Let $(X, d)$ be a separable metric space and $S:=\{x_1,x_2,\dots,\}$ be a countably dense subset of $X$. Let $\mu$ be a Borel probability measure of $X$. A proof I am reading states the following which I don't follow:
For each $j = 1,2,\dots$, the spheres $\partial B(x_j,r)$ are pairwise disjoint. Hence for arbitrarily small $r>0$ it holds that $\mu(\partial B(x_j,r))=0$ for every $j=1,2,\dots$.
What I don't understand is 1.) why are the spheres $\partial B(x_j,r)$ pairwise disjoint for a fixed $r$? If $x_j$ is fixed, then by density of $S$ there exists $x_k\in B(x_j, r/2)\setminus \{x_j\}$ and I do not see how $\partial B(x_j,r)$ and $\partial B(x_k,r)$ are pairwise disjoint. But this might not matter since the proof itself really relies more on the latter claim that
2.) for an arbitrarily small $r>0: \mu(\partial B(x_j,r))=0$. Why is this then true? Is there some cover argument that I just don't see? We only know that $\mu$ is a Borel pr. measure, not that it is shift invariant.
$\partial B(x,r)\subset \{y: d(y,x)=r\}$. From this 1) is obvious.
By 1) you can have at most countably many $r$ for which $\mu (\partial B(x,r)) >0$. So there exists $r_n$(strictly) decreasing to $0$ such that $\mu (\partial B(x,r_n)) =0$ for all $n$.