Q:

Measurable sets in $[0,1]$

Suppose $E$ is a Lebesgue measurable sets in $[0,1]$ such that $m(E)$ is positive.
  Show that there exist two points $x,y$ in $E$ such that $x-y$ is irrational.

I tried to prove that $f(x)=m(E\cap [0,x])$ is continuous, but I don't know how to show that E has two points whose difference is irrational. Can someone give me some hint?
Thanks

A:

If $x-y$ is rational for all $x, y\in E$, then take any $x\in E$ and note that $E\subseteq S := \{q+x : q\in \mathbb{Q}\cap [-1, 1]\}$. Therefore, as $m(S) = m(\mathbb{Q}\cap [-1, 1]) = 0$, we have $m(E) = 0$. The contrapositive of this is that if $m(E) > 0$, then $x-y$ cannot be rational for all $x, y\in E$.

