Let $X$ be a topological space, equipped with its Borel $\sigma$-algebra $\mathcal B(X)$, and let $\mathbb P$ be a Radon probability measure on $(X, \mathcal B(X))$. Recall that the support of the measure $\mathbb P$ is the smallest closed set of full measure. 


Is the support necessarily separable? If so, why? If not, what is a counterexample?



When is the support of a Radon measure separable?
Let $I$ be a set of cardinality larger than the continuum. Then the product topology $[0,1]^{I}$ is compact but not separable. Give the interval $[0,1]$ the Lebesgue measure, then give $[0,1]^{I}$ the product measure $\mu$. If $U$ is a non-empty open subset of $[0,1]^{I}$, then $U$ contains a basic open set $\prod\_{i\in I}U\_{i}$ where $|\{i\in I|U\_{i}\neq[0,1]\}|$ is finite. Therefore $0<\mu(\prod\_{i\in I}U\_{i})\leq \mu(U)$. Said differently, if $C$ is a closed subset of $[0,1]^{I}$ with $\mu(C)=1$, then $C=[0,1]^{I}$.


If you replace $[0,1]$ with the circle $S$, then $S^{I}$ is a compact non-separable group which does not have separable support as jbc mentioned.


