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I know this is true for separable metric spaces, and locally compact metric spaces, but is it true in general?

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It is usually better to include your question in the body rather than just in the title. To clarify, I think you are asking: if you have a Borel probability measure on a metric space, must its support have full measure? (Where "support" is the intersection of all closed sets of full measure.) – Nate Eldredge Nov 1 '10 at 4:47
Bjørn Kjos-Hanssen points out obliquely in his comment to my answer that it seems unlikely that you actually have a proof that the result is true for locally compact spaces, as such a proof would also establish that there are no measurable cardinals, as the space in my counterexample is discrete and therefore locally compact. Could you explain? – Joel David Hamkins Nov 7 '10 at 23:54
up vote 10 down vote accepted

Following Pietro's lead, let me observe that if there is a measurable cardinal, then there is a counterexample.

Suppose that $\kappa$ is a measurable cardinal. Then there is a $\kappa$-additive 2-valued measure $\mu$, measuring all subsets of $\kappa$, giving them either measure $0$ or $1$, giving measure $1$ to the whole space and giving measure $0$ to any set of size less than $\kappa$ (among others). If we give $\kappa$ the discrete topology, then every set is closed (and hence Borel), and the support is empty.

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Combine this with Autoleech's proof that there is no counterexample for locally compact spaces, and we will have a proof that there are no measurable cardinals - finally! – Bjørn Kjos-Hanssen Nov 1 '10 at 15:56
More generally, we can do this with a real-valued measurable cardinal. And (if it is consistent that 2-valued measurable cardinal exists) it is consistent that a real-valued measurable cardinal exists but no 2-valued measurable cardinal exists. – Gerald Edgar Nov 1 '10 at 17:27

Every $\sigma$-smooth measure is $\tau$-smooth. This is what we need. As noted, if there is a (real-valued) measurable cardinal, then this may fail for a metric space. A space is called "measure-compact" iff every $\sigma$-smooth measure is $\tau$-smooth.

The reference for all of this (up to 1965) is: V. S. Varadarajan, "Measures on Topological Spaces". In a completely regular space we would use "zero sets" (a set where some continuous real-valued function vanishes). But in a metric space these are the same as the closed sets. A (finite, Borel) measure $\mu$ on a metric space is $\sigma$-smooth iff it is coutably additive, but this means if $A_n$ is a decreasing sequence of closed sets, then $\mu(A_n)$ converges to $\mu(\bigcap_n A_n)$. A stronger condition on $\mu$ is $\tau$-smooth: if $A_t$ is a decreasing net of closed sets, then $\mu(A_t)$ converges to $\mu(\bigcap_t A_t)$. The "support" of a probability measure $\mu$ is the intersection of all closed sets of measure $1$. And (assuming $\mu$ is $\tau$-smooth) this intersection again has measure $1$.

As I recall, a metric space is measure-compact if and only if there is no discrete subset with real-valued measurable cardinal. So, in particular, if there are no real-valued measurable cardinals, then the answer to the question in the title is YES. Joel has provided the converse. Thus this question is presumably independent of ZFC.

The term "measure-compact" is due to Moran, 1965. By analogy with "real-compact" which may be characterized in the same way using only $\{0,1\}$-valued measures.

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excuse me, it 's not clear to me whether the definition of $\sigma$-smooth and $\tau$-smooth are the one you wrote after "iff", or are those theorems ? – Pietro Majer Nov 2 '10 at 10:44
Probably these in terms of closed sets (or zero sets) are theorems, and the definitions involve continuous functions instead: A positive linear functional $\Lambda$ on $C_b(X)$ is $\tau$-smooth iff for any net $f_t$ in $C_b(X)$ that decreases pointwise to zero, we have $\lim \Lambda(f_t) = 0$. – Gerald Edgar Nov 2 '10 at 15:48
Reference corrected. – Gerald Edgar Aug 7 '13 at 14:10

Consider an uncountable discrete metric space $X $ (i.e., metrized by the Kronecker delta). Define a measure on $X$ putting for any $A\subset X,\\ $ $\mu(A)=1$ or $\mu(A)=0$ according whether $A$ belongs to a given non-principal ultrafilter $\mathcal{F}$, or not (sigma-additivity holds, for there are no disjoint subsets of positive measure). Then $\mu$ is a Borel probability measure with empty support.

[edit] Actually, this is additive, but to ensure sigma-additivity it would be needed that $\mathcal{F}$ be closed under countable intersections.

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Since there are disjoint uncountable sets, this isn't a measure (it's not additive). It is a measure on the sigma-algebra of countable/co-countable sets, but that algebra does not contain all the Borel sets in your space. – Joel David Hamkins Nov 1 '10 at 15:02
ops I was initially thinking of the restricted sigma algebra indeed – Pietro Majer Nov 1 '10 at 15:10
Thank you! I've made a very quick tentative of rescue. – Pietro Majer Nov 1 '10 at 15:20
About your revised answer: the filter should be countably closed to have an additive measure, and it needs to measure all the Borel sets, which in the discrete case means all sets. So it has to be an ultrafilter. And this takes a measurable cardinal. – Joel David Hamkins Nov 1 '10 at 15:23
That is, if there is a countably complete non-principal ultrafilter on a set, then there is a measurable cardinal, and your new answer is the same as my answer. – Joel David Hamkins Nov 1 '10 at 15:27

Here's a simple argument for why large cardinals are really needed here:

Suppose $\kappa$ is the least cardinal such that there is a collection of size $\kappa$ of open null sets with non null union. Let $I$ be the sigma-ideal of those subsets of $\kappa$ over which the union of these open sets is null. Then the boolean algebra $\mathcal{P}(\kappa)/I$ satisfies the countable chain condition since otherwise, there would be uncountably many pairwise disjoint non null open sets. Cardinals which admit such ideals are sometimes called quasi-measurable. Using Ulam's matrix, it can be verified that the least quasi-measurable is weakly inaccessible.

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Rather than a counterexample, here is an equivalent condition for topological spaces (and hence metric spaces).

Let $(\Lambda,\mathcal{T})$ be a topological space with the Borel $\sigma$-algebra and $\mu$ a probability measure on it. Say that $\mathcal{T}$ outlines $\mu$ if given an arbitrary collection of open sets $\mathcal{U}\subset \mathcal{T}$ with $\mu(U)=0$ for every $U \in \mathcal{U}$, $$ \mu\left(\bigcup_{U\in \mathcal{U}}U\right) = 0 $$ Let $S := \mathrm{supp}(\mu)$. Then $\mathcal{T}$ outlines $\mu$ if and only if $\mu(S) = 1$.


($\Rightarrow$) Observe that $y \in \Lambda \backslash S \iff$ there exists an open neighbourhood $U_y$of $y$ with $\mu(U_y) = 0$, in which case $U_y \subset \Lambda \backslash S$. Taking the union, $\Lambda \backslash S = \bigcup_{y\in \Lambda \backslash S}U_y$, so that $\mu(\Lambda \backslash S)=0$ by assumption. Hence $\mu(S) = 1$ since $\Lambda = S \cup \Lambda \backslash S$ and both sets are certainly Borel-measurable.

($\Leftarrow$) Suppose that $\mathcal{U}$ is a collection of open subsets of measure zero. Then $S \subset \Lambda \backslash \bigcup \mathcal{U}$, so that bya similar argument to above, $\Lambda = S \cup \Lambda \backslash S \implies \mu(U) \leq 0$, and hence $\mu(U) = 0$ by non-negativity. So $\mathcal{T}$ outlines $\mu$. This concludes the proof.

Note that the condition ``$\mathcal{T}$ outlines $\mu$'' is always satisfied for a measure on a separable metric space, since such spaces are second countable, so the union in $(\Rightarrow)$ can be reduced to a countable union, which gives a measure of zero by the assumed countable additivity of a measure.

[I came up with this proof myself, sorry I don't have further sources for you to look at.]

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