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.

To show superadditivity of the Lebesgue-Stieltjes premeasure, use finite additivity (on pairwise disjoint semiring elements) and monotonicity of the premeasure - take a limit (supremum) when bounded above.  Otherwise, the result is trivial.  Superadditivity is actually easier to verify than subadditivity.
Subadditivity is shown by a compactness argument.  It is not that tricky, because you "follow your nose" with the semiring elements involved by "adding" and "subtracting" endpoints as needed to use compactness followed by right continuity.
