Let's say I have two fixed countable sets, $A=\{a_n:n\in\mathbb{N}\}\subset\mathbb{R}$ where the $a_n$ are all distinct, and $\{b_n:n\in\mathbb{N}\}\subset\mathbb{R}$. Here $b_n\geq 0$ and $\sum_n b_n<\infty$. I define $F\colon\mathbb{R}\to\mathbb{R}$ by $ F(x)=\sum \{b_n: a_n\leq x\}. $
It's not hard to see that this function is nondecreasing, and can be shown the function is continuous on the right.
It's a standard theorem of measure theory that the function $\mu((a,b])=F(b)-F(a)$ on the collection of half-open intervals $(a,b]$ for $a\leq b$ can be extended to a unique measure (also denoted $\mu$) on the Borel $\sigma$-algebra of $\mathbb{R}$.
But what's the reason why this measure takes values on singletons by $\mu(\{a_n\})=b_n$, and if I remove the set $A$, then I get a set with measure $0$, i.e., $\mu(\mathbb{R}\setminus A)=0$? Thanks.