2025 Fall Real Analysis

Proof.

For each , define

Since for every , we have

Also, the sets are increasing:

Therefore, by continuity from below of Lebesgue measure,

Choose such that

Now let be measurable with . Then

Since , we have . Hence

On , we have

Thus

Since this lower bound is independent of , it follows that

This proves the claim.

Proof.

Use polar coordinates

The region becomes

Also,

and . Therefore

Thus

Since , the -interval has finite length. Hence the only possible singularities occur when

or when

that is, when

First consider the behavior near . We need to study

As , set

Then and

If , then the integral becomes comparable to

which is finite if and only if

If , then the factor is integrable near , and the logarithmic factor cannot destroy integrability. Hence the integral is finite for all when .

If , then is too singular near , and the logarithmic factor cannot restore integrability. Hence the integral diverges for all when .

So near , integrability requires

or

Now consider the behavior near . Since

and as , we have

Also, near , the factor is bounded above and below by positive constants. Therefore the relevant integral is comparable to

This integral is finite if and only if

Therefore we must have

Combining the two conditions, the case is impossible, because near it would require , while near it requires . Thus the only possible condition is

Conversely, if

then the integral is finite near , finite near , and bounded away from both singularities. Hence

Therefore,

Proof.

First, applying the assumption to and to , we get

for every .

We will show that

for every .

Fix , and let

Since is compact, it is contained in some cube . Let . Cover by finitely many cubes of side length , where

Choose a continuous partition of unity subordinate to slightly enlarged cubes, such that

and

Then

Each belongs to . Therefore,

Using the assumed estimate, we get

Since ,

Thus

Hence

Using the estimates on the supports,

Since , we have

Therefore,

It follows that

Thus

for every .

Since and its integral against every continuous compactly supported test function is zero, we conclude that

almost everywhere on .

Proof.

Since , is continuous and has compact support. Therefore attains its maximum. Choose such that

If , the conclusion is immediate. So assume and set

Thus and

Choose a nonnegative function such that

Define

Then

and concentrates at as .

Now define

Then is absolutely continuous, hence , and

Therefore

Finally,

Since is continuous at and is an approximate identity concentrating at , we have

Hence

Thus there exists a sequence with such that

Proof.

First suppose that is Lebesgue measurable. Since every open interval is measurable, the sets

are measurable and disjoint. Therefore, by additivity of Lebesgue measure,

Since and measurable sets have measure equal to outer measure, we get

Conversely, assume that for every bounded open interval ,

We prove that is Lebesgue measurable by verifying Carathéodory's criterion.

Let be arbitrary. We need to show that

By subadditivity of outer measure, we always have

Thus it remains to prove the reverse inequality.

If , there is nothing to prove. So assume

Let . By the definition of outer measure, there exist bounded open intervals such that

and

For each , by assumption,

Since

we have

Similarly,

so

Therefore,

Using the assumed equality on each interval , we get

Since was arbitrary,

Together with the opposite inequality from subadditivity, this gives

for every .

Hence, by Carathéodory's criterion, is Lebesgue measurable.