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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.

leo
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Snsd
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  • I am suspicious about your first hypothesis : How can you take a box such that $\lambda(E \cap J) > 3 \lambda(J)/4$? I must say though, this is a very interesting question. – Patrick Da Silva Nov 22 '11 at 07:13
  • See, e.g: http://en.wikipedia.org/wiki/Lebesgue_differentiation_theorem – gary Nov 22 '11 at 07:21
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    @PatrickDaSilva: $E$ has a subset $E'$ with positive and finite measure. There is an open set $U$ such that $E'\subseteq U$ and $\lambda(U)<\frac{4}{3}\lambda(E')$. Since $U$ is a countable union of nonoverlapping boxes, $U=\cup_k B_k$, it follows that $\sum_k\lambda(E'\cap B_k)>\sum_k\frac{3}{4}\lambda(B_k)$. Hence there exists $k$ such that $\lambda(E'\cap B_k)>\frac{3}{4}\lambda(B_k)$, and you can take $J=B_k$. – Jonas Meyer Nov 22 '11 at 07:25
  • Sorry, I had written a proof that such a box exists, but didn't include it here for brevity. – Snsd Nov 22 '11 at 07:39
  • I guess that Jonas's argument should be added to yours Snsd, unless you have a better idea to keep your $E$ instead of Jonas' $E'$. Although his clarification changes nothing to the rest of your argument. I think Jonas didn't need to use $E'$ in his argument, he could've just kept $E$ all along. – Patrick Da Silva Nov 22 '11 at 07:39
  • @PatrickDaSilva: Finite measure is used because $\sum a_k \leq \sum b_k$ does not imply there exists $k$ such that $a_k\leq b_k$ if both sums are infinite. One loses no generality in assuming $E$ has finite measure to begin with (as long as one knows how to give justification that $E$ has a subset with positive and finite measure). – Jonas Meyer Nov 22 '11 at 07:43
  • Your argument would be perfect if I could understand why $\lambda(J \cap (J+x) ) \le 3 \lambda(J)/2$. Sorry if I'm just tired and not able to fill in the blanks by myself but your question is very interesting and so is the proof... heh – Patrick Da Silva Nov 22 '11 at 07:43
  • @Jonas : I didn't notice that was the issue that you were taking care of. Good point – Patrick Da Silva Nov 22 '11 at 07:45
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    You may want to have a look at this thread and also at this one as well as the links therein. – t.b. Nov 22 '11 at 08:09
  • I'm with @PatrickDaSilva, why is valid the inequality: $\lambda(J\cap (J+x))\leq 3\lambda (J)/2$? Patrick, I notify you to see if you discover it. – leo Jan 31 '12 at 01:02
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    I think this can be fixed if either OP justifies why this inequality holds, or just by taking a smaller epsilon. It might depend on $n$ though, but it seems to be just geometry at this point. I now accept that "this argument holds by the handwaving theorem". =P – Patrick Da Silva Jan 31 '12 at 05:07
  • $\mathbb Z$ is measureble (measure 0) in $\mathbb R$ but the length of differences are unbounded. Done. – Ross Millikan Jun 22 '12 at 04:29

1 Answers1

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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 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 proceeding in this way is taken from Robert Bartle, I don't remember right now the book where this comes from.

leo
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