Does the support of a Borel probability measure always have full measure in a metric space?
I know this is true for separable metric spaces, and locally compact metric spaces. Is it true in general?
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Sign up to join this communityDoes the support of a Borel probability measure always have full measure in a metric space?
I know this is true for separable metric spaces, and locally compact metric spaces. Is it true in general?
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.
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.
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.
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.
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 $\mu(S) = 1$.
Proof:
($\Rightarrow$) 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.]