Show that the Riemann Stieljes integral of fd$\gamma$ exists Now we have that $\beta$ is continuous on [a,b] and of bounded variation. Now we have another function $g \in R(\beta)$ i.e it is Riemann-Stieljes integrable with respect to $\beta$. 
Now we have $\gamma(x)$ = $\int_a^x gd\beta$ for $x\in[a,b]$. Now if $f$ is a continuous function we have to show that: $$\int_a^b fd\gamma = \int_a^b fgd\beta$$
My work so far: 
For existence:
So far I have shown that $\gamma$ is continuous on [a,b] and of bounded variation. There is a theorem in rudin that states that if your integrator is continuous and f is bounded then $f\in R(\gamma)$ but it also assumes $\gamma$ is monotonically increasing which is not necessarily the case here. Any help or hints would be appreciated
To show that  $\int_a^b fd\gamma = \int_a^b fgd\beta$ :
It is enough to show that $\int_a^b fd\gamma = \int_a^b fgd\beta$ in the case where $\beta$ is monotonically increasing ( this was a hint given to me i don't really know why it is enough to just consider it when it is monotonically increasing)
Now $f \in R(\beta)$ since it is continuous (theorem in rudin). Now I have shown that the riemann sum $$S(f,\gamma,P) = \sum_{j=1}^n f(t_j)[(\gamma(x_j)-\gamma(x_{j-1})]$$ = $$\sum_{j=1}^n [f(t_j)(\int_{x_j-1}^{x_j} gd\beta)]$$ for some $t_j \in [x_{j-1}, x_j]$ and P is any parition.
Now i know that $fg$ are both Riemann Stieljes integrable with respect to $\alpha$ by thoerem 6.13 in Rudin. Hence for all $\epsilon > 0$ there exists a partition $P_\epsilon$ such that for any partition P finer than $P_\epsilon$ we have that:
$$ \sum_{j=1}^n f(t_j)g(t_j)[(\beta(x_j) - \beta(x_{j-1})] - \int_a^b fgd\beta < \epsilon $$
Now I can't seem to combine these two facts to get the desired result, any help or hints would be very appreciated thank you!
 A: Any function of bounded variation is the difference of two monotonically increasing functions, so it is enough to prove this in the case that $\beta$ is monotonically increasing. 
To finish, note that
$$\left| S(f,\gamma,P) -  \int_a^b fg \, d\beta 
\right| \\= \left|\sum_{j=1}^n f(t_j)\int_{x_j-1}^{x_j} g \,d\beta
- \sum_{j=1}^n\int_{x_j-1}^{x_j} f(y)g(y)\, \,d\beta \right| \\ \leqslant \sum_{j=1}^n \int_{x_j-1}^{x_j}|f(t_j)- f(y)| |g(y)| \,d\beta
 .$$
Since $g \in R(\beta)$ is bounded there exists $M > 0$ such that, $|g(y)| \leqslant M $ for all $y \in [a,b]$.
Defining $M_j(f) = \sup_{x \in [x_{j-1},x_j]}\, f(x)$ and  $m_j(f) = \inf_{x \in [x_{j-1},x_j]}\, f(x)$ we have
$$\left| S(f,\gamma,P) -  \int_a^b fg \, d\beta 
\right|   \leqslant M \sum_{j=1}^n \int_{x_j-1}^{x_j}|f(t_j)- f(y)| \,d\beta
 \\ \leqslant  M \sum_{j=1}^n (M_j(f) - m_j(f))\int_{x_j-1}^{x_j} \,d\beta \\ =  M \sum_{j=1}^n [\,M_j(f) - m_j(f)\,]\, [\, \beta(x_j)-\beta(x_{j-1}) \, ] \\ = M\left(U(f,\beta,P) - L(f,\beta,P) \right).$$
Given the hypotheses, we also have $f \in R(\beta)$. Thus, for any $\epsilon >0$ there is a partition $P_\epsilon$ such that if $P$ is any refinement, then $U(f,\beta,P) - L(f,\beta,P) < \epsilon/M$ and
$$\left| S(f,\gamma,P) -  \int_a^b fg \, d\beta 
\right| < \epsilon,$$
proving
$$\int_a^b f \, d\gamma = \int_a^b fg \, d\beta 
$$
