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

My approach is this. Take some $J$, 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 $x\in\mathbb{R}^n$ 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 $E-E$ contains the box of $x$ such that $|x|\leq \epsilon$.

Is this valid? If not, can it be fixed? Many thanks.

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    $\mathbb Z$ is measureble (measure 0) in $\mathbb R$ but the length of differences are unbounded. Done.2012-06-22

1 Answers 1

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This is answered here, I will just post another approach that can be useful.

Lemma. If $K$ is a compact subset of $\mathbb{R}^n$ with positive measure, then the set $D:=\{x-y:x,y\in K\}$ contains an open ball centered at the origin.

Proof. Since $0\lt \lambda(K)\lt\infty$, there exist an open set $G$ in $\mathbb{R}^n$ such that $K\subset G\text{ and } \lambda (G)\lt 2\lambda(K)$ and as its complement $G^c:=\mathbb{R}^n\setminus G$ is closed and $K$ is compact, we have $\delta:=d(K,G^c)\gt 0.$ We claim that if $x\in\mathbb{R}^n$ with $||x||\lt\delta$ then $K+x\subseteq G$. If not, there exist a $y\in K$ s.t. $y+x\not\in G$ and then $\delta=d(K,G^c)\leq ||y-(x+y)||=||x||,$ which is an absurd if we assume that $||x||\lt\delta$ to begin with. Now $K+x\subseteq G$ and $K\subseteq G$, thus $(K+x)\cup K\subseteq G$. If $(K+x)\cap K=\emptyset$, then $\lambda(G)\geq \lambda((K+x)\cup K)=\lambda(K+x)+\lambda(K)=2\lambda(K)$ which contradicts the choice of $G$. So, for all $x\in B(0,\delta)$ we have $(K+x)\cap K\neq\emptyset$, therefore $B(0,\delta)\subset D$. QED

Theorem. Let $E\subseteq\mathbb{R}^d$. If if $E$ is measurable and $\lambda(E)\gt 0$ then the set $E-E:=\{x-y:x,y\in E\}$ contains an open ball centered at the origin.

Proof. Since $\mathbb{R}^d=\bigcup_{n\in\mathbb N}B(0,n)$ we have $E=E\cap\mathbb{R}^d=\bigcup_{n\in\mathbb N}E\cap B(0,n),$ then $0\lt\lambda(E)\leq \sum_{n\in\mathbb{N}} \lambda(E\cap B(0,n))$ and then, there exist a $n\in\mathbb{N}$ such that $\lambda(E\cap B(0,n))\gt 0$. Let $F=E\cap B(0,n)$. Thus $F\subseteq E$ is bounded and measurable, therefore there exist a closed set $K\subseteq F$ such that $0\lt \frac{\lambda(F)}{2}\lt \lambda(K).$ Notice that $K$ is closed and bounded, that means it is a compact set with positive measure. By the Lemma the set $K-K$ contains an open ball centered at the origin say $B$. Then $B\subset K-K\subseteq E-E$ as we wanted.

Note. The idea of proceed in this way is taken from Robert Bartle, I don't remember right now the book where this come from.