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

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

If not, what is a counterexample?

• Given a $\sigma$-finite measure $\mu$, it is always possible to construct a probability measure which is equivalent to $\mu$, i.e has the same null-sets, so for your question, it would suffice to show that you can find a $\sigma$-finite measure with full support. I don't know about that though. Commented Mar 11, 2023 at 8:36
• No, it is not always possible, but an example is not straightforward. Commented Mar 11, 2023 at 8:44

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.
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].