# Showing that the measurable sets $A,B$ satisfy $|A \times B| = |A||B|$

Show that the Lebesgue measurable sets $A,B$ satisfy $|A \times B| = |A||B|$. The approach I was going to use, was to use Tonelli's theorem which says $$\int \limits_{A \times B}f = \int \limits_{A} \left[ \int \limits_{B}f(x,y) dy \right] dx$$ Then use a theorem that says $$\int \limits_{A}f = \sum \limits_{i=1}^m a_j |A_{j}|$$ This will get me $$\sum \limits_{i=1}^m a_j |A_{j} \times B_{j}| = \sum \limits_{j=1}^m \left( \sum \limits_{i=1}^n b_i |B_{i}| \right) _j |A_{j}|$$ but I am not sure how I could manipulate this to remove the summation. The only method I could think of would be to say $f$ is a constant for a certain interval such that $$\int \limits_{A}f = a |A|$$ then, when $a=b$ $$a |A\times B| = b |B||A| \Rightarrow |A\times B| = |B||A|$$ Is that a stretch?

Edit: $$A \times B = \lbrace (x,y)\in A \times B : x\in A , y\in B \rbrace$$

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How are you defining $|A \times B|$? – Christopher A. Wong Oct 31 '12 at 0:12
I never understand how it can seem appealing to write "thm" and "const" instead of "theorem" and "constant" when it saves you three or four keystrokes each while forcing several dozen prospective readers to break their reading flow and think about the intended meaning of the abbreviation -- especially when the people at whose expense you're skimping on keystrokes are the ones you're asking for help. – joriki Oct 31 '12 at 0:26
Is that better:)? – rioneye Oct 31 '12 at 0:28
Yes, thanks :-) – joriki Oct 31 '12 at 0:30
Maybe you know that $|A\times B|=|A|\cdot |B|$ if $A$ and $B$ are Borel measurable. Do you see why this will be true for _Lebesgue measurable sets? – Davide Giraudo Oct 31 '12 at 10:43