Constructing a Measure from a Function A well-known result is that we can always construct a countably additive function $\mu$ from a nondecreasing and right-continuous function $G$. More specifically, we define on the semiring $\mathcal{C}$ of all intervals $(a,b]$,
$$\mu((a,b])=G(b)-G(a),$$
where $\mu$ is the Lebesgue measure when $G$ is the identity mapping. I'm curious if the following function gets this property as well:

Fix a countable set $C=\{c_n:n\in\mathbb{N}\}\subset\mathbb{R}$ where each $c_n$ is distinct, and a countable set $\{a_n:n\in\mathbb{N}\}\subset\mathbb{R}$ where each $a_n$ is non-negative and $\sum_na_n<\infty$.
Define $G:\mathbb{R}\rightarrow\mathbb{R}$ such that
$$x\longmapsto\sum\{a_n:c_n\leq x\}.$$
That is, is this function right-continuous and non-decreasing? Furthermore, does the measure $\mu$ constructed from this function satisfy $\mu(\{c_n\})=a_n$ and $\mu(\mathbb{R}\backslash C)=0?$

EDIT: $G$ being right-continuous and non-decreasing seems pretty easy to see. Non-decreasing is rather obvious from the property of the $a_n$'s and I basically argued right-continuity to myself in the comment box below.
However for the second part, I'm not sure how to approach either property. How do you interpret $\mu(\{c_n\})$ to get only $a_n$ remaining in the subtraction? I appreciate any help!
 A: Some hints: Using indicator functions  write 
$$G(x)=\sum_n a_n1_{[c_n,\infty)}(x).$$ We don't assume that the $c_n$s are ordered, indeed they may well be dense in $\mathbb{R}$.
Nevertheless $G$ is right continuous since it is the uniform limit
of $G_N(x)=\sum_{n=1}^N a_n1_{[c_n,\infty)}(x)$, which are obviously right continuous.
The uniform convergence uses the summability of the sequence $a_n$.
For $\varepsilon>0$, we have $G(c_n)-G(c_n-\varepsilon)=\sum_m a_m$, where the sum is over all $m$ with $c_n-\varepsilon<c_m\leq c_n$. Even though the set of such $c_m$ may be infinite for every $\varepsilon>0$, the value of $\sum_m a_m$ decreases to $a_n$ as 
$\varepsilon\downarrow 0$.
Thus, we have $\mu(\{c_n\})=G(c_n)-G(c_n-)=a_n$ for each $n$. Since $\mu(C)=\mu(\mathbb{R})=\sum_n a_n$, we conclude that  $\mu(\mathbb{R}\backslash C)=0$.
A: Let $a=\sum\limits_na_n$. Since $\{c_n\}$ is the decreasing intersection of the sets $]c_n-1/k,c_n]$ and the measure of these is finite, $\mu(\{c_n\})$ is the decreasing limit of
$$
G(c_n)-G(c_n-1/k)=\sum\limits_ia_i\cdot[c_n-1/k\lt c_i\leqslant c_n].
$$
When $k\to\infty$, $a_i\cdot[c_n-1/k\lt c_i\leqslant c_n]\to a_n\cdot[i=n]$ hence Lebesgue convergence theorem shows the RHS converges to $a_n$.
Likewise, $\mu(\mathbb R)=a$ by definition and, by countable additivity,
$$
\mu(C)=\sum\limits_n\mu(\{c_n\})=\sum\limits_na_n=a,
$$
hence $\mu(\mathbb R\setminus C)=0$.
