The product of a measurable subset of $\mathbb{R}^n$ and a measurable subset of $\mathbb{R}^m$ I'm very new to undergraduate Lebesgue measure and am still having problems with it.
I'm trying to prove the following:
Let $A\subset\mathbb{R}^m$ and $B\subset\mathbb{R}^n$ and $A\times B\subset\mathbb{R}^{m+n}$.
If $A$ and $B$ are both measurable, then $A \times B$ is measurable.  
If $A$ and $B$ are both measurable, $|A \times B|=|A|\cdot|B|$.
 A: Okay, given your definition:
If $G$ is open in $\mathbb{R}^n$ and $H$ is open in $\mathbb{R}^m$, is $G\times H$ open in $R^{m+n}$?
If $A\subseteq G\subseteq\mathbb{R}^n$ and $B\subseteq H\subseteq \mathbb{R}^m$, is $A\times B\subseteq G\times H$ true?
If $A\subseteq G$ and $B\subseteq H$, what is $(G\times H)\setminus(A\times B)$? (Careful here...) Can you make this arbitrarily small, if you can make $G-A$ and $H-B$ arbitrarily small?
A: This is an elaboration of Arturo's answer; it is too long to be a comment. It usus your definition of measurable set and Fubini's theorem. Proceed in steps:


*

*We may assume tha $A$ and $B$ are of finite measure, say $m(A),m(B)\le M$.

*Let $\epsilon>0$ be given. Choose open sets $G\subset\mathbb{R}^m$, $H\subset\mathbb{R}^n$ such that $A\subset G$, $B\subset H$ and $m(G\setminus A)<\epsilon$, $m(H\setminus B)<\epsilon$; then $m(G),m(H)\le M+\epsilon$.

*Choose open sets $G^*\subset\mathbb{R}^m$, $H^*\subset\mathbb{R}^n$ such that $G\setminus A\subset G^*$, $H\setminus B\subset H^*$, $m(G^*\setminus(G\setminus A))<\epsilon$, $m(H^*\setminus(H\setminus B))<\epsilon$. Then $m(G^*),m(H^*)<2\epsilon$.

*$G\times H\setminus A\times B=A\times(H\setminus B)\cup(G\setminus A)\times H\subset G\times H^*\cup G^*\times H$.

*$G\times H^*$ and $G^*\times H$ are open, and hence measurable.

*Use some form of Fubini's theorem to show that $m(G\times H^*)=m(G)\cdot m(H^*)$ and $m(G^*\times H)=m(G^*)\cdot m(H)$.

*Deduce that $m^*(G\times H\setminus A\times B)\le4(M+\epsilon)\epsilon$ ($m^*$ is the exterior measure; we still do not know that $G\times H\setminus A\times B$ is measurable.)

*Once you know $A$ and $B$ are measurable, again Fubini's theorem will show that $m(A\times B)=m(A)\cdot m(B)$.

