

Because

and

we must have

Since is semi-finite, there exists such that

and

Then

and

Also,

This contradicts the definition of as the supremum of finite measures of measurable subsets of .

Therefore, for every , there exists with

Problem 3.

Assume that . Compute

Justify your answer.

Proof.

For each fixed , since , we have for almost every .

If , then

for every . Therefore,

for almost every .

Now we need a dominating function. Since , for every ,

Hence

Because and has finite measure,

Thus, by the Dominated convergence theorem,

Therefore,

Problem 4.

Let be uncountable. Prove that there exists such that, for every ,

is uncountable.

Proof.

Suppose not. Then for every , there exists such that

is countable.

For each , choose an open interval with rational endpoints such that

Then

is countable, because

Now let be the collection of all open intervals with rational endpoints. This collection is countable. Also, by construction,

Therefore,

The right-hand side is a countable union of countable sets, hence is countable. This contradicts the assumption that is uncountable.

Therefore, there must exist some such that for every ,

is uncountable.

Problem 5.

Let be a sequence of measurable real-valued functions defined on a measurable set with

Suppose that, for each ,

for some . Show that for every , there exist a closed set and a finite constant such that

and

Proof.

Define

Since each is measurable, is measurable. By assumption, for every ,

Hence

for every .

For each , define

Then each is measurable, and

Also, since for every ,

Because , by continuity from below,

Therefore,

Choose such that

Since is measurable and has finite measure, by inner regularity of Lebesgue measure, there exists a closed set such that

Then , and

Finally, since , for every we have

Thus, for every and every ,

Taking

we obtain the desired uniform bound on .

Problem 6.

Let be locally absolutely continuous with

Prove that the limit

exists.

Proof.

For each , define

Then

Thus it is enough to show that

converges.

Since is absolutely continuous on every compact interval, for we have

Therefore

Hence

By changing the order of integration,

Thus

Summing over , we get

Therefore

converges absolutely. Hence the sequence of partial sums

has a finite limit as .

Therefore,

exists.