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
for all .
To link to , we use the Cauchy-Schwarz inequality:
Hence from (1), we get
Integrating over (Fubini’s theorem), we get
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
(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
(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
Finally, we want to show that (2) implies that
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