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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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    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.2011-11-22
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    See, e.g: http://en.wikipedia.org/wiki/Lebesgue_differentiation_theorem2011-11-22
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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 $J=B_k$.2011-11-22
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    Sorry, I had written a proof that such a box exists, but didn't include it here for brevity.2011-11-22
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    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.2011-11-22
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    @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).2011-11-22
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    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... heh2011-11-22
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    @Jonas : I didn't notice that was the issue that you were taking care of. Good point2011-11-22
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    You may want to have a look at [this thread](http://math.stackexchange.com/q/38902/) and also at [this one](http://math.stackexchange.com/questions/59769/theorem-of-steinhaus) as well as the links therein.2011-11-22
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    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.2012-01-31
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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". =P2012-01-31
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    $\mathbb Z$ is measureble (measure 0) in $\mathbb R$ but the length of differences are unbounded. Done.2012-06-22

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