In Appendix B of his Uniform Central Limit Theorems (1999), Dudley writes:

It is consistent with the usual axioms of set theory (including the axiom f choice) that there are no measurable cardinals, in other words all cardinals are of measure 0; see, for example, Drake (1974, pp. 67-68, 177-178). It is apparently unknown whether existence of measurable cardinals is consistent (Drake, 1974, pp. 185-186). So, for practical purposes, a probability measure defined on the Borel sets of a metric space is always concentrated in some separable subspace.

The continuum hypothesis implies that the cardinality $c$ of the continuum (that is, of $[0,1]$) is of measure 0 (RAP, Appendix C).

Has there been any progress on the existence of measurable cardinals (assuming ZFC) in the 14 years since this book was published? i.e., do they exist? What makes the problem difficult?

The Wikipedia article talks about measurable cardinals but doesn't point out the open problem (this should be fixed). You can prove that measurable cardinals do exist if you assume ZF+AD.

From a few paragraphs prior in Appendix B, here are the definitions:

A cardinal number $\zeta$ is said to be measurable if for a set $S$ of cardinality $\zeta$, there exists a probability measure $P$ defined on all subsets of $S$ which has no point atoms; in other words, $P(\{x\}) = 0$ for all $x \in S$. If there is no such $P$, $\zeta$ is said to be of measure 0.

In the absence of any news, can somebody formulate the problem of existence of measurable cardinals in the structural set theory of ETCS? I would accept that as answer.

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    $\begingroup$ You cannot prove in first-order logic that the existence of measurables is consistent with $\mathsf{ZFC}$. I would not consider this an open problem, just as I do not consider open the consistency of arithmetic. (There is nothing to edit about open problems in the Wikipedia article, and $\mathsf{AD}$ is incompatible with choice, so it is really the wrong thing to focus on in this case.) $\endgroup$ Jul 8, 2013 at 21:36
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    $\begingroup$ The formulation by Dudley is confusingly imprecise and incorrect in all interpretations I can think of. $\endgroup$ Jul 8, 2013 at 21:54
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    $\begingroup$ I see now (after reading François's comment) that Dudley is also using "measurable" in a nonstandard fashion, meaning what we more commonly call (atomlessly) real-valued measurable. It is consistent that the continuum is real valued measurable, that is, there is a probability measure on $\mathcal P(\mathbb R)$ that is zero on singletons. This is equivalent to the existence of a (necessarily, not translation invariant) extension of Lebesgue measure to all subsets of $[0,1]$. ("Consistent:" It is equiconsistent with the existence of measurables, relative to $\mathsf{ZFC}$.) $\endgroup$ Jul 8, 2013 at 22:06
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    $\begingroup$ A more careful discussion of measurable cardinals from an analyst point of view is in Federer's Geometric measure theory. And for a more modern treatment, Fremlin's treatise on Measure theory. (Set theoretic issues are mostly in volume 5). $\endgroup$ Jul 8, 2013 at 22:08
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    $\begingroup$ @Tom: The first issue is that measurable cardinals are defined in terms of two-valued probability measures. Also, the way I read Dudley's definition, every cardinal from the first real-valued measurable on is "Dudley measurable" since there is no restriction that says that the measure couldn't concentrate on a smaller set. Finally, the phrase "measure 0 cardinal" is hopelessly confusing. $\endgroup$ Jul 8, 2013 at 22:14

1 Answer 1


This is not really a problem.

If $\kappa$ is a measurable cardinal then $V_\kappa$, or the set of sets which are hereditarily have size smaller than $\kappa$, is a model of $\sf ZFC$. This means that $\sf ZFC$ cannot even prove the consistency of $\sf ZFC+\exists\kappa\text{ measurable}$, because of the incompleteness theorem. Well, at least if $\sf ZFC$ is consistent to begin with.

Furthermore, if $\kappa$ is the least measurable cardinal, and if there is a measurable cardinal then there is a least measurable cardinal, then in $V_\kappa$ there are no measurable cardinal, therefore $V_\kappa\models\sf ZFC+\lnot\exists\kappa\text{ measurable}$.

Regarding the assumption on determinacy, two points are relevant here: $\sf ZF+AD$ proves that the axiom of choice fails. And moreover the consistency strength of $\sf ZF+AD$ is much higher than that of $\sf ZFC$, or even $\sf ZFC+\exists\kappa\text{ measurable}$. This means that if we assume that $\sf ZF+AD$ is consistent then we can prove a lot more than we can from $\sf ZFC$ or $\sf ZFC+\exists\kappa\text{ measurable}$.

So we know that $\sf ZFC$ cannot prove that a measurable cardinal exist, or even the consistency of one, and we know that if $\sf ZFC+\exists\kappa\text{ measurable}$ is consistent, then so is $\sf ZFC+\lnot\exists\kappa\text{ measurable}$.

I am not too familiar with structural set theory or $\sf ETCS$, but Misha Gavrilovich has a characterization of measurable cardinals using his model categorical construction $\rm QtNaamen$. You can find the papers in Misha's homepage, in particular "Exercises de style: A homotopy theory for set theory. Part II" with Assaf Hasson.

  • $\begingroup$ Excellent dissolution of the problem, Asaf Karagila and @Andres Caicedo. Thanks. $\endgroup$ Jul 8, 2013 at 21:49
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    $\begingroup$ @Tom: No problems. I've added something which is more related to categories, but I don't know how well it works with $\sf ETCS$ or structural set theory. $\endgroup$
    – Asaf Karagila
    Jul 8, 2013 at 21:54
  • $\begingroup$ @Tom: The measure problem - is there a non-atomic (two-valued) probability measure defined on all subsets of a given set - makes perfect sense in ETCS. $\endgroup$ Jul 8, 2013 at 22:10

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