Suppose that $X$ is a locally compact Hausdorff space, and let $\mu$ be a positive measure defined on the $σ$-algebra of Borel subsets of $X$, with $μ(X)<∞$.
Definition 1. One says that $μ$ is regular if, for every Borel set $E⊆X$, one has that $$ μ(E) = \sup\big\{μ(K): K ⊆ E, \text { and $K$ is compact} \big\} = $$ $$ =\inf\big\{μ(U): U\supseteq E, \text { and $U$ is open} \big\}. $$
Definition 2. An atom for $\mu$ is any Borel subset $A\subseteq X$, such that $\mu(A)>0$, and such that, for every Borel subset $B\subseteq A$, one has that $\mu(B)=0$, or $\mu(A\setminus B)=0$.
Every singleton $A=\{x_0\}$ is an atom, provided its measure is nonzero, and so is $\{x_0\}\cup N$, when $N$ has measure zero.
I can prove that, under the hypothesis that $μ$ is regular and $X$ is second-countable, these are the only atoms, but I could not find this result anywhere in the literature. Can anyone help me locate a reference?
PS 1. This is related to Must atoms of a Borel measure space be singletons?, where the result is proved, but no reference is given.
PS 2. It is known that every finite Borel measure on a second-countable, locally compact Hausdorff space, is regular.
