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