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As title goes, without assuming continuum hypothesis, is there a subset of $\mathbb{R}$ with positive measure but not being continuum? That is, $$\textrm{Does there }\exists A\subseteq \mathbb{R}, \quad \textrm{such that }\quad \mu(A)>0, \quad \aleph_0<|A|<\aleph_1$$ I believe that maybe in some system without assuming continuum hypothesis, there exists.

And by the regularity of Lebesgue measure, it suffices to deal with compact set.

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marked as duplicate by Asaf Karagila cardinals Feb 28 '18 at 6:06

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ mathoverflow.net/questions/86948/… $\endgroup$ – Moishe Kohan Feb 28 '18 at 3:28
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    $\begingroup$ Note that there is an important subtlety here: the assumption of measurability. There are, provably in ZFC, sets of reals of cardinality $\aleph_1$ which are not measurable; since null sets are measurable, this means such sets have positive outer measure, and if CH fails they have cardinality $<2^{\aleph_0}$. So if we drop the implicit measurability assumption, we get independence. $\endgroup$ – Noah Schweber Feb 28 '18 at 3:49
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There's a standard exercise in measure theory that says that if $A$ has positive measure, then $A-A$ contains an interval. It follows from this that $A$ must have the same cardinality as the reals.

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I find a solution due to This question. Due to the regularity of Lebesgue measure, it suffices to deal with compact set. And a standard result is a closed subset of $\mathbb{R}$ is either at most countable or continuum.

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    $\begingroup$ For readers not familiar with it, that result is the Cantor-Bendixson theorem. Its analogue for open sets is trivial, but when the sets are closed things are quite interesting. And the result continues to hold for more complicated sets (e.g. Borel, and even analytic); such results are part of descriptive set theory. $\endgroup$ – Noah Schweber Feb 28 '18 at 3:52
  • $\begingroup$ Thank you, I didn't known it is a theorem titled. I saw this result in Hewitt and Ross's Abstract harmonic analysis Volume 1 P31(4.26) $\endgroup$ – Cube Bear Feb 28 '18 at 4:02

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