We know that if A in R^n and B in R^m are (lebesgue) measurable sets, then so is the set AxB in R^(n+m).
Prove that the converse is also true. That is, if AxB is measurable, then so is A and B.
As stated, that result is false. For example, if S is a non-measurable subset of $\displaystyle \mathbb{R}$ then $\displaystyle S\times\{0\}$ is a null set (hence measurable) in $\displaystyle \mathbb{R}^2$.
To make the result correct, you need to add the hypothesis that $\displaystyle A\times B$ has measure greater than 0. Then (from the arguments used to prove Fubini's theorem) almost all the slices of $\displaystyle A\times B$ are measurable. But in a product set all the slices are equal to A (for horizontal slices) or B (for vertical slices). Therefore A and B are measurable.