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.