Ex3-11

Solution to Chapter 3, Exercise 11

Question: Prove Theorem 3.29.

Theorem 3.29 ''Suppose that $$|E|_e<+\infty$$. Then $$E$$is measurable if and only if given $$\epsilon>0$$,$$E=(S \cup N_1)-N_2$$, where $$S$$is a finite union of nonoverlapping intervals and $$|N_1|_e,|N_2|_e<\epsilon$$.''

First, suppose that $$E$$is measurable. Given $$\epsilon>0$$, we can find an open set $$G$$so that we have $$|G-E|_e < \epsilon$$.

Since $$G$$is open, by Theorem 1.11 $$G$$can be written as a countable union of nonoverlapping intervals $$\{ I_k \}_{k=1}^\infty$$.

$$G=\bigcup_{k=1}^{\infty}I_k$$

Since $$|E|_e<\infty$$, we can choose the nonoverlapping intervals so that

$$|G|_e=\left|\bigcup_{k=1}^{\infty}I_k\right|_e=\sum_{k=1}^\infty |I_k|_e <\infty$$Then since $$\sum_{k=1}^\infty |I_k|_e$$is finite, there exists an $$N\in\mathbb{N}$$such that$$\left| \sum_{k=1}^\infty |I_k|_e - \sum_{k=1}^N |I_k|_e \right| <\epsilon$$$$\left| \sum_{k=N+1}^\infty |I_k|_e \right|=\sum_{k=N+1}^\infty |I_k|_e<\epsilon$$

Let

$$\begin{align} S&=\bigcup_{k=1}^N I_k \\ N_1&=\bigcup_{k=N+1}^\infty I_k \\ N_2&=G-E \end{align}$$

Then

$$\begin{align} |N_1|_e&=\sum_{k=N+1}^\infty |I_k|_e <\epsilon \\ |N_2|_e&=|G-E|_e<\epsilon \end{align}$$

and since $$E\subset G$$,

$$E=G-N_2=(S \cup N_1)-N_2.$$

Conversely, given any $$\epsilon>0$$. There exists $$N_1,N_2$$and $$S$$ such that $$E=(S \cup N_1)-N_2$$, where $$S$$is a finite union of nonoverlapping intervals and $$|N_1|_e,|N_2|_e<\frac{\epsilon}{4}$$.

We know that $$S$$is measurable, then there exists an open set $$G$$such that $$S \subset G$$and $$|G-S|<\frac{\epsilon}{4}$$.

By Theorem 3.6, we can find an open set $$G_1$$such that $$N_1 \subset G_1$$and $$|G_1|_e \leq |N_1|_e + \frac{\epsilon}{4} < \frac{\epsilon}{2}$$.

Since $$S \subset G$$and $$N_1 \subset G_1$$, we have $$S \cup N_1 \subset G \cup G_1$$.

Then $$E=(S \cup N_1)-N_2 \subset S \cup N_1 \subset G \cup G_1$$. Also, we have

$$\begin{align} |(G \cup G_1)-E|_e &= |(G-E)\cup(G_1-E)|_e\\ &\leq |G-E|_e + |G_1-E|_e \\ &\leq |G-E|_e + |G_1|_e \\ &\leq |G-E|_e + \frac{\epsilon}{2} \end{align}$$

since $$G_1-E \subseteq G_1$$. Furthermore,

$$\begin{align} G-E &= G-((S \cup N_1)-N_2) \\ &= G-((S \cup N_1)\cap N_2^C) \\ &= G\cap((S \cup N_1)\cap N_2^C)^C \\ &= G\cap((S \cup N_1)^C\cup N_2) \\ &= (G\cap(S \cup N_1)^C)) \cup (G\cap N_2) \\   &= (G\cap S^C \cap N_1^C) \cup (G\cap N_2) \\    &\subseteq (G\cap S^C) \cup N_2 \\    &= (G-S) \cup N_2 \end{align}$$

We used the fact that $$A-B=A\cap B^C$$. Then we have

$$\begin{align} |G-E|_e &\leq |G-S|_e + |N_2|_e \\ &<\frac{\epsilon}{4} + \frac{\epsilon}{4} \end{align}$$

Therefore,

$$|(G \cup G_1)-E|_e < \frac{\epsilon}{4}+\frac{\epsilon}{4}+\frac{\epsilon}{2} = \epsilon$$

Hence, we have found an open set $$G'=G\cup G_1$$ such that $$E \subset G'$$and $$|G'-E|_e<\epsilon$$.