2025 Spring Real Analysis

Proof.

Since , we have

Fix . Because is integrable, there exists such that

Now for any measurable set , we split

On the set , we have , so

Also,

Therefore,

Choose

Then whenever , we have

Hence

Thus

Proof.

Since for every and pointwise almost everywhere, we have

almost everywhere on .

For real numbers , we have

where

Applying this with and , we get

Therefore,

Now, since almost everywhere,

almost everywhere. Also, because and almost everywhere, we have

Since , the Dominated Convergence theorem gives

Therefore,

Proof.

Let

Then and . We will show that if is not zero almost everywhere, then

as , which contradicts the hypothesis.

Suppose, toward a contradiction, that is not zero almost everywhere. Then there exist and such that the set

has positive measure. Hence, for every ,

We claim that

as .

Indeed, by Tonelli's theorem and the change of variables ,

Since , translations are continuous in . Therefore

as . Since , there exists such that whenever ,

Thus

But

which tends to as . Hence

Therefore,

contradicting the assumption that the limsup is finite. Thus almost everywhere, which means

almost everywhere.

Proof.

Define

We will show that this extension is absolutely continuous on . This will imply that is absolutely continuous on .

For , differentiating gives

Indeed,

Therefore,

Since for , we get

But

Hence

Now, for , the ordinary fundamental theorem of calculus gives

Letting , we have

because . Also, since ,

Therefore,

Thus the extension of to satisfies

with . Hence is absolutely continuous on . In particular, is absolutely continuous on .

Proof.

We first prove (i) . If , then

Now assume . Suppose, for contradiction, that

Then there exists such that, for a subsequence,

for all sufficiently large .

Set

Then

Moreover, since and , we have

Also, the assumption implies that, for some ,

for all sufficiently large .

Since , the space is uniformly convex. Hence there exists such that whenever

and

we have

Applying this to and , we get

for all sufficiently large .

But since , we have

By weak lower semicontinuity of the norm,

This contradicts . Therefore,

This proves (i) .

Now we disprove (ii) . Let . Define the Rademacher functions on by setting on the first half of each dyadic interval of length and on the second half. Extend by outside .

Then

Indeed, if , then . Approximate in by a dyadic step function . For sufficiently large,

because has average zero on each dyadic interval on which is constant. Hence

Thus in .

Now set

Then

Also, on , the function is equal to on half of the interval and on the other half. Therefore

for every , while

Hence

However,

for every . Thus

Therefore, the statement is false when .

Proof.

(a) Let be the Cantor function on .

Recall that the Cantor function is non-decreasing, satisfies

and is constant on every connected component of the complement of the Cantor set .

Since the Cantor set has Lebesgue measure , the complement has full measure. For every , there exists an open interval containing on which is constant. Therefore

for every . Hence

for almost every . Also,

This proves (a).

(b) Since has Lebesgue measure , for each there exists an open set such that