Riemann Steiltjes Integral I have been stuck with this problem for a few hours. So far I have tried to prove that b- 1/n tends to b when n tends to infinity. 
Let be $\alpha,\ f,\:[a,b]\to\mathbb{R}$ continous, such that f is Riemann Stieltjes Integrable
Show $[\int_a^{b-1/n}f\ d\alpha]_{n\in\mathbb{N}}\to \int_a^b f\, d\alpha$.
 A: The easy case is where $\alpha$ monotone. Assume  WLOG that $\alpha $ is non-decreasing.  The integrand $f$ is continuous and, hence, bounded with $|f(x)| \leqslant M$ for $x \in [a,b]$.  Hence,
$$0 \leqslant \left|\int_{b-1/n}^b f \, d\alpha  \right| \leqslant M[\alpha(b) - \alpha(b-1/n)]$$
Since $\alpha $ is continuous we have by the squeeze theorem,
$$\lim_{n \to \infty}\int_{b-1/n}^b f \, d\alpha  = 0,$$
and the desired result follows from the additivity property of the integral,
$$\int_{b-1/n}^b f \, d\alpha  + \int_a^{b-1/n} f \, d\alpha   = \int_a^b f \, d\alpha  $$
A more general and difficult case is where we only have that $\alpha$ is of bounded variation.  For any $\epsilon > 0$ there exists $\delta > 0$ such that for any partition $P = (x_0,x_1, \ldots,x_m)$  with $\|P\| < \delta$ we have for any corresponding RS-sum 
$$\left| \int_{b-1/n}^b f \, d\alpha - \sum_{j=1}^mf(\xi_j)[\alpha(x_j) - \alpha(x_{j-1})] \right| < \epsilon,  $$
which implies
$$\left| \int_{b-1/n}^b f \, d\alpha \right| < \epsilon + \sum_{j=1}^m|f(\xi_j)|\,|\alpha(x_j) - \alpha(x_{j-1})| \leqslant \epsilon +M V_{b-1/n}^b(\alpha). $$
Since $\alpha $ is continuous it can be shown with some effort (proved here) that the total variation function $x \to V_{x}^b(\alpha) $ is also continuous. Hence, $V_{b-1/n}^b(\alpha) \to V_b^b(\alpha) = 0$ as $n \to \infty$, and for every $\epsilon > 0$,
$$\lim_{n \to \infty}\int_{b-1/n}^b f \, d\alpha  \leqslant \epsilon,$$ 
which implies that again
$$\lim_{n \to \infty}\int_{b-1/n}^b f \, d\alpha  = 0$$
The most difficult case is where $\alpha$ has unbounded variation yet the RS-integral exists, but the first two cases cover what is usually considered for RS-integrals.
