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 $\int_a^b g\ d\beta = \int_a^b gf\, d\alpha$.
 A: 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 $$\int_a^bgd\beta = \lim\int_a^bs_nd\beta = \lim\int s_nfd\alpha = \int gfd\alpha.$$
A: Proof:
With $I = \int_a^b gf \, d\alpha$, apply the mean value theorem for integrals (since $f$ is continuous and $\alpha$ is non-decreasing) to a Riemann-Stieltjes sum.  There exists $\eta_j \in (x_{j-1},x_j)$ for all $j$ such that  
$$\beta(x_j) - \beta(x_{j-1}) = \int_{x_{j-1}}^{x_j} f \, d \alpha = f(\eta_j)(\alpha(x_j) - \alpha(x_{j-1})) ,$$
and
$$\left|\sum_{j=1}^n g(\xi_j)(\beta(x_j) - \beta(x_{j-1})) - I\right| \\ = \left|\sum_{j=1}^n g(\xi_j)f(\eta_j)(\alpha(x_j) - \alpha(x_{j-1}))- I \right|\\ \leqslant \left|\sum_{j=1}^n g(\xi_j)f(\xi_j)(\alpha(x_j) - \alpha(x_{j-1}))- I \right| + \left|\sum_{j=1}^n g(\xi_j)(f(\eta_j)-f(\xi_j))(\alpha(x_j) - \alpha(x_{j-1}))\right|. $$
For all sufficiently fine partitions the first term on the right-hand side is smaller than $\epsilon/2$ since $I = \int_a^b gf d\alpha$ exists. 
The second term on the right-hand side is also smaller than $\epsilon/2$ with sufficiently fine partitions since $f$ is uniformly continuous on $[a,b]$, $g$ is bounded, and $\alpha$ is non-decreasing.  We have $|g(x)| \leqslant M$ for $x \in [a,b]$ and $\delta > 0$ such that if $|x-y| < \delta$ then $|f(x) - f(y)| < \epsilon/(2M (\alpha(b) - \alpha(a)))$ for all $x,y \in [a,b]$. Hence, if the partition norm is less than $\delta$, then
$$\left|\sum_{j=1}^n g(\xi_j)(f(\eta_j)-f(\xi_j))(\alpha(x_j) - \alpha(x_{j-1}))\right| \leqslant \sum_{j=1}^n |g(\xi_j)||f(\eta_j)-f(\xi_j)||\alpha(x_j) - \alpha(x_{j-1})| \\ \leqslant M(\alpha(b) - \alpha(a))\frac{\epsilon}{2M(\alpha(b)-\alpha(a))} \\ = \frac{\epsilon}{2}.$$
Thus,
$$\int_a^b g d \beta = I = \int_a^b gf d \alpha.$$
