Let $\left(\alpha_n \right)_{n\in \mathbb{N}}$ a succesion in $BV[a,b]$ and $f:[a,b] \rightarrow \mathbb{R}$ such that $f \in R_{\alpha_n} [a,b]$. If $\alpha \in BV[a,b]$ and $V_a^b(\alpha_n - \alpha) \rightarrow 0$ when $n \rightarrow \infty$, show that $f \in R_\alpha[a,b]$ and $$\lim_{n\rightarrow \infty}\int_a^b f d\alpha_n = \int_a^b f d\alpha $$
What I tried is :
$$\left|S(\alpha_n, P, T)- S(\alpha, P, T)\right| < \epsilon$$ for any $\epsilon$
But I think i need to prove, that $\lim_{n\rightarrow \infty}\int_a^b f d\alpha_n$ exist and then $S(\alpha, P, T)$ converge to the integral. Or something like that. Is this righ? How to show the existence of the integral? Is there other way to attack this problem?
First, note that if $f$ is Riemann-Stieltjes integrable, then it must be bounded on $[a,b]$.
Also, note that since $|\alpha_m(x) - \alpha_n(x)|\leqslant |\alpha_m(x) - \alpha(x)|+ |\alpha_n(x) - \alpha(x)|$, we have
$$V_a^b(\alpha_m - \alpha_n) \leqslant V_a^b(\alpha_m - \alpha) + V_a^b(\alpha_n - \alpha)$$
Thus, given $m \geqslant n$,
$$\begin{align}\left|\int_a^b f \, d\alpha_m - \int_a^b f \, d\alpha_n\right| &= \left|\int_a^b f \, d(\alpha_m - \alpha_n)\right| \\ &\leqslant ( \, \sup_{x \in [a,b]} f(x) \, ) V_a^b(\alpha_m - \alpha_n) \\ &\leqslant ( \, \sup_{x \in [a,b]} f(x) \, )( \,V_a^b(\alpha_m - \alpha) + V_a^b(\alpha_n - \alpha)\, ) \end{align}$$
The second inequality is proved here using a proposition proved here.
Since the RHS converges to $0$ as $n \to \infty$, it follows that $\left(\int_a^b f \, d\alpha_n\right)$ is a Cauchy sequence and , hence, convergent.
Now you can proceed to show that $f$ is Riemann-Steiltjes integrable with respect to $\alpha$ using $I = \lim_{n \to \infty}\int_a^b f \, d\alpha_n$ and
$$\left|S(P,f,\alpha) - I \right| \leqslant \left|S(P,f,\alpha) - S(P,f,\alpha_n) \right| + \left|S(P,f,\alpha_n) - \int_a^b f \, d\alpha_n \right| + \left|\int_a^b f \, d\alpha_n - I \right| $$