Cartesian product of Lebesgue measurable sets is Lebesgue measurable Let $A \subset \mathbb{R^m}$, $B \subset \mathbb{R^n}$ and $\lambda_n$ the Lebesgue measure on $\mathbb{R^n}$.

How to prove that if $A$ and $B$ are Lebesgue measurable sets then also $A \times B$ is Lebesgue measurable.

For showing that the cartesian product is Lebesgue measurable, I tried to use:

$A \times B$ is Lebesgue measurable if $\varepsilon&gt;0$, and an open set $U$ and a closed set $V$ exist such that $V \subset A \times B \subset U$ and $\mathcal{L^{n+m}}(U$\ $V)&lt;\varepsilon$.

I don't know how to continue from here and how to conclude that the cartesian product is also Lebesgue measurable.
As a prelude, let me first say that $A\times B$ is clearly in the product $\sigma$ algebra, since the product sigma algebra $\mathcal{F}\otimes \mathcal{G}$ of two measurable spaces is the sigma algebra generated by measurable 'rectangles'. i.e: 

$\mathcal{F}\otimes \mathcal{G}= \sigma\Big( \{ E\times F: E\in \mathcal{F}, F\in \mathcal{G} \} \Big)$

Going by this definition, your set $A\times B$ is Lebesgue measurable in $\mathbb{R}^{n+m}$. However going by your criterion, recall first that for $A\in \mathcal{L}^n$ and $B\in \mathcal{L}^m$, we have:

$\lambda_n \otimes \lambda_m (A\times B)= \lambda_n(A)\times \lambda_m(B)$

Since $A,B$ are Lebesgue measurable, there exist $U_1,U_2$ open and $V_1,V_2$ closed such that: 

$U_1\supseteq A \supseteq V_1$ , $U_2\supseteq B \supseteq V_2$ while $\lambda_n(U_1\setminus V_1)\leq \frac{1}{2} \tilde{\epsilon}$ and $\lambda_m(U_2\setminus V_2)\leq \frac{1}{2} \tilde{\epsilon}$

Notice that $U_1\times U_2\supseteq A\times B \supseteq V_1\times V_2$ while $U_1\times U_2$ is open and $V_1 \times V_2$ is closed. Since you can write:

$U_1\times U_2= \Big(  U_1\times V_2 \Big) \sqcup \Big( U_1 \times (U_2 \setminus V_2) \Big)=  \Big( V_1\times V_2 \Big) \sqcup \Big( (U_1\setminus V_1)\times V_2 \Big) \sqcup \Big( U_1 \times (U_2\setminus V_2) \Big) $

Then:

$\lambda_{n+m}\Big( (U_1\times U_2) \setminus (V_1 \times V_2) \Big)= \lambda_{n+m}\Big( (U_1\setminus V_1)\times V_2 
 \Big) + \lambda_{n+m}\Big(  U_1 \times (U_2\setminus V_2)
 \Big)= $

$= \lambda_n(U_1 \setminus V_1)\cdot \lambda_m(V_2)+ \lambda_n(U)\cdot \lambda_m(U_2 \setminus V_2)$

If you assume $\lambda_n(U_1), \lambda_m(U_2)\leq M$, then for $\tilde{\epsilon}= \dfrac{\epsilon}{2M}$, you've shown that the difference is of measure less than $\epsilon$. Otherwise, there is a standard argument (which I will elaborate if you request) of going from the finite measured case to the $\sigma$-finite case.
As a prelude, let me first say that $A\times B$ is clearly in the product $\sigma$ algebra, since the product sigma algebra $\mathcal{F}\otimes \mathcal{G}$ of two measurable spaces is the sigma algebra generated by measurable 'rectangles'. i.e: 

$\mathcal{F}\otimes \mathcal{G}= \sigma\Big( \{ E\times F: E\in \mathcal{F}, F\in \mathcal{G} \} \Big)$

Going by this definition, your set $A\times B$ is Lebesgue measurable in $\mathbb{R}^{n+m}$. However going by your criterion, recall first that for $A\in \mathcal{L}^n$ and $B\in \mathcal{L}^m$, we have:

$\lambda_n \otimes \lambda_m (A\times B)= \lambda_n(A)\times \lambda_m(B)$

Since $A,B$ are Lebesgue measurable, there exist $U_1,U_2$ open and $V_1,V_2$ closed such that: 

$U_1\supseteq A \supseteq V_1$ , $U_2\supseteq B \supseteq V_2$ while $\lambda_n(U_1\setminus V_1)\leq \frac{1}{2} \tilde{\epsilon}$ and $\lambda_m(U_2\setminus V_2)\leq \frac{1}{2} \tilde{\epsilon}$

Notice that $U_1\times U_2\supseteq A\times B \supseteq V_1\times V_2$ while $U_1\times U_2$ is open and $V_1 \times V_2$ is closed. Since you can write:

$U_1\times U_2= \Big(  U_1\times V_2 \Big) \sqcup \Big( U_1 \times (U_2 \setminus V_2) \Big)=  \Big( V_1\times V_2 \Big) \sqcup \Big( (U_1\setminus V_1)\times V_2 \Big) \sqcup \Big( U_1 \times (U_2\setminus V_2) \Big) $

Then:

$\lambda_{n+m}\Big( (U_1\times U_2) \setminus (V_1 \times V_2) \Big)= \lambda_{n+m}\Big( (U_1\setminus V_1)\times V_2 
 \Big) + \lambda_{n+m}\Big(  U_1 \times (U_2\setminus V_2)
 \Big)= $

$= \lambda_n(U_1 \setminus V_1)\cdot \lambda_m(V_2)+ \lambda_n(U)\cdot \lambda_m(U_2 \setminus V_2)$

If you assume $\lambda_n(U_1), \lambda_m(U_2)\leq M$, then for $\tilde{\epsilon}= \dfrac{\epsilon}{2M}$, you've shown that the difference is of measure less than $\epsilon$. Otherwise, there is a standard argument (which I will elaborate if you request) of going from the finite measured case to the $\sigma$-finite case.