In Appendix B of his [Uniform Central Limit Theorems](https://rads.stackoverflow.com/amzn/click/com/0521052211) (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](https://rads.stackoverflow.com/amzn/click/com/0521007542), 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](https://en.wikipedia.org/wiki/Measurable_cardinal) talks about measurable cardinals but doesn't point out the open problem (this should be fixed). You can prove that [measurable cardinals do exist](https://mathoverflow.net/questions/33468/measurable-cardinals-under-axiom-of-determinacy) 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](http://arxiv.org/abs/1212.6543) of [ETCS](http://www.cwru.edu/artsci/phil/Lawvere%20ETCS.pdf)? I would accept that as answer.



Do Measurable Cardinals Exist? (assuming ZFC)
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](http://mishap.sdf.org/), in particular "**Exercises de style: A homotopy theory for set theory. Part II**" with Assaf Hasson.


