# non Lebesgue measurable set with empty interior?

Let E a set contained in R with non-empty interior. Then i know that it can be seen as the countable union of open disjoint intervals and every intervals has positive measure. So, there exist a subset of E s.t. is not Lebesgue measurable (Vitali-set like).

But if E has EMPTY INTERIOR? I think that, due to the fact that Lebesgue measure has no atom, necessarily the measure of E is zero, so there is no non-measurable subset. Is it correct my idea?

Every set of positive measure in $$\mathbb R$$ contains a non-measurable subset. [See https://math.stackexchange.com/questions/2079436/does-every-non-null-lebesgue-measurable-set-contain-a-non-measurable-subset?rq=1 ]. If $$C$$ is fat Cantor set then $$C$$ has no interior but there is a non-measurable subset.
• @maru0032 That is right. Fat Cantor sets have empty interior but they don't have measure $0$. Commented Aug 20, 2021 at 11:27
• @maru0032 If $E$ has non-empty interior then it has positive measure so it contains a non-measurable subset. But your argument is not correct. Just because $E$ contains an interior point you cannot say that $E$ is an open set. If it is not open you cannot express it as a union of open intervals. Commented Aug 20, 2021 at 11:30