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Is there a set $X \subseteq \mathbb{R}$ such that Lebesgue outer measure is countably additive on subsets of $X$? Of course we also require $X$ to have positive outer measure.

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  • $\begingroup$ No. I'll try to write it up soon.Outer measure is translation-invariant,which may play a role in the proof. $\endgroup$ – DanielWainfleet Sep 26 '15 at 1:43
  • $\begingroup$ This (math.stackexchange.com/questions/206618/…) might help. $\endgroup$ – PhoemueX Sep 26 '15 at 8:51
  • $\begingroup$ If $X$ has no closed subsets of cardinality strictly between $\aleph_0$ and $\mathfrak c$, then the usual Bernstein set construction works. This resolves the problem under CH. A similar argument should work if you assume Martin's axiom (as then all closed sets of cardinality under $\mathfrak c$ are null, I guess). I wouldn't be surprised if it turned out to be independent of ZFC. Are you sure it's true? Is it an exercise, or a problem of your own? $\endgroup$ – tomasz Sep 26 '15 at 11:22

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