$B_x = \{y \in [0,1]: x-y \in \Bbb{Q}\}$, $ \varepsilon=\{C \subset [0,1]: \exists x \quad C=B_x\} $. By the axiom of choice we can choose exactly one element of each equivalence class $\varepsilon$ and make a set $P$. Let $m$ be the Lebesgue measure
I could prove that if $E \subset P$ is measurable then it must be that $m(E) = 0$. Now I have to prove that if $E \subset [0,1)$ and $m^*(E)>0$ then there is a subset of $E$ that is not Lebesgue measurable.
I am confused with proving that this is not Lebesgue measurable with the outer measure. I was trying to use subadditivity, but I could not find a solution. Any help is appreciated!