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I'm really stuck on the following problem.

Let $E\subset[0,1]\times[0,1]$ be a Lebesgue measurable subset of positive measure. Show that there exists $x,y\in E$, with $x\neq y$, such that $x-y\in\mathbb{Q}\times\mathbb{Q}$.

I was thinking of doing a proof by contradiction, but I'm not sure what I would try to establish as my contradiction (perhaps using density/countability of $\mathbb{Q}$?). Any help is greatly appreciated!

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You need to use the Steinhaus theorem from which you get that $E-E$ contains an open set.

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  • $\begingroup$ We've never gone over the Steinhaus Theorem in any of my lectures. Is there another way to tackle this? $\endgroup$ – Sir_Math_Cat Oct 4 '18 at 17:48
  • $\begingroup$ @Sir_Math_Cat yes, I post another answer. $\endgroup$ – Yanko Oct 4 '18 at 20:09
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As a request from the OP I post another answer to this question:

Suppose by contradiction that for all $x,y\in E$ we have $x-y\in \mathbb{Q} \Rightarrow x=y$.

Lemma 1: Let $r=(p,q)\in\mathbb{Q}^2$ be non-zero then $E,E+r$ are disjoint.

Indeed if by contradiction $e,e'\in E$ are such that $e=e'+r$ then $e-e' = (p,q)$ and so $e=e'$ which is absurd. $\blacksquare$

Lemma 2: Let $r$ as before, and let $E_r:=(E+r)\cap [0,1]\times [0,1]$ then $\mu(E_r)\geq \mu(E)-pq$

The square $[1-p,1]\times [1-q,1]$ is pushed out after translating by $r$. The size of the square is $pq$ and so the measure of $E$ inside this square must be less than that. $\blacksquare$

Let $c>0$ denote the measure of $E$. Choose infinitely many rational pairs $r_1,r_2,...$ (in fact we only need finitely many depending on $c$) such that $E_{r_i}$ is of measure at least $c/2>0$. This is possible by Lemma 2, also by Lemma 1 all $E_{r_i}$ are disjoint. Hence

$\mu(\bigsqcup_{i=1}^{\infty} E_i) = \sum_{i=1}^\infty \mu(E_i)(= c/2\cdot \infty)=\infty$

Which is absurd as all $E_i$ lies in $[0,1]^2$.

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