Problem
$E\subseteq\mathbb R$ is a (not-necessarily measurable) set. $a+E=\{a+x\colon x\in E\}$. If the Lebesgue-measure $m(E\cap(a+E))=0$ for all $a\in\mathbb R\setminus\{0\}$, is it true that $m(E)=0$?
Discussion
The answer is certainly yes if $E$ is measurable. Suppose $a,b$ are different Lebesgue points of $E$, then $m(E\cap(a-b+E))=m((-a+E)\cap(-b+E))>0$, just like the proof of Steinhaus' theorem. I need to know whether the condition could be weakened.
Any ideas? Thanks!
Suppose $E$ is a Hamel basis for $\mathbb R$ over the rationals $\mathbb Q$. Since $e_1 = a + e_2$ implies $a = e_1 - e_2$, which for any given $a \ne 0$ can be true for at most one pair $(e_1, e_2)$ of basis elements, $E \cap (a + E)$ has cardinality at most $1$. But there exist Hamel bases that are not Lebesgue measurable (indeed there are Hamel bases that are Bernstein sets, and therefore have full outer measure).