So, the problem, as in the title, is:
If $f$ is $g$-Riemann-Stieltjes integrable on $[a,b]$, $g \in BV[a,b]$, prove that it's $g$-RS-integrable on every subinterval $[a,c] \subset [a,b]$, where $a < c< b$, as well as on $[c,b]$, and that $$\int_{a}^{b} fdg = \int_{a}^{c} fdg + \int_{c}^{b}fdg.$$
Some of the things I've tried: I know that if I have a partition $P$ on $[a,c]$, I can extend the partition to be a partition $\tilde{P}$ on $[a,b]$ so that $\Delta(P)=\Delta(\tilde{P})$, by retaining the $[a,c]$ part and adding points $c+\Delta(P)/2$, $c+\Delta(P)$, ... , $c+k\Delta(P)/2$, as long as these numbers are in $[a,b]$. However, I'm not sure how to use that to derive $$\textrm {diam}(\{\sigma(P,\xi, f, g): P \textrm{ is a partition on } [a,c] \textrm{ and }\Delta(P)<\delta \}) \leq \textrm{diam}(\{\sigma(P_{1},\xi_{1}, f, g): P_{1} \textrm{ is a partition on } [a,b] \textrm{ and }\Delta(P_{1})<\delta \}),$$
which I have a hint might be true. However, even if I were to derive this, I don't know how that would effectively imply the existence of the RS-integral $$ \int_{a}^{c} fdg. $$
I've also tried comparing this to the proof on Riemann-integrability, but the ones I found and could think of use Darboux sums, which aren't as powerful in the Riemann-Stieltjes case.
In case some people use different notation, here are the definitions of the notation I used:
If $P:a=x_{0},...,x_{n-1},x_{n}=b$, then: $\Delta(P) = \max_{1 \leq i \leq n} (x_{i}-x_{i-1})$, $\sigma(P, \xi, f, g) = \sum_{i=1}^{n} g(\xi_{i})(f(x_{i})-f(x_{i-1}))$, $\xi = (\xi_{1},...,\xi_{n}),$ where $\xi_{i} \in [x_{i-1},x_{i}]$.
EDIT: After taking a look at Rudin's "Principlies of mathematical analysis", I found the proof of this claim. However, Rudin uses a different definition from what I consider the Riemann-Stieltjes integral, namely through the analogue of Darboux sums; the definition I'm using can be found on Wikipedia. Also, I'd like to note that there's an exercise in my textbook that states an alternative definition of the Riemann-Stieltjes integral, the same one that can be found in Rudin, and asks the reader to prove that these definitions are not equivalent.
If the integrator $g$ were monotonically increasing then this could be proved in the same way it is done for the Riemann integral using Darboux sums or other another approach.
Here we have $g \in BV([a,b])$. As a function of bounded variation, $g = \beta_1 - \beta_2$, a difference of two increasing functions. There are many such decompositions and it is not always the case that $f$ which is Riemann-Stieltjes integrable with respect to $g$ is also Riemann-Stieltjes integrable with respect to $\beta_1$ and $\beta_2$. However, there is one decomposition where the integrability always extends, and that is where $\beta_1(x) = V_a^x(\beta)$, the total variation on $[a,x]$, and $\beta_2 = \beta_1 - g$. A proof is given here.
With $\beta_1$ and $\beta_2$ defined in this way, it is straightforward to show that
$$\tag{*}\int_a^b f \, dg = \int_a^b f \, d\beta_1 - \int_a^b f \, d\beta_2 .$$
Since $\beta_1$ and $\beta_2$ are increasing it follows from the statement in the first paragraph that $f$ is $\beta_1-$ and $\beta_2-$RS integrable on every subinterval $[a,c] \subset [a,b]$, and for $j = 1,2$,
$$\tag{**}\int_a^b f \, d\beta_j = \int_a^c f \, d\beta_j + \int_c^b f \, d\beta_j.$$
The argument leading to (*) can be reversed to show that the RS-integrability of $f$ with respect to $\beta_1$ and $\beta_2$ on $[a,c]$ implies the RS-integrability of $f$ with respect to $g$ on $[a,c]$, and, finally,
$$\int_a^b f \, dg = \int_a^c f \, dg + \int_c^b f \, dg.$$