Moving from the Borel sigma algebra $\mathfrak B(\mathbb R)$ to the Lebesgue sigma algebra $\mathfrak L(\mathbb R)$ increases the number of mearurable sets substantially: $\mathfrak B(\mathbb R)$ has the same cardinality as $\mathbb R$ while $\mathfrak L(\mathbb R)$ has the same cardinality as $\mathcal P(\mathbb R)$ (see for example here).

I know that the Vitali set is an example of a set in $\mathcal P(\mathbb R)$ that is not Lebesgue-measurable. Now I am wondering how many such sets there are in $\mathcal P(\mathbb R)$:

What is the cardinality of $\mathcal P(\mathbb R) \setminus \mathfrak L(\mathbb R)$?


1 Answer 1


Pick a Vitali subset $S\subseteq [0,1)$. For any Lebesgue-measurable subset $H\subseteq (2,3)$, the set $S\cup H$ is not Lebesgue-measurable. And the function $H\mapsto S\cup H$ is injective.

This gives you at least (hence exactly) $2^{2^{\aleph_0}}$ Lebesgue non-measurable subsets of $\Bbb R$.

  • $\begingroup$ You could pick any subset of $(2,3)$. The set $S\cup H$ would be not measurable, because its intersection with $[0,1]$ is not measurable. $\endgroup$
    – orangeskid
    Sep 17, 2017 at 18:36
  • $\begingroup$ @orangeskid True. $\endgroup$
    – user228113
    Sep 17, 2017 at 18:43

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