# How to prove product of measurable set is measurable using dynkin's $\pi-\lambda$ theorem?

My question:

If $E_1$ and $E_2$ are measurable subsets of $R^1$, I want to show that $E_1 \times E_2$ is a measurable subset of $R^2$ and $|E_1 \times E_2|=|E_1||E_2|$.

My attempt:

First I tried to use the Dynkin's $\pi-\lambda$ theorem to show this holds for $E_1$ and $E_2$ are Borel sets.

The proof is as follows.

(First, let $C$={$E_1|$$E_2is closed interval,E_1 \times E_2 is a measurable set, |E_1 \times E_2|=|E_1||E_2|}). Let D={closed interval}, then D\in C, and D is a \pi system. Now prove C is a \lambda system. First, R\in C. Second, E_1,F_1\in C, then E_1 \ F_1\in C. Third, E_{1n} \uparrow E_1, E_{1n}\in C, then E_1 \in C.So by dynkin's \pi \lambda theorem, Borel sets=\sigma(D)\in C. Now, let C^1={E_1|$$E_2$is Borel sets,$E_1 \times E_2$ is a measurable set, $|E_1 \times E_2|=|E_1||E_2|$}). By the last step, $D$={closed interval}, then $D\in C$, iterating the last step, by dynkin's $\pi \lambda$ theorem, Borel sets=$\sigma(D)\in C^1$.

So I think I proved(not sure): If $E_1$ and $E_2$ are Borel subsets of $R^1$, $E_1 \times E_2$ is a measurable subset of $R^2$ and $|E_1 \times E_2|=|E_1||E_2|$.

My question:

1.How to extend this result form Borel sets to Lebesgue measurable sets?

2.About proving $C$ is a $\lambda$ system, I don't know how to justify the second and third conditions:

(Second, $E_1$,$F_1\in C$, then $E_1$ \ $F_1\in C$. Third, $E_{1n} \uparrow E_1, E_{1n}\in C$, then $E_1 \in C$.So by dynkin's $\pi \lambda$ theorem, Borel sets=$\sigma(D)\in C$.)

Thanks for your help!