Riemann-Stieltjes integral problem: $\int_{a} ^{b} g\, d\beta=\int_{a} ^{b} fg\, d\alpha$ Help, I've been stuck with this for hours, so far I've tried expanding the $\alpha$ integral using the definition of upper and lower integrals U and L but it doesn't seem to be a good way.
Let be $\alpha,\ f,\ g\ :[a,b]\to\mathbb{R}$ continous, $\alpha$ non- decreasing and $f(x) \ge 0$.
Let be $\beta(x) = \int_a^x f\, d\alpha$.
Show $\int_a^b g\ d\beta = \int_a^b gf\, d\alpha$.

First suppose $g$ is a step function, that is, ones can write $g=\sum_{i=1}^k\alpha_i\chi_{A_i}$, where $A_i=[a_i, b_i]$ and $[a,b]=\bigcup_{i=1}^kA_i$. So
$$\int_a^b g d\beta=\sum_{i=1}^k\alpha_i[\beta(b_i)-\beta(a_i)]=\sum_{i=1}^k\alpha_i\int_{a_i}^{b_i}fd\alpha=$$ 
$$=\sum_{i=1}^k\int_{a_i}^{b_i}\alpha_ifd\alpha=\int_a^b\sum_{i=1}^k\alpha_i\chi_{A_i}fd\alpha=\int_a^b gfd\alpha.$$
So the statement is valid for step functions.
Now let $g$ continuous and let $(s_n)$ a sequence of step functions that converges uniformly to $g$ (this can be done since $g$ is continuous on the compact $[a,b]$). Since the convergence is uniform, you can pass the limit under the integral sign and get $$\lim\int_a^bs_nd\beta = \int_a^bgd\beta$$ and $$\lim\int_a^bs_nfd\alpha = \int_a^bgfd\alpha.$$ So it follows from what we just proved for step functions that $$\int_a^bgd\beta = \lim\int_a^bs_nd\beta = \lim\int s_nfd\alpha = \int gfd\alpha.$$
