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.


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.

  • $\begingroup$ 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. $\endgroup$ – Ilya Dec 6 '14 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 '14 at 16:25

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