# 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, $|A \times B|=|A|\cdot|B|$.

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So... how do you define "measurable subset" of $\mathbb{R}^k$, and how do you define the measure of a measurable subset of $\mathbb{R}^k$? – Arturo Magidin Apr 20 '11 at 2:52
A subset E in $\mathbb{R^n}$ is measurable if given any $\epsilon$ > 0 there exists an open set G $\subseteq$ $\mathbb{R^n}$ with E $\subseteq$ G such that |G\E|$_e$ < $\epsilon$ – emka Apr 20 '11 at 2:58
A related question: math.stackexchange.com/q/13876 – Jonas Meyer Apr 20 '11 at 3:20

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?

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Am I using the idea of G$_\delta$ sets? – emka Apr 20 '11 at 3:08
@josh: I didn't say anyting about $G_{\delta}$ sets. I only talked about open sets. Of what I wrote, what made you think about $G_{\delta}$ sets? – Arturo Magidin Apr 20 '11 at 3:20
Well, in my textbook I have the following: E is measurable is and only if E = H -Z, where H is a G$_\delta$ and |Z|=0. – emka Apr 20 '11 at 3:30
@josh: I used the definition you gave. I don't know what your textbook is, nor am I taking (or teaching your class), so I cannot guess what theorems you may or may not be able to use (the Government doesn't like it when I try to read minds of students without a warrant, so I try not to do it). Nothing I said refered to $G_{\delta}$ sets, so I do not see why you think my questions above had anything to do with $G_{\delta}$ sets. If you prefer to ignore my response and go with theorems in your textbook, I'll be more than happy to delete the answer so as not to distract you. – Arturo Magidin Apr 20 '11 at 3:34
come now, don't be so harsh. @Josh: while this may or may not be a homework problem, it will benefit you to read and follow the advice here, since the question you asked is similar in scope to one that is typically assigned for exercise. Pay especially attention to the section on what information you should include when asking a question. – Willie Wong Apr 21 '11 at 11:58

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:

1. We may assume tha $A$ and $B$ are of finite measure, say $m(A),m(B)\le M$.
2. 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 $m(G),m(H)\le M+\epsilon$.
3. 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 $m(G^*),m(H^*)<2\epsilon$.
4. $G\times H\setminus A\times B=A\times(H\setminus B)\cup(G\setminus A)\times H\subset G\times H^*\cup G^*\times H$.
5. $G\times H^*$ and $G^*\times H$ are open, and hence measurable.
6. Use some form of Fubini's theorem to show that $m(G\times H^*)=m(G)\cdot m(H^*)$ and $m(G^*\times H)=m(G^*)\cdot m(H)$.
7. 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.)
8. Once you know $A$ and $B$ are measurable, again Fubini's theorem will show that $m(A\times B)=m(A)\cdot m(B)$.
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