This post is a sequel to Exercises in Real Analysis.

4. Let be the triangle , and be the restriction of the planar Lebesgue measure on . Suppose that . Prove that

.

*Solution*. Assume on the contrary that

.

Then for some , we have

(1)

for all .

To link to , we use the *Cauchy-Schwarz inequality*:

.

Hence from (1), we get

, .

Integrating over (Fubini’s theorem), we get

.

Contradiction.

**********

5. Let be a compact metric space, and be a finite positive Borel measure on . Suppose that for every . Prove that for every there is a such that, if is any Borel subset of having diameter less than , then .

*Solution*. Since , by *continuity of measure* we have

for any . Thus, given , there exists such that

.

Clearly is an open cover of , and by compactness there is a finite subcover, say

.

Recall that the finite cover has some **Lebesgue number** , i.e., every subset of with diameter less than is contained entirely in some . Hence (assuming is Borel)

.

**********

6. Let and be compact metric spaces, and let and denote the Banach spaces of continuous, real-valued functions on and respectively (with sup norm). Suppose that is a continuous surjective map. Let

.

(a) Show that is a closed subspace of , and that

.

(b) Let be a finite positive Borel measure on . Prove that there is a finite positive Borel measure on such that for all Borel subsets of .

*Solution.* (a) It is easy to show that is a closed subspace.

It is also clear that . To show the reverse inclusion, consider any . Since is surjective, for each there is some such that . We define a function on by setting . The function is well defined, for if , then since .

It remains to show that is continuous. Fix and suppose . Fix such that . Then

.

Since is compact, some subsequence converges to some . By continuity of ,

.

Finally, by continuity of , we have

.

Hence .

(b) To *construct a measure* on a compact metric space, a good idea is to use the **Riesz representation theorem**.

Define a positive linear functional on the *subspace* by setting

if .

(You should verify that this definition makes sense.) The linear functional is bounded on , since

.

(You should verify the last inequality.) To use the Riesz representation theorem, we need a linear functional defined on the *whole* . To get such a functional, we simply use the **Hahn-Banach theorem** to extend to a *bounded* linear functional on . (Note that the question only requires us to construct *a* measure. Uniqueness is not guarenteed.)

Now, the Riesz representation theorem provides us with a positive Borel measure on such that

, .

And when , we have

. (2)

Finally, we want to show that (2) implies that

(3)

for all Borel subsets of .

It suffices to prove (3) where is open. Fix such a set . Then, there exists a sequence such that .

.

By the monotone convergence theorem, we get

.

Here is another solution for Real analysis prelim provided by Wongting.

Question:(a) For a nonnegative function on , show that there is a nonnegative strictly increasing function with and .

(b) For a sequence of nonnegative functions on with , find a nonnegative strictly increasing function with and .

Solution:(a) Let . We can choose with such that

We then define and is linear between ‘s. Clearly we only have to check .

To see this, let and , . Observe that , and so . We then have

where . Thus .

(b) We modify the above proof. Let . There is such that

Inductively, we can find with such that for all ,

We then choose exactly as before, i.e. and linear in between. The argument in (a) then shows that , noting that is independent of .

Nice proof!

In the last sentence, it should be . [Yes, thanks! -KKK]