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let's assume $\neg CH$, then there's a set $X$ such that $|\mathbb N|<|X|<|\mathbb R|$. i'm wondering about the lebesgue measure of such set... is it even possible to measure it? would it be possible that its measure is more than $0$? i think not, because all the subset of cantor set are measure $0$ sets, and there would be a set $K$ such that $|X|=|K|$ and $K\subset Cantor$. is it correct?

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Such sets are not Borel, but can be measurable. In case it is, it is consistent to assume Martin's axiom plus non-CH and then all such measurable X have measure 0. But it also possible to have models of non-CH where some measurable X of intermediate size has positive measure. The last remark about the Cantor set only shows that in all models we have measure 0 sets of largest possible size, which is not relevant to the question, only insofar that it shows there are always in a non-CH world intermediate size sets of measure 0. – Henno Brandsma Jan 29 '12 at 10:50
If a set of small size is measurable, its measure is 0. The Lebesgue measure of a measurable $A$ is the supremum of the measures of the compact subsets of $A$, and compact sets are either countable or of the same size as the reals. Now, if by "measure" you only mean "outer measure", things are different. As Henno points out, the answer is independent of the usual axioms of set theory, even assuming $\lnot CH$. – Andrés E. Caicedo Jan 29 '12 at 16:56
This is a great question, but one which has already arisen on MO at the question to which Ricky Demer links, which has some informative answers. So I have voted to close this one as a duplicate. – Joel David Hamkins Jan 29 '12 at 23:15

That set only have two options: it would be measurable (of $0$ measure) or it will be non-measurable. You can prove that all the measurable sets of positive measure have the same cardinality as the Continuum. In case that the set is non-measurable it's outer measure would be positive.

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The proof shows any measurable set of positive measure has, inside it, a compact set of positive measure. And therefore has cardinal of the continuum. – Gerald Edgar Feb 21 '13 at 18:47
(Which had already been pointed out in the comments.) – Andrés E. Caicedo Feb 21 '13 at 18:51

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