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

measurableif 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 beof 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.

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 $\mathsf{ZFC}$.) $\endgroup$ – Andrés E. Caicedo Jul 8 '13 at 22:06Geometric measure theory. And for a more modern treatment, Fremlin's treatise onMeasure theory. (Set theoretic issues are mostly in volume 5). $\endgroup$ – Andrés E. Caicedo Jul 8 '13 at 22:08two-valuedprobability 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. $\endgroup$ – François G. Dorais♦ Jul 8 '13 at 22:14