2024 Fall Real Analysis

Proof.

Let

Since is positive and continuous on , we have .

For any , let

By continuity of , . We also have

Thus we have

Letting , we get

Since is arbitrary, we complete the proof.

Proof.

It is enough to prove the statement on each interval , where , and then take a countable union.

Fix . Consider

By Tonelli's theorem

Using the change of variables , we get

Hence

Since , the last integral is finite. Therefore

for almost every .

For such , the series

converges, so its terms must tend to . Thus

and therefore

for almost every .

Finally, since

and this is a countable union, the conclusion holds for almost every . Hence

Proof.

Assume that for every ,

We use the layer-cake identity

Indeed,

Therefore, by Tonelli's theorem

Interchanging the order of integration gives

Using the hypothesis,

Now split the integral at :

Since ,

and

Hence

Therefore,

Proof.

First observe that is periodic with period , and

Define

Then is also periodic with period , and

It is enough to prove that

We first prove this when . Define

Since has period and mean zero over one period, is bounded on . Therefore,

Since is bounded, it follows that

By linearity, the same conclusion holds for every step function on :

Now let . Since step functions are dense in , for every there exists a step function such that

Also, since is bounded,

Hence

Therefore,

Since is arbitrary,

Finally, because , we have

Thus

Taking , we obtain

Proof.

Suppose, for contradiction, that such a function exists.

Take

Then , so by assumption,

However, since both and belong to , the convolution is a continuous function on .

Thus is continuous and equals almost everywhere. We claim this is impossible.

Indeed, if a continuous function equals almost everywhere, then on and on . This follows from continuity: if, for example, for some , then by continuity would differ from on an interval of positive measure, contradicting equality almost everywhere.

Therefore would have to satisfy

and

But then continuity at is impossible, since the left limit would be while the right limit would be .

Applying this to , we get a contradiction. Hence no such exists.

Proof.

Fix , and set

For each , by the definition of , there exists such that

Let

Then

and the previous inequality becomes

Equivalently,

We now use the following Vitali's covering lemma from any finite collection of intervals, one can choose a pairwise disjoint subcollection such that the union of the original intervals is contained in

where denotes the interval with the same center as and five times the length.

Now let be compact. The intervals cover , so by compactness there is a finite subcover. Applying the covering lemma to this finite collection, we obtain pairwise disjoint intervals such that

Therefore

Using the estimate for each , we get

Since the intervals are pairwise disjoint,

Hence

Since this holds for every compact set , by inner regularity of Lebesgue measure,

Thus