Riemann Stieltjes integral. Prove that $\lim_{|P| \to 0} S(f,P,T) = \int_a^bfd\alpha$ Problem:
Let $\alpha \in BV[a,b]$ and let $f$ be continuous on $[a,b]$. Prove that $\lim_{|P| \to 0} S(f,P,T) = \int_a^bfd\alpha$
Background: BV stands for bounded variation. I am using $|P|$ to denote the norm of the partition $P$ of $[a,b]$, where $|P|=\max\{|x_i-x_{i-1}| : x_i,x_{i-1}\in P\}$. The set $T$ is the set of numbers where the function is evaluated for the sum.
Attempt: I know that when $\alpha$ is increasing, we can essentially split the integral up over the partition and apply the mean value theorem a bunch of times, which gives the result nicely. With this idea, I was thinking we could let $\alpha = \beta - \gamma$, where $\beta$ and $\gamma$ are increasing functions. Then we could once again use MVT. My attempts in this direction have not worked out so far. 
Any hints would be appreciated!
 A: If $f$ is continuous and $\alpha$ is increasing then $\int_a^b f \, d\alpha$ exists by a well-known theorem. It is also true that if the Riemann-Stieltjes integral exists and we have $\alpha$ increasing and either $f$ or $\alpha$ continuous, then for any set of tags $T$ we have
$$\lim_{\|P\| \to 0}   S(P,f,\alpha,t) = \int_a^b f \, d\alpha.$$
This is somewhat difficult to prove using only continuity of $\alpha$ but easy to prove when $f$ is continuous using your argument described in Attempt along with the fact that $f$ is uniformly continuous.
When $\alpha$ is of bounded variation there exist increasing functions $\beta$ and $\gamma$ such that $\alpha = \beta - \gamma$. 
Since $f$ is continuous  we can be sure that $\int_a^b f \, d\beta$ and $\int_a^b f \, d\gamma$ exist, where $\beta$ and $\gamma$ are increasing. By the result discussed at the beginning of this answer, for any $\epsilon > 0$ there exist $\delta_1, \delta_2 > 0$ such that, if $\|P\| < \delta_1$ then for any set of tags $T$ we have 
$$\left|S(P,f,\beta, T) - \int_a^b f \, d\beta\right| < \frac{\epsilon}{2},$$
and if $\|P\| < \delta_2$ we have 
$$\left|S(P,f,\gamma, T) - \int_a^b f \, d\gamma\right| < \frac{\epsilon}{2}.$$
Hence, if $\|P\| < \delta = \min(\delta_1,\delta_2)$, then we have
$$\left|S(P,f,\alpha, T) - \left(\int_a^b f \, d\beta - \int_a^b f \, d\gamma \right)\right|\\ = \left|S(P,f,\beta, T) - S(P,f,\gamma, T) - \left(\int_a^b f \, d\beta - \int_a^b f \, d\gamma \right)\right|\\ \leqslant \left|S(P,f,\beta, T) - \int_a^b f \, d\beta\right| + \left|S(P,f,\gamma, T) - \int_a^b f \, d\gamma\right| \\ < \epsilon. $$
Therefore, 
$$\lim_{\|P\| \to 0} S(P, f, \alpha, t) = \int_a^b f \, d\beta  - \int_a^b f \, d\gamma .$$
We finish using a well-known proposition (proved below) that states if $f$ is continuous and $\alpha$ is of bounded variation, then $\int_a^b f \, d\alpha$ exists and 
$$\int_a^b f \, d\alpha = \int_a^b f \, d\beta  - \int_a^b f \, d\gamma. $$
Q.E.D.
Proof of Proposition.
Since $f$ is integrable with respect to $\beta$ and $\gamma$ it follows from integration-by-parts that $\beta$ and $\gamma$ are integrable with respect to $f$.  By linearity it follows that $\beta - \gamma$ is integrable with respect to $f$ and
$$\int_a^b (\beta - \gamma) \, df  = \int_a^b \beta \, df - \int_a^b \gamma \, df.$$
Once again, using integration-by-parts we have that $f$ is integrable with respect to $\alpha = \beta - \gamma$ where 
$$\int_a^b f \, d\alpha = \int_a^b f \, d(\beta - \gamma) \\ = f(b)[\beta(b) - \gamma(b)] - f(a)[\beta(a) - \gamma(a)] - \int_a^b (\beta - \gamma) \, df \\ = \left(f(b) \beta(b) - f(a)\beta(a) - \int_a^b \beta \, df \right) - \left(f(b) \gamma(b) - f(a)\gamma(a) - \int_a^b \gamma \, df \right) \\ = \int_a^b f \, d\beta - \int_a^b f \, d\gamma.$$
