Title:
Suppose $E\subset\mathbb{R}^{n}$ with $m(E) = 0$. Is $m(\{(x,y)\in\mathbb{R}^{n}: x-y\in E \})=0$?
(Where $m$ is the Lebesgue measure)
I believe this set is measurable, but even that is not so easy for me to see. We can restrict to the case where $E$ is Borel (closely contain $E$ in a $G_{\delta}$ set) and get everything we want though: then because $f(x,y) = x - y$ is continuous, so $f^{-1}(E) = \{(x,y)\in\mathbb{R}^{n}: x-y\in E \}$ is measurable.
It is enough to prove this when $E$ is a Borel set: there exists a Borel set, in fact a $G_{\delta}$ set, $F$ such that $E \subset F$ and $m(F)=0$. Proving the result for $F$ gives the result for $E$.
When $E$ is a Borel set this is an easy consequence of Fubini's Theorem: $m\{(x,y): x-y \in E\}=\int I_{\{(x,y): x-y \in E\}} dm(x)dm(y)$ $=\int I_{y+E}(x) dm(x)dm(y)=\int m(E+y) dm(y)=0$ since $m(E+y)=0$ for all $y$.