(Steinhaus theorem) Let ${E \subset {\bf R}^d}$ be a Lebesgue measurable set of positive measure, then the difference set $E - E$ contains an open neighbourhood of the origin.
Use the following result to prove the theorem: Given $E \subset {\bf R}^d$ a measurable set of positive measure, and ${\varepsilon > 0}$, there exists a cube ${Q \subset {\bf R}^d}$ of positive sidelength such that ${m(E \cap Q) > (1-\varepsilon) m(Q)}$.
Proof:
Suppose for contradiction that $0$ is not an interior point of $E - E$. Then any neighborhood of $0$ contains an $x$ such that $x \notin E - E$.
$\forall 1 \leq i \leq d$, let $E_i := \{ x_i \in {\bf R}^d: (x_1, \ldots ,x_i, \ldots ,x_d) \in E \}$ be the projection of $E$ onto the $i$^the dimension. Let $\varepsilon > 0$, and $Q$ be the cube induced by the above result, with sidelength $s > 0$. Define $c > 0$ to be such that $c := \min(\{ m((Q \cap E)_j): 1 \leq j \leq d\})$. Let $\varepsilon' > 0$ be such that $\varepsilon' < c - \varepsilon^{1/d}s$. Then for any $x = (x_1, \ldots ,x_d) \in \bf{R}^d$ with $|x_j| < \varepsilon'$ for all $j$, we have $m((x + Q) \cap E) = m(Q \cap (E-x)) \geq (c -\varepsilon')^d > \varepsilon s^d$. On the other hand, since $(x + Q) \cap E \subset x + Q \setminus E$, we also have $m((x + Q) \cap E) \leq m(x + Q \setminus E) = m(Q \setminus E) < \varepsilon s^d$, a contradiction.
Some hint on the correct direction will be greatly appreciated should it turns out that the proof is a total non-sense.