Suppose that $E$ is measurable and $E+q \subseteq E$ for all $q \in \mathbb Q$. Show that either $m(E)=0$ or $m(E^c)=0$.
I wanted to try proving the statement using the Lebesgue Density theorem. This is my attempt:
If $m(E)=0$, there is nothing to show. So suppose that $m(E)>0$. For almost every $x\in E$, we have that $$\lim_{r\to 0} \frac{m(E\cap [x-r,x+r]}{2r} = 1$$ In particular, for any $\epsilon>0$ there is some $\delta>0$ such that for any $q< \delta$, $$\frac{m(E \cap [x,x+q])}{2q} -1\leq \frac{m(E\cap [x-q,x+q])}{2q} -1 <\epsilon$$ which implies that $$\frac{m(E \cap [x,x+q])}{2q} \geq 1-\epsilon \text{ which further implies that } m(E\cap[x,x+q]) \geq 1-\epsilon$$ We can suppose that $q\in \mathbb Q$ and that $\delta<1$, so that we further get that $$m(E\cap [x, x+1]) \geq 1-\epsilon$$ and since $\epsilon$ is arbitrary, $$m(E\cap [x,x+1])=1$$ On the other hand, by hypothesis for all $q\in \mathbb Q$, $$1=m(E+q \cap [x+q, x+1+q]) \leq m(E\cap [x+q,x+1 +q]) \leq1$$ which further gives us that $m(E\cap[x+q,x+q+1])=1$ for all $q\in \mathbb Q$, i.e. $E$ is of full measure.
I don't know if I am overlooking something, as I have been told to try using other methods instead?
I appreciate any revisions or suggestions!
