Let $(X, \tau)$ be a metrizable space and $\mathcal{B}(\tau)$ the Borel $\sigma$-field generated by the open sets in $\tau$. Let $\mu$ be a probability measure on $\mathcal{B}(\tau)$. Then, its support, say $\text{supp}(\mu)$, is defined as the smallest closed set $C$ such that $\mu(C)=1$. If $(X, \tau)$ is separable, then the support exists.
Let's now assume that $(X,\tau)$ is a Polish metrizable space. Let's consider an open subset $S \subset X$ and let's endow it with the relative metric topology, $\tau_S$. Clearly, $(X, \tau_S)$ is still separable and a probability measure $\mu'$ on $\mathcal{B}(\tau_S)$ admits a support. Considering the closure $\bar{S}$ of the set $S$, we can also endow it with a relative topology $\tau_{\bar{S}}$ and observe that $(\bar{S}, \tau_{\bar{S}})$ is still Polish. A measure $\mu''$ on the Borel $\sigma$-field $\mathcal{B}(\tau_{\bar{S}})$ admits a support once more, but: does the support in this case have some better properties than that in the case of $\mu'$? Which is the effect of adding completeness on the properties of the support of a probability measure?
More precisely: in the case of $\bar{S}$, the support of $\mu''$ can be equivalently defined as the set of elements of the subspace for which every open neighborhood (w.r.t. the relative topology) has positive probability. Is such an equivalent definition also valid for $S$ and $\mu'$?
Best Answer
Completeness doesn't really have anything to do with it. The equivalence in your last paragraph is valid in any topological space as long as the support exists. In particular it is valid in any separable metrizable space, such as $(S, \tau_S)$ for any subset $S \subset X$ whatsoever.
Indeed, let $Y$ be any topological space, $\mu$ a Borel probability measure, and assume that the support of $\mu$ exists; call it $E$. Suppose every open neighborhood of $y \in Y$ has positive measure. Then if $F$ is any closed set not containing $y$, we have that $F^c$ is an open neighborhood of $y$, thus $\mu(F^c) > 0$ and so $\mu(F) < 1$. Hence $F \ne E$. We conclude $y \in E$.
Conversely, suppose $y \in E$ and let $U$ be any open neighborhood of $y$. Then $E \setminus U$ is a proper closed subset of $E$; since $E$ is by definition the smallest closed set of measure $1$, we must have $\mu(E \setminus U) < 1$. Therefore $\mu(U) > 0$.
As an aside, when your set $S$ is open, then $(S, \tau_S)$ actually is Polish - indeed, a subset of a Polish space is completely metrizable iff it is $G_\delta$. See Theorem 3.11 of Kechris, Classical Descriptive Set Theory. However, as noted above, we don't need that here.