Why is the Lebesgue-Stieltjes measure a measure? I'm having difficulty convincing myself the Lebesgue-Stieltjes measure is indeed a measure. The Lebesgue-Stieltjes measure is defined as such:

Given a nondecreasing, right-continuous function $g$, let $\mathcal{H}_1$ denote the algebra of half-open intervals in $\mathbb{R}$. We define the Lebesgue-Stieltjes integral to be $\lambda: \mathcal{H}_1 \rightarrow[0,\infty]$, with $\lambda(I)=0$ if $I=\emptyset$, $
\lambda(I)=g(b)-g(a)$ if $I=(a,b]$, $-\infty\leq a < b < \infty$ and $\lambda(I)=g(\infty)-g(a)$ if $I=(a,\infty)$, $-\infty\leq a < \infty$.

Showing countable subadditivity is done by a careful application of the $\epsilon2^{-n}$ trick, and while I couldn't do this on my own, this can be found in most analysis textbooks. What about the other inequality to show $\sigma$-additivity? Does anyone know of a resource that proves this or could share how to do this? I suspect this requires quite a bit more trickery than the proof of subadditivity.

Firstly, note that the measure defined here is a Radon measure (that is $\lambda(B)<\infty$ for any bounded borel set $B$). hence it is also $\sigma$-finite (Because $\mathbb{R}=\bigcup_{n\in\mathbb{Z}}(n,n+1]$).
So if I can only show that the measure $\lambda$ is $\sigma$-additive on the semifield $\{(a,b]:-\infty\leq a\leq b\leq\infty\}$ (showing this is trivial), then it would be so over $\mathcal{B}$ by Caratheodory Extension Theorem.    
