The Support of a Borel Measure is Closed, and Carries Full Measure on a Separable Space

Let $(X,d)$ be a metric space, let $\mathcal{T}_{d}$ be the collection of subsets of $X$ that are open in $(X,d)$, which is a topology on $X$ by The Open Subsets of a Metric Space Form a Topology, and let $\mathcal{B}(X,d)$ be the Borel $\sigma$-algebra of $(X,d)$. Let $\mu$ be a Borel measure on $(X,d)$ and let $\operatorname{supp}\mu$ be its support. Then the following hold.

1. (The support is closed)(i) The set $X\setminus\operatorname{supp}\mu$ belongs to $\mathcal{T}_{d}$. Consequently $\operatorname{supp}\mu$ is closed in $(X,\mathcal{T}_{d})$ and belongs to $\mathcal{B}(X,d)$, a $\sigma$-algebra containing the complement of each of its members by σ-Algebra and Measurable Space, so that $\mu(\operatorname{supp}\mu)$ and $\mu(X\setminus\operatorname{supp}\mu)$ are defined.

2. (Full measure on a separable space)(ii) If $(X,d)$ is separable, then

$$\mu\bigl(X\setminus\operatorname{supp}\mu\bigr)=0 .$$