Let $\mu$ be a Borel probability measure on $X$ and $A$ its support. Which conditions on $X$ that makes $[A \neq \emptyset \implies \mu(A)=1]$ hold?

Question

Let $X$ be a topological space and $\mu$ a Borel probability measure on $X$. Let $A := \operatorname{supp} \mu$ be the intersection of all closed sets with full measure. Then $A$ is closed and thus measurable. If $X$ is a separable or a locally compact metric space, then $\mu (A) = 1$.

Now we only assume $A \neq \emptyset$. Could we prove that $\mu(A)=1$? If not, which conditions on $X$ that makes $\mu(A)=1$ hold?