Say $\kappa$ is small if any set of cardinality $\kappa$ has outer-Lebesgue measure zero. We know that, in the Cohen model of ZFC where CH is false, there is a Borel partition of the unit interval of small cardinality whose every member is null (cf. this paper by Dubins and Prikry). Of course, this doesn't show that ~CH entails the existence of such a partition, and I am curious if this question has been settled, i.e. given ZFC, does ~CH entail that there is a Borel partition of the unit interval of small cardinality whose every member is null, or is there a model of ZFC+~CH where no such partition exists? Any pointer would be greatly appreciated.

  • 1
    $\begingroup$ I don’t think you mean to restrict to Lebesgue measurable sets in the smallness definition. Any Lebesgue measurable set of cardinality less than continuum is measure zero. $\endgroup$ Commented Oct 18, 2023 at 0:25
  • $\begingroup$ @ElliotGlazer Ah you are right! Just edited it. Thanks! $\endgroup$
    – Y.Z.
    Commented Oct 18, 2023 at 1:07

1 Answer 1


If you adjoin many random reals (with the usual measure-algebra forcing) to a model of GCH, you can make the cardinal $\mathfrak c$ of the continuum large while no family of fewer than $\mathfrak c$ sets of outer measure zero can cover the unit interval.

In the terminology of cardinal characteristics of the continuum, this result says that, in the random real model, the covering number for Lebesgue measure equals $\mathfrak c$. In that form, it can be found in Section 11.4 of my chapter in the Handbook of Set Theory; a preprint version of the chapter is available at https://dept.math.lsa.umich.edu/~ablass/hbk.pdf . The result is also in the book "Set Theory: On the structure of the real line" by Bartoszynski and Judah, and probably in many other places.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.