Ex3-12

Question 3.12
If E1 and E2 are measurable subsets of R^1, show that E1 × E2 is a measurable subset of R2 and |E1 × E2|=|E1||E2|.
Solution:
Lemma 3.1: If A ⊆ R and Z ⊆ R with |Z| = 0, then |A×Z| = 0. Similarly,|Z × A| = 0. 
Proof:
Let \epsilon>0. Since |Z| = 0, there exists intervals \{I_k\}such that Z\subseteq\cup I_k and\cup |I_k|<\epsilon.
Write A_{n} = A ∩ [−n, n]. Then A=\cup A_n.n. Note that 
A_n \times Z\subseteq [-n,n] \times \cup I_k = \cup( [-n,n]\times I_k) 
so\left|A_{n} \times Z\right|_{e} \leq \sum_{k=1}^{\infty} 2 n\left|I_{k}\right|=2 n \epsilon,
Since\epsilon>0 is arbitrary,\left|A_{n} \times Z\right|=0 for each  n . So
|A \times Z|_{e}=\left|\bigcup_{n=1}^{\infty}\left(A_{n} \times Z\right)\right|_{e} \leq \sum_{n=1}^{\infty}\left|A_{n} \times Z\right|_{e}=0
Thus  | |A \times Z|=0   as desired.
Now, since E_{1}is measurable, E_{1}=H_{1} \cup Z_{1}, where H_{1} is of type Fσ and| Z_{1}| = 0. Similarly, write E_{2}=H_{2} \cup Z_{2}where H_{2} is of type Fσ and |Z_{2}| = 0.
Then
E_{1} \times E_{2}=\left(H_{1} \times H_{2}\right) \cup\left(H_{1} \times Z_{2}\right) \cup\left(Z_{1} \times H_{2}\right) \cup\left(Z_{1} \times Z_{2}\right) .
Note that H_1\times H_2 is of type Fσ, while the other terms have measure zero by the previous lemma. Thus E_1\times E_2 is measurable. 
Case 1) Suppose |E_1| and |E_2| are both finite. 
Since E_1, E_2are measurable, for each k ∈ N there are open sets S_k ⊇ E_1, T_k ⊇ E_2 such that |S_k \ E_1| < 1/k, | T_k \ E_2| < 1/k. We may assume
S_{k+1} ⊆ S_k,T_{k+1} ⊆ T_k (if S_{k+1} \nsubseteq S_{k} , then define S_{k+1}^{\prime}=S_{k+1} \cap S_{k}instead).
SinceS_k is open,S_k = S_{k}=\bigcup_{k \in \mathbb{N}} I_{i} for some nonoverlapping closed intervals.
Similarly, T_{k}=\bigcup_{j \in \mathbb{N}} J_{j}for some nonoverlapping closed intervals. 
So
\begin{aligned}
\left|S_{k} \times T_{k}\right| &=\left|\bigcup_{(i, j) \in \mathbb{N} \times \mathbb{N}}\left(I_{i} \times J_{j}\right)\right| \\
&=\sum_{i, j \in \mathbb{N}}\left|I_{i} \times J_{j}\right| \\
&=\sum_{i, j \in \mathbb{N}}\left|I_{i}\right|\left|J_{j}\right| \\
&=\left(\sum_{i \in \mathbb{N}}\left|I_{i}\right|\right)\left(\sum_{j \in \mathbb{N}}\left|J_{j}\right|\right) \\
&=\left|S_{k}\right|\left|T_{k}\right| .
\end{aligned}
Write S=\bigcap_{k=1}^{\infty} S_{k}, T=\bigcap_{k=1}^{\infty} T_{k}.Then\left|S \backslash E_{1}\right|=\left|T \backslash E_{2}\right|=0.Hence,
\begin{aligned}
\left|E_{1} \times E_{2}\right| &=|S \times T| \\
&=\lim _{k=\infty}\left|S_{k} \times T_{k}\right| \\
&=\lim _{k \rightarrow \infty}\left|S_{k}\right|\left|T_{k}\right| \\
&=\left|E_{1}\right|\left|E_{2}\right|,
\end{aligned}
where the second equality follows by MCT for measure, since Sk×Tk & S×TS_{k} \times T_{k} \searrow S \times T
and\left|S_{k} \times T_{k}\right|<\infty for some k since |E_1|, |E_2| are both finite. The last equality also follows by MCT for measure.
Case 2) Suppose one of |E_1|, |E_2| are infinite
If  \left|E_{1}\right|=\inftyand \left|E_{2}\right|>0 , then write E_{1}^{n}=E_{1} \cap[-n, n] .
\begin{aligned}
\left|E_{1} \times E_{2}\right| &=\lim _{n \rightarrow \infty}\left|E_{1}^{n} \times E_{2}\right| \\
&=\lim _{n \rightarrow \infty}\left|E_{1}^{n}\right|\left|E_{2}\right| \\
&=\left|E_{1}\right|\left|E_{2}\right| \\
&=\infty
\end{aligned}
\text{where the first equality follows by MCT for measure, since} E_{1}^{n} \times E_{2} \nearrow E_{1} \times E_{2}.
\text { If }\left|E_{1}\right|=\infty \text { and }\left|E_{2}\right|=0,\left|E_{1} \times E_{2}\right|=0 \text { by our first lemma. }
(proved)