Ex3-11

Solution to Chapter 3, Exercise 11
Question: Prove Theorem 3.29.
Theorem 3.29 Suppose that |E|_e<+\infty. Then Eis measurable if and only if given \epsilon>0,E=(S \cup N_1)-N_2, where Sis a finite union of nonoverlapping intervals and |N_1|_e,|N_2|_e<\epsilon.
First, suppose that Eis measurable. Given \epsilon>0, we can find an open set Gso that we have |G-E|_e < \epsilon.
Since Gis open, by Theorem 1.11 Gcan 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 <\inftyThen since \sum_{k=1}^\infty |I_k|_eis 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_2and S such that E=(S \cup N_1)-N_2, where Sis a finite union of nonoverlapping intervals and |N_1|_e,|N_2|_e<\frac{\epsilon}{4}.
We know that Sis measurable, then there exists an open set Gsuch that S \subset Gand |G-S|<\frac{\epsilon}{4}.
By Theorem 3.6, we can find an open set G_1such that N_1 \subset G_1and |G_1|_e \leq |N_1|_e + \frac{\epsilon}{4} < \frac{\epsilon}{2}.
Since S \subset Gand 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.