Solovay shows that the existence of a measurable cardinal is equiconsistent with the existence of a countably additive extension of Lebesgue measure that is defined on all sets of real numbers. Given such an extension, does there exist a set of real numbers whose symmetric difference with each Borel set of real numbers has positive measure?

  • $\begingroup$ A related question: mathoverflow.net/questions/204409/… $\endgroup$
    – Ashutosh
    Jun 14, 2015 at 16:52
  • 2
    $\begingroup$ Interestingly, if you are willing to ditch the axiom of choice, you can get this extension without large cardinals! (Also due to Solovay) $\endgroup$
    – Asaf Karagila
    Jun 14, 2015 at 17:29

1 Answer 1


Yes. This follows from a theorem of Gitik and Shelah which says that the measure algebra of any such extension is nowhere separable. Hence by Maharam's theorem, there is some set which divides every borel set into two pieces of equal measure.

An immediate way to see this is as follows: Assume for some total extension $m$ of Leb. measure, this fails. Then $\mathcal{P}(\mathbb{R}) / Null(m) \cong Borel / Null(Leb.)$ which contradicts a theorem of Gitik and Shelah.

A "forcing-free" proof of Gitik-Shelah theorem can be found in: A. Kamburelis, A new proof of the Gitik-Shelah theorem, Israel Journal of Mathematics, October 1990, Volume 72, Issue 3, pp 373-380

  • $\begingroup$ Thank you for the answer. Could you suggest references for these two theorems? $\endgroup$ Jun 14, 2015 at 17:11
  • $\begingroup$ The rough idea is to break the space into countably many homogeneous pieces and apply Gitik-Shelah + Maharam's theorem to obtain an independent event (over the restrictions of Borel sets) on each one of them. See page 84 and the references mentioned here: math.wisc.edu/~akumar/IND_IDEALS.pdf $\endgroup$
    – Ashutosh
    Jun 14, 2015 at 17:21
  • $\begingroup$ I also added a Maharam's theorem free reason. $\endgroup$
    – Ashutosh
    Jun 14, 2015 at 17:26
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    $\begingroup$ theorem 2.6, M. Gitik and S. Shelah, Forcings with ideals and simple forcing notions $\endgroup$ Jun 14, 2015 at 18:33

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