Evaluating Riemann-Stieltjes integral $\int_a^bf(x)g(x)d\alpha(x)$ I am looking to evaluate the following Riemann-Stieltjes integral:
$$\int_a^bf(x)g(x)d\alpha(x)$$
and I am unsure how to deal with the product on the inside. I have considered the following:
$$\int_a^bf(x)g(x)\,d\alpha(x)=\int_a^bf(x)\,dG(x)$$
Where $G(x):=\int_a^xg(x)\,d\alpha(x)$, but I don't think this really helps me out. I also tried doing integration by parts, but this left me with another integral and an $\alpha(x)$ not in a differential so I'm not sure if that helps either. Is there a good method of evaluating this?
Assuming $g\ge 0$, we can define a Borel measure on $[a,b]$ by setting $\mu(a,x]=G(x)$. For a simple $\phi=\sum_{i=1}^n a_i \chi_{A_i}$, the integral $\int \phi\text{d}G=\int\phi \text{d}\mu=\sum a_i\int_{A_i}g\text{d}\alpha=\int\sum a_i\chi_{A_i}g\text{d}\alpha=\int \phi g\text{d}\alpha$. For $f\in L^+$, find a sequence of simple $\phi_n \to f$ and apply the MCT to obtain $\int f\text{d}G=\int fg\text{d}\alpha$.
$\left| \int_a^bf(x)g(x)d\alpha(x) \right| \leq F.G.A'.(b-a)$ 
where $|f(x)|\leq F \ ; \  |g(x)| \leq G \ ; \  |\alpha'(x)| \leq A'$.