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Is it consistent with ZFC that there is a subset of $[0,1]$ whose cardinality is less than that of the continuum but which has positive Lebesgue measure?

Obviously not given CH. And, given ZFC, there is such a subset iff there is a subset of full measure that has cardinality less than that of the continuum. Moreover, I think it follows from the consistency of ZFC with the non-existence of Sierpinski subsets of $[0,1]^2$ that it is consistent with ZFC that there is a subset of $[0,1]$ whose cardinality is less than that of the continuum and which has positive outer Lebesgue measure.

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Inner regularity of Lebesgue measure implies that every set with positive measure has a Cantor subset of positive measure. This is contained in undergraduate courses in real analysis, so I vote to close. –  Bill Johnson Mar 25 '13 at 15:34
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@Bill, in fairness, not in everyone's course on measure theory (although perhaps I just didn't do that particular homework exercise that week) –  Yemon Choi Mar 25 '13 at 16:45
    
I wonder now: Is it consistent with ZFC that there is a translation-invariant extension of Lebesgue measure that assigns non-zero measure to some set of cardinality less than the continuum? Maybe I should make that a separate question. –  Alexander Pruss Mar 25 '13 at 18:32
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up vote 14 down vote accepted

No. It is a famous exercise that if $X\subset\mathbf{R}$ has positive measure then $X-X$ contains an interval. It follows that $X$ has cardinality continuum.

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Thanks! I was about to delete my question in light of <a href="mathoverflow.net/questions/8972/…; (which has good information on related stuff) and then I saw your very quick answer. –  Alexander Pruss Mar 25 '13 at 14:48
    
Link was bad: mathoverflow.net/questions/8972/… –  Alexander Pruss Mar 25 '13 at 14:48
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Alternatively, if $X\subset[0,1]$ has positive measure, find $x\in[0,1]$ such that $[0,x]\cap X$ and $[x,1]\cap X$ both have positive measure. Then find $x_0$ and $x_1$ such that $0<x_0<x<x_1<1$ and $X$ has positive measure in each of the intervals $[0,x_0]$, $[x_0,x]$, $[x,x_1]$, $[x_1,1]$. Continue subdividing, and find continuum-many points in $X$. –  Sean Eberhard Mar 25 '13 at 14:49
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Aha, yes, that question includes information about the non-measurable case, which I ignored. –  Sean Eberhard Mar 25 '13 at 14:51
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@Sean: Concerning your comment about repeated subdivision, did you intend the "continuum-many points" in the last sentence to be the limits of the subdivision points? If so, how do you know they're in $X$. If not, which continuum-many points did you mean? –  Andreas Blass Mar 25 '13 at 17:24
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