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?


1 Answer 1


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.

  • $\begingroup$ yes, you are right. I hadn't thought of the generalised Cantor set. But the first part (E with non-empty interior) is true right? $\endgroup$
    – maru0032
    Commented Aug 20, 2021 at 10:35
  • $\begingroup$ Just to specify, the error in my proof is that empty interior does not imply measure zero if i correct understand $\endgroup$
    – maru0032
    Commented Aug 20, 2021 at 10:37
  • $\begingroup$ @maru0032 That is right. Fat Cantor sets have empty interior but they don't have measure $0$. $\endgroup$ Commented Aug 20, 2021 at 11:27
  • $\begingroup$ @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. $\endgroup$ Commented Aug 20, 2021 at 11:30

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