Existence of distinct points with rational difference in lebesgue measurable set Let $X \subset [0,1]$ be Lebesgue measurable with $\mu(X)>0$. Show that there exist two (distinct) points $a, b \in X$ with $a-b \in \mathbb{Q}$.
I've thought about this for a while but can't seem to get anywhere. Can anyone show me how this can be done?
Edit: Can I use the fact that if $(E_n)_{n \in \mathbb{N}}$ is a sequence of sets in $[0,1]$ with $\sum_{n \in \mathbb{N}} \mu(E_n) < \infty$ then almost all points in $[0,1]$ lie in at most finitely many $E_n$?
 A: Choose an integer $n\gt\frac2{\mu(X)}$, and choose distinct rational numbers $t_1,t_2,\dots,t_n\in[0,1]$.
If the sets $X+t_1,\dots,X+t_n$ were pairwise disjoint, then we would have
$$\mu\left(\bigcup_{i=1}^n(X+t_i)\right)=\sum_{i=1}^n\mu(X+t_i)=n\mu(X)\gt2,$$
which is impossible since $\bigcup_{i=1}^n(X+t_i)\subseteq[0,2].$
Hence there are two distinct rational numbers $t,t'$ such that $(X+t)\cap(X+t')\ne\emptyset$. Choose $x,x'\in X$ so that $x+t=x'+t'$; then $x-x'=t'-t\in\mathbb Q\setminus\{0\}$.
A: Define $x\sim y$ if $x-y\in\mathbb{Q}$, We see $\sim$ is an equivalent relation on $[0,1]$(pick one element from each equivalent class to form a set is the usual way of constructing a non-measurable set)
Suppose the conclusion if false, then $X$ contains at most one element in each equivalent class, we complete $X$ to $X'$ such that $X'$ contains exactly one element of each equivalent class.
Then the inner Lebesgue measure $\mu_{*}(X') \geq \mu_{*}(X) >0$.
Now copy the proof of $X'$ is non-measurable to see this implies $\mu_{*}([0,1]) = +\infty$
