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I have pretty no knowledge in set theory, so likely the question has a trivial answer. All countable subsets of $[0,1]$ have Lebesgue measure of zero, thus all sets of positive Lebesgue measure are uncountable. Does it yet mean that all these sets have same cardinality as $[0,1]$? Clearly, the answer is yes under the continuum hypothesis, but I wonder whether CH is crucial here and what would be the answer without CH.

I guess there is no difference whether we consider only Borel sets, or all Lebesgue measurable ones.

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We can prove the continuum hypothesis for Borel sets. Namely every Borel set of positive measure has the cardinality of the continuum. We can do this by finding a perfect subset inside a Borel set.

But there's an easier solution. Recall the theorem of Steinhaus saying that if $A$ is a measurable subset, then $A-A=\{a-b\mid a,b\in A\}$ contains an open interval around $0$.

With the help of some basic cardinal arithmetic it's easy to show that $A$ has the cardinality of the continuum.

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  • $\begingroup$ Thanks, I was thinking of the first approach, as Cantor space is one of the benchmark Borel spaces. The second solution is interesting, nice to know of it. $\endgroup$
    – SBF
    Dec 6, 2014 at 16:10
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    $\begingroup$ Without Steinhaus's theorem, the result can be deduced for all measurable sets from the first paragraph, essentially by definition, as the inner Lebesgue measure of a set $A$ is (defined as) the supremum of the measures of its compact subsets. $\endgroup$
    – Andrés E. Caicedo
    Dec 6, 2014 at 16:25

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