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

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

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

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

Is this valid? Thank you.