At most countably many balls are $\mu$-continuity sets in a Polish space

Question

Let $\mu$ be a locally finite Borel measure on a Polish space $X$. A $\mu$-continuity set $A$ is such that $\mu(\partial A) = 0$, such sets are important when working with Portmanteau's theorem. A few years ago I remember seeing that for a fixed $x \in X$, all but at most countably many balls with centre $x$ in $X$ are $\mu$-continuity sets. However, I can't seem to recreate that proof.

Does anyone have any idea of a proof for this? I remember that Polish requirement was important but that doesn't seem to be helping me.

Answer 1

1) Locally finite measure on the separable metric space $X$ is $\sigma$-finite. Indeed, fix a dense countable set $Z\subset X$ and call a ball with center in $Z$ and rational radius good, if it has finite measure. Let $Y\subset X$ be a union of all good balls. Clearly $\mu$ is $\sigma$-finite on $Y$. Assume that $Y\ne X$, take a point $x\notin Y$. Since $\mu$ is locally finite, there is a ball $B(x,r)$ of finite measure. Choose $a\in Z\cap B(x,r/3)$ and rational number $q\in (r/3,r/2)$. The ball $B(a,q)\subset B(x,r)$ is good and contains $x$. A contradiction.

So, we may write $\mu=\sum_{i=1}^\infty \mu_i$, where each $\mu_i$ is finite.

2) For a ball $A=B(x,r)$ the boundary $\partial A$ is contained in the sphere $S(x,r)=\{y:d(x,y)=r\}$. Note that for each $\mu_i$ at most countably many spheres $S(x,r)$ have positive measures (because these spheres are disjoint). Therefore, for $\mu$ also all but countably many spheres have zero measure.