Why does this measure map one sequence to another? 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.
 A: If you can prove the result on the singletons, the measure of $\mathbb{R}\setminus A$ will follow, since: 
$$\mu(\mathbb{R})=\sum_{n\in\mathbb{Z}}\mu(n,n+1] = \sum_{n\in\mathbb{N}}b_n,$$ and by $\sigma$-additivity we have:
$$\mu(\mathbb{R}\setminus A) = \mu(\mathbb{R}) - \sum_{n\in\mathbb{N}}\mu(\{a_n\})= \sum_{n\in\mathbb{N}}b_n - \sum_{n\in\mathbb{N}}b_n = 0.$$
To see that $\mu\bigl(\{a_n\}\bigr) = b_n$, note that if $a_n\in (a,b]$, then 
$$\begin{align*}
\mu((a,b]) &= F(b)-F(a)\\
&= \sum\{b_m:a_m\leq b\} - \sum\{b_m: a_m\leq a\}\\
 &= \sum\{b_m: a\lt a_m\leq b\}\\
& \geq b_n.\end{align*}$$
Therefore, since $\mu(\{a_n\})$ is the infimum of the measure of any set $(a,b]$ that contains it, it follows that $\mu(\{a\}) \geq b_n$. 
On the other hand, let $\epsilon\gt 0$. 
Then there exists $N\gt 0$ such that $\sum_{m\geq N}b_m\lt \epsilon$. Let $\delta\gt0$ be such that $(a_n-\delta,a_n]$ does not contain any of $a_1,\ldots,a_{N-1}$ (except possibly $a_n$ itself). This is possible, since we only have to avoid finitely many points of $\mathbb{R}$. Then:
$$\mu(\{a_n\}) \leq \mu\bigl((a_n-\delta,a_n]\bigr) = \sum_{a_n-\delta\lt a_m\leq a_n+\delta}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!b_m \ \ \ \ \ \leq b_n+\sum_{m\geq N}b_m = b_n+\epsilon.$$
Thus, $b_n\leq \mu(\{a_n\}) \leq b_n+\epsilon$ for every $\epsilon\gt 0$, which means that $\mu(\{a_n\}) = b_n$, as desired.
