Topological Conditions that imply Non-measurability I gave a short presentation on Baire spaces, and one of the cute results of the theory I showed is that a Vitali set cannot be nowhere dense. This led me to think that a subset of the reals $A$ which satisfied:


*

*$A$ is uncountable

*$A$ has empty interior

*$\overline{A}$ contains an interval


is not Lebesgue measurable. However, a couple more minutes of thought and the rationals with the Cantor set occurred as a counter example.
Thus my question is: are there any topological conditions on a subset of the reals which implies it is not measurable? I am aware that every measurable subset of the reals is a measure zero set away from being a $G_\delta$ or an $F_\sigma$ subset, but this condition is not purely topological.
 A: Bernstein sets cannot be Lebesgue measurable. They also cannot have the Baire property, or the perfect set property (they are constructed as a counterexample for the last one).
Recall that a Bernstein set is a set which does not contain a perfect set, but its intersection with any perfect set is non-empty. This means that it meets every open interval, so its closure is the unit interval, and that it has an empty interior and it is of size continuum.
Moreover, every open set $U$ which contain a Bernstein set $B$ must have the property $|[0,1]\setminus U|=\aleph_0$, otherwise $|[0,1]\setminus U|=2^{\aleph_0}$, and since its closed it contains a perfect set. In particular it means that $B\cap[0,1]\setminus U$ is non-empty, but $B\subseteq U$ so that is impossible. 
Therefore the outer measure of a Bernstein set is $1$. On the other hand, the inner measure of a Bernstein set is clearly $0$ because every closed subset of a Bernstein set has to be countable, otherwise it would contain a perfect set.
Therefore a Bernstein set cannot be Lebesgue measurable.
