# Lebesgue Measurable Subset of $[0,1]\times[0,1]$

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!

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

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

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 $$[0,1]^2$$.