2011 Fall Real Analysis

Problem 1.

Let a measurable bounded set has the property that every continuous is uniformly continuous. Show that X is compact.

Proof.

is compact if and only if is closed. If and is a limit point of . We consider a continuous function

we shall prove that is not uniform continuous. Let be a sequence of such that . Since , we can inductively choose a subsequence such that

Therefore is not uniformly continuous, a contradiction.

Problem 2.

(a). For any construct an open set so that and . Here and later stands for the Lebesgue measure.

(b). Let . Find . Show that .

Proof.

(a). Let be the sequence of rational numbers in . for any , we define

Then

(b). We have

Thus for any , . Thus and . Since contains all the reational numbers, we have .

Problem 3.

Show that either a -algebra of subsets of a set is finite, or else it has uncountably many elements.

(Hint: if the -algebra is not finite, how many pairwise disjoint elements does it necessarily contain?)

Proof.

If the -algebra on a measure space contains a collection of infinite disjoint subsets, then the power set of this collection would be uncountable. Otherwise, let be a finite collection of subsets of the -algebra. Assume that is the maximal number of such collections. the we claim that

because otherwise would provide one more. Let be a subset. Then if is not one of the 's, we must have an such that but . Thus splits into two disjoint subsets and , contradicting to the fact that is the maximum number of disjoint collections.

Problem 4.

Let be absolutely continuous and assume that and that . Show that the following limit exists and compute its value.

Proof.

The limit is equal to . Since is absolutely continuous and , we have

Therefore by Cauchy inequality we have

Since , we must have

by the Lebesgue's Dominated Convergence Theorem. This completes the computation.

Problem 5.

Let be a closed set and let . Assume that for all measurable with , we have

Prove that for a.e. .

Proof.

We prove by contradiction. Let be a positive measure set such that for all , . Since is a closed set, is open. As a result, we are able to write

of disjoint open intervals. Since

there exsits an such that . Since

there exists an such that , where . Thus we have a contradiction

which is not in .

Problem 6.

If is a Lebesgue measurable real-valued function on such that the function is Lebesgue integrable over the square . Show that is Lebesgue integrable over .

Proof.

Let be a constant so that

where . We thus have

Therefore we have

Thus is integrable.