Non-separable metric probability space
Let us say a metric probability space $(X,\rho,\mu)$ has property (*) if: the support of $\mu$ is contained in a separable subspace of $X$.
Questions: 1. Is there a standard name for this property?
Is it true that continuum+choice implies property (*) -- is there a reference?
Is it true that if (either? both?) continuum+choice don't hold, (*) can fail? Again, reference please!
fa.functional-analysis set-theory real-analysis mg.metric-geometry measure-theory
Iosif Pinelis
Aryeh KontorovichAryeh Kontorovich
Apparently, a relevant source here is H. J. Keisler and A. Tarski, From accessible to inaccessible cardinals. Fundamenta mathematicae, vol. 53 (1964), pp. 225--308, a review of which is given at https://philpapers.org/rec/KEIFAT .
I don't have the paper by Keisler and Tarski at this point with me. I found a reference to Keisler and Tarski in Добавление III (Appendix III) to the Russian translation of Convergence of Probability Measures by Billingsley; apparently, this Добавление does not appear in the English original of the Billingsley book.
So, this is what I have gathered:
The condition that the support of $\mu$ is contained in a separable subspace of $X$ is equivalent to the support itself being separable; see e.g. https://math.stackexchange.com/questions/516886/prove-that-a-subset-of-a-separable-set-is-itself-separable . Measures with a separable support were simply called separable in that Добавление.
Theorem 2 stated and proved in that Добавление says that all probability measures on a metric space $X$ are separable iff every discrete subset of $X$ is of non-measurable cardinality.
A subset $S$ of a metric space is called discrete if every point of $S$ is in some open ball not containing other points of $S$. A cardinality $k$ is called measurable if for some set $U$ of cardinality $k$ (or, equivalently, for all sets $U$ of cardinality $k$) there is a non-atomic probability measure on the power set $2^U$.
Obviously, $\aleph_0$ is non-measurable. If the continuum hypothesis holds, then the continuum cardinality is also non-measurable. It is also noted in that Добавление that the existence of measurable cardinalities is a famous unsolved problem, called the measure problem.
Iosif PinelisIosif Pinelis
$\begingroup$ Many thanks for the detailed reply, Iosif. $\endgroup$ – Aryeh Kontorovich Jul 30 '18 at 5:26
$\begingroup$ Appendix III, the reference to Keisler and Tarski, and Theorem 2 do appear in my copy of the 1968 original from Wiley and Sons. Which English original were you looking at? $\endgroup$ – Robert Furber Jul 30 '18 at 21:00
$\begingroup$ Additionally, the terminology Billingsley uses for measurable cardinals, although it is a direct translation of Ulam's terminology in the original paper, is out of date. Nowadays we say that a cardinal $\kappa$ is real-valued measurable if there is a $\kappa$-additive probability measure $\mu$ on $\kappa$, considered as a discrete metric space, such that $\mu(\{x\}) = 0$ for all $x \in \kappa$. It is then a theorem that the smallest cardinal with a countably additive probability measure vanishing at every point is the first real-valued measurable cardinal. $\endgroup$ – Robert Furber Jul 30 '18 at 21:10
$\begingroup$ Billingsley also undersells the following point (as do all analysis books of that era when measurable cardinals come up, such as Gillman & Jerison's Rings of Continuous Functions and Schaefer's Topological Vector Spaces) -- it is not really an "unsolved problem" as to whether real-valued measurable cardinals exist, rather the existence of real-valued measurable cardinals proves the consistency of ZFC, so they cannot be proven to exist, nor even to be relatively consistent, if we just start with ZFC. A nice textbook reference for these facts is Jech's "Set Theory: Third Millennium Edition". $\endgroup$ – Robert Furber Jul 30 '18 at 21:14
$\begingroup$ One more thing -- the first measurable cardinal is weakly inaccessible, so the continuum is measure-free not only if $2^{\aleph_0} = \aleph_1$, but also if it is $\aleph_2$, $\aleph_3$, $\aleph_{\epsilon_0 + 1}$, $\aleph_{\omega_5}$, etc. $\endgroup$ – Robert Furber Jul 30 '18 at 21:24
Iosif Pinelis has given an answer to question 1 and partial answers to 2 and 3. Since he advised me to turn my comments into an answer, here it is.
I will deal with the case where the axiom of choice holds, and mention the case where it does not only at the end.
I will start with some definitions. A measure $\mu$ on $(X,\mathcal{P}(X))$ is called $\kappa$-additive if for all pairwise disjoint families of sets $(S_i)_{i \in I}$ such that $|I| < \kappa$ (or, speaking loosely, such that the cardinality of the family is $< \kappa$), we have $\mu\left(\bigcup_{i \in I} S_i\right) = \sum_{i \in I}\mu(S_i)$. So "countably additive" is equivalent to $\aleph_1$-additive (Not $\aleph_0$-additive, notice the strict inequality. Finite additivity is $\aleph_0$-additivity.).
A real-valued measurable cardinal is a cardinal $\kappa$ such that there exists a $\kappa$-additive probability measure $\mu$ on $(\kappa, \mathcal{P}(\kappa))$ that vanishes on singletons, i.e. $\mu(\{x\}) = 0$ for all $x \in \kappa$. There are two kinds of real-valued measurable cardinals. Measurable cardinals are real-valued measurable cardinals such that there exists a probability measure taking only the values $0$ and $1$ (this can be defined equivalently in terms of the existence of an ultrafilter closed under intersections of cardinality $< \kappa$). Atomlessly measurable cardinals are those such that the measure $\mu$ is atomless, i.e. if $\mu(S) > 0$, there exists a $T \subseteq S$ such that $0 < \mu(T) < \mu(S)$.
The following facts were essentially proved by Ulam in his paper Zur Masstheorie in der allgemeinen Mengenlehre, although with different terminology. They are proved in a modern way in Jech's Set Theory: Third Millennium Edition, Chapter 10. A real-valued measurable cardinal is never a successor cardinal, and never the union of a strictly smaller family of strictly smaller sets, i.e. a real-valued measurable cardinal is weakly inaccessible. A measurable cardinal is not embeddable in the powerset of a strictly smaller cardinal, i.e. measurable cardinals are strongly inaccessible. Atomlessly measurable cardinals embed in the continuum, so have cardinality less than or equal to it.
A key fact is that the smallest real-valued measurable cardinal is the smallest cardinal having a countably-additive measure $\mu$ vanishing on singletons. This is how Ulam originally phrased things, so if $\kappa$ is the smallest real-valued measurable, he called cardinals $< \kappa$ unmeasurable and cardinals $\geq \kappa$ measurable. However, in the modern terminology, $\kappa^+$ is not real-valued measurable. Because the notion of "cardinals not admitting a countably-additive measure vanishing on singletons" is such a useful notion, and we can't use the term "unmeasurable" for it, Fremlin introduced the name measure-free for cardinals smaller than the first real-valued measurable cardinal.
Now we can approach the problem of whether non-separable measures exist. Any set $X$ can be given the discrete metric, and the Borel sets of the topology determined by this metric are exactly $\mathcal{P}(X)$. If $X$ has a countably additive measure $\mu$ vanishing on singletons, then $\mu$ does not have separable support, because separable subsets of a discrete space are countable sets, and $\mu$ vanishes on every countable set by countable additivity. So if there is a real-valued measurable cardinal, then there is a metric space without the property (*) in the question. As pointed out in Iosif's answer, Billingsley proves the converse as Theorem 2 of Appendix 3 in the first edition of Convergence of Probability Measures and its Russian translation, making nontrivial use of the paracompactness of metric spaces. All together, a metric space has no inseparable measures iff all its discrete subspaces have measure-free cardinality.
The continuum hypothesis enters as follows. If there exists an atomlessly measurable cardinal $\kappa$, we have $\kappa \leq 2^{\aleph_0}$, and as real-valued measurable cardinals are weakly inaccessible, we not only have cardinals in between $\aleph_1$ and $2^{\aleph_0}$, but a weakly-inaccessible-cardinalful of them (in fact later results showed that there are even more than this). So if the continuum hypothesis holds, or even if the continuum is any smallish sort of cardinal, like $\aleph_3$ or $\aleph_{\epsilon_0 + 1}$ or $\aleph_{\omega_9}$, there are no atomlessly measurable cardinals. But there may still be measurable cardinals, which will necessarily be strongly inaccessible. So the continuum hypothesis implies that metric spaces that have cardinals smaller than the first inaccessible (in fact, again, we can actually go larger than the first inaccessible too, but it is hard to describe exactly how high we can go before reaching the first measurable cardinal), such as $2^{\aleph_0}$, $2^{2^{2^{\aleph_5}}}$ and even $\beth_{\omega_1}$, do not have inseparable measures.
However, Silver, Solovay and Kunen showed that if the existence of measurable cardinals is consistent with ZFC, then they are compatible with the generalized continuum hypothesis. Therefore the answer to question 2 is no. And for 3, of course, in the absence of the continuum hypothesis, the reals, equipped with the discrete metric, may be a counterexample (if atomlessly measurable cardinals exist).
Billingsley, like many other authors of his era (such as Schaefer in Topological Vector Spaces and Gilman and Jerison in Rings of Continuous Functions), describes the question of whether real-valued measurable cardinals exist as an "unsolved problem". However, the existence of real-valued measurable cardinals is better understood as an axiom that exceeds the consistency strength of ZF(C), thanks to set-theoretic results that have been proven since then. That is to say, one can show that the nonexistence of real-valued measurable cardinals is consistent with ZFC (by taking V=L), but the consistency of real-valued measurable cardinals cannot be proven, assuming only the consistency of ZFC.
The reason for this is that if $\kappa$ is measurable, $V_\kappa$ is set-sized model of ZFC, so measurable cardinals prove the consistency of ZFC. Solovay showed in his article Real-Valued Measurable Cardinals that the existence of atomlessly measurable cardinals and the existence of measurable cardinals are equiconsistent, so the existence of real-valued measurable cardinals also implies the consistency of ZFC.
I will now briefly discuss what can happen if choice does not hold. In Solovay's model (defined in an much earlier article) where the axiom of choice does not hold, but dependent choice does, and all subsets of $\mathbb{R}$ are Lebesgue-measurable, Lebesgue measure itself defines a probability measure on $([0,1], \mathcal{P}([0,1]))$ vanishing on singletons, which is therefore a non-separable measure on $[0,1]$, equipped with the discrete metric. So in some sense, the absence of choice makes things worse (we only need an inaccessible cardinal, not a measurable one to build this model).
Robert FurberRobert Furber
$\begingroup$ Many thanks for the detailed and very helpful answer, @Robert! I've upvoted and would upvote again several times if I could. $\endgroup$ – Aryeh Kontorovich Aug 2 '18 at 9:11
$\begingroup$ I made an edit, please confirm. $\endgroup$ – Aryeh Kontorovich Aug 2 '18 at 10:02
$\begingroup$ @AryehKontorovich The $\aleph_1$ is intentional. It's a strict inequality. Otherwise the definition of $\kappa$-additivity doesn't work in the definition of a measurable cardinal (the measure is only additive for disjoint families strictly smaller than $\kappa$). Sets of cardinality $< \aleph_1$ are exactly countable sets. $\endgroup$ – Robert Furber Aug 2 '18 at 10:06
$\begingroup$ @AryehKontorovich On my side it looks like my counter-edit won, so there is no problem with the answer as it is now. $\endgroup$ – Robert Furber Aug 2 '18 at 10:29
$\begingroup$ @AryehKontorovich That was loose wording by me (I have edited the answer). What I mean is the existence of such a cardinal being consistent. $\endgroup$ – Robert Furber Aug 2 '18 at 10:56
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