$\sigma$-field generated by the continuity sets of a measure

Question

Let $\mu$ be a probability measure on the Borel subsets of a topological space $X$ (a compact metric space if necessary). A Borel set $B$ is a $\mu$-continuity set if $\mu(\partial B)=0$, where $\partial B$ is the boundary of $B$. Is the $\sigma$-field generated by the continuity sets of $\mu$ equal to the Borel $\sigma$-field?

Answer 1

If the space is metric and separable, then yes. To see this, it suffices to show (by separability) that a base of the topology is contained in this class of $\mu$ continuity sets.

To see this, note for $x \in X$ that $\partial B_r (x) \subset \{y \mid d(x,y)=r\}$. In particular, $\partial B_r (x) \cap \partial B_s (x)=\emptyset$ for $r\neq s$. But in a space of finite measure, there can only be countably many pairwise disjoint sets of positive measure. Hence, $B_r (x)$ is a $\mu$ continuity set for all but at most countably many $r$. In particular, the $\mu$ continuity sets contain a neighborhood base of $x$.