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Is this seems so odd to me, can someone please help? I think it should be some Cantor type construction but can't figure out.

SBF
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jeff
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    A first observation: it can't be closed, because then it would not contain some interval $I$ and so $m(A \cap I)$ would be zero. It can't be open either, for then it would contain an interval, and the measure of the intersection would be equal. So it must be at least one level up in the hierarchy. In light of the fact that it must be dense and the Baire category theorem, perhaps you can use a $G_\delta$ set. For instance the complement of a countable union of appropriately chosen Cantor sets might work. – Ian Dec 11 '14 at 15:06
  • I believe the first observation is not correct. As I understand, you claim that a closed set which does not contain an interval must have measure zero. This cannot be true: take an enumeration $r_1,r_2,\dots$ of the rationals and put $r_n$ inside an open interval of length $\frac{\epsilon}{2^{n+1}}$. The union of these intervals has measure at most $\epsilon$, contains all rationals and is open. Its complement is closed, cannot contain an interval (since it contains no rationals) and cannot have measure $0$. – user 59363 Dec 11 '14 at 15:56
  • OK, you're right, now I understand what you mean. – user 59363 Dec 11 '14 at 16:10
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    Almost duplicate of this question – Alex Ravsky Dec 11 '14 at 16:23

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