If $f$ is continuous and $g$ is of bounded variation then $f$ is integrable with respect to $g$. Suppose $f$ is continuous on $[a,b]$ and $g$ is of bounded variation on $[a,b]$.
Prove that $f$ is integrable with respect to $g$ on $[a,b]$.
How can I prove this? 

You can prove this in the easier case when $g$ is monotone on $[a,b]$, and then use that $g=V-(V-g)$ is a difference of monotone functions, where $V$ is the total variation of $g$ over $[a,x]$, i.e $V_g(a,x)$. You'll have to use that $f$ is uniformly continuous, and that $g$ has variation bounded by $g(b)-g(a)$. Then, choose a partition  with mesh $<\delta$ such that for each $|x-y|<\delta$, we have $$\tag 1 |f(x)-f(y)|<\frac{\epsilon}{A}$$ where $A=2(g(b)-g(a))$.
You'll have to use that being Riemann Steiltjes integrable is equivalent to the Riemann condition (for monotone integrators), that is, that given $\epsilon >0$ there exists a partition $P_\epsilon$  such that $P\supset P_\epsilon$ implies $$U(P,f,\alpha)-L(P,f,\alpha)<\epsilon$$
To this end, you'll have to sue that, because of $(1)$, $$M_k-m_k\leq \epsilon/A$$
where $M_k,m_k$ are the supremum and infimum of $f$ over the $k$th interval. Good luck.
