Is it always possible to construct a distribution with full support?

Question

Does any (arbitrary) measurable topological space admit a probability distribution with full support?

If not, what is a counterexample?

Answer 1

Under some reasonable assumption on the space, the result is true. Recall that a Polish space is a topological space homeomorphic to a complete separable metric space. In that post, it is shown that any uncountable Polish space without isolated point has a non-atomic Borel probability measure of full support.

On the other hand, the support of a Borel probability measure on a metric space has a separable support. The support contains a dense countable subset. As a consequence, you can't have a Borel probability measure of full support on a non separable metric space. So the simplest counterexample is ${\bf R}$ endowed with the discrete topology.

Answer 2

Any non-separable Banach space serves as a counter-example. I will assume Zorn's Lemma for the proof.

Let $\mu $ be a Borel probability measure on a real Banach space $B$. If $ \mu $ has a support in the sense there is a smallest closed set $S$ with $ \mu (S)=1$ then $S$ is necessarily separable.

Let $\varepsilon >0$ and $A_{\varepsilon }=\{E\subseteq B:\left\Vert x-y\right\Vert \geq \varepsilon $ $\forall x,y\in E\}$. For each $n$ the collection $A_{1/n}$ has a maximal element $E_{n}$ with respect to set inclusion (by Zorn's Lemma). If $0<\delta <\frac{1}{n}$ and $x\in S$ then $% S=\bigcup_{\{x\in E_{n}\}}S\cap B(x,\delta )$ because if $y\in S\backslash B(x,\delta )$ for every $x\in E_{n}$ then $E_{n}\cup \{y\}$ would be in $A_{n}$ contradicting maximality of $E_{n}$. Now $\mu (S\cap B(x,\delta ))>0$ for each $x\in E_{n}$ and the sets $S\cap B(x,\delta ),x\in E_{n}$ are disjoint. [ If $x\in S$ and $\mu (S\cap B(x,\delta ))=0$ then $% \mu (S\backslash B(x,\delta ))=1$ contradicting the definition of support]. It follows that $E_{n}$ is at most countable. The countable set $ E=\bigcup_{_{n}}E_{n}$ is dense in $S$ since $S=\bigcup_{x \in E_{n}}S\cap B(x,\delta )$ if $0<\delta <\frac{1}{n}$ and $x\in S$. [ The following elementary fact is used here: if $\{x_{n}\}$ is a sequence in a metric space $(X,d),S$ is a subset of $X$ such that each point of $S$ is a limit of a subsequence of $\{x_{n}\}$ then $S$ is separable in its own right: the closure of $\{x_{n}\}$ in $X$ is a separable metric space and $S$ is a subset of this space so it is separable].

EDIT: Norm is not needed. The proof works with just a metric.