## E-E for a Lebesgue measurable set E contains some box?

I want to show that if $E\subset \mathbb{R}^d$ is a Lebesgue measurable set where $\lambda(E)>0$, then $E-E=\{x-y:x,y\inE\}\supseteq\{z\in\mathbb{R}^d:|z|<\delta\ }$ for some $\delta>0$, where $|z|=\sqrt{\sum_{i=1}^d z_i^2}$.

MY approach is this. I take $J\in\mathcal{E}$, a box in $\mathbb{R}^n$ with equal side lengths such that $\lambda(E\cap J)>3\lambda(J)/4$. Setting $\epsilon=3\lambda(J)/2$, take $x\in\mathbb{R}^d$ such that $|x|\leq\epsilon$. Then $E\cap J\subseteq J$ and $((E\cap J)+x)\cup(E\cap J)\subseteq J\cup(J+x)$. Since Lebesgue measure is translation invariant, it follows that $\lambda((E\cap J)+x)=\lambda(E\cap J)$, and so $((E\cap J)+x)\cap(E\cap J)\neq\emptyset$. If it were empty, then $2\lambda(E\cap J)=\lambda(((E\cap J)+x)\cup(E\cap J))\leq\lambda(J\cup(J+x))\leq 3\lambda(J)/2$, thus $\lambda(E\cap J)\leq 3\lambda(J)/4$, a contradiction.

Then $((E\cap J)+x)\cap (E\cap J)\neq\emptyset$, and so $x\in (E\cap J)-(E\cap J)\subseteq E-E$. Thus the box centered at 0 is contained in $E-E$.

Is this valid? Thank you.