(Proof) $g\in BV[a,b]$ and $f\in\mathcal{R}_g[a,b]$ and $f$ bounded $\Rightarrow 1/f\in\mathcal{R}_g[a,b]$ As presented in the title, $g\in BV[a,b]$, $f\in\mathcal{R}_g[a,b]$ with $1/f$ bounded by some positive $M$. How do I imply that $1/f$ is also Riemann-Stieltjes integrable with respect to the integrator $g$?
My thoughts: I already know that one $BV$ + one continuity implies R-S integrality, but it does not seem to work since I don't know if $1/f$ is continuous or not. So I don't know how to start.
 A: If $g$ were monotone increasing then we have that $1/f$ is Riemann-Stieltjes integrable with respect to $g$, if for any $\epsilon >0$ there is a partition $P =(x_0,x_1,\ldots,x_n)$ of $[a,b]$ such that 
$$U(P,1/f,g) - L(P,1/f,g) < \epsilon,$$
Note that $U(P,1/f,g) - L(P,1/f,g) = \sum_{j=1}^n\omega_j(1/f)(g(x_j) - g(x_{j-1})),$ where the oscillation $\omega_j(1/f)$ is given by
$$\omega_j(1/f) = \sup_{x,y\in [x_{j-1},x_j]}\left|\frac{1}{f(x)} - \frac{1}{f(y)}\right|$$
Since $f$ is Riemann-Stieltjes integrable it must be bounded above and below on $[a,b]$, and  since we are given that $1/f$ is bounded we have constants $\alpha,\beta$ such that $0 < \alpha \leqslant  |f(x)| \leqslant \beta$ for $x \in [a,b]$. 
Thus for all $x,y \in [a,b]$ we have
$$\left|\frac{1}{f(x)} - \frac{1}{f(y)}\right| = \frac{|f(x) - f(y)|}{|f(x)||f(y)|} \leqslant \frac{|f(x) - f(y)|}{\alpha^2},$$
and, hence, $\omega_j(1/f) \leqslant \omega_j(f)/\alpha^2$ and $U(P,1/f,g) - L(P,1/f,g) \leqslant (U(P,f,g) - L(P,f,g))/\alpha^2$.
Since $f$ is Riemann-Stieltjes integrable with respect to $g$, there exists a partition $P$ such that $U(P,f,g) - L(P,f,g) < \epsilon \alpha^2$  and it follows that
$$U(P,1/f,g) - L(P,1/f,g) \leqslant \frac{U(P,f,g) - L(P,f,g)}{\alpha^2} < \epsilon$$
Therefore, $1/f \in \mathcal{R}_g([a,b])$ when $g$ is monotone increasing.
We can extend this proof  for $g \in BV([a,b])$ since there always exists monotone increasing functions $h_1,h_2$ such that $g = h_1-h_2$ and such that $f \in \mathcal{R}_{h_1}([a,b]) \cap \mathcal{R}_{h_2}([a,b]) $.  By the above argument we then have $1/f \in \mathcal{R}_{h_1}([a,b]) \cap \mathcal{R}_{h_2}([a,b]) $.  By linearity of the Riemann-Stieltjes integral this imples that $1/f$ is integrable with respect to $g = h_1-h_2$.
One choice for $h_1$ and $h_2$ that works is $h_1(x) = V_a^x(g)$, the total variation of $g$ on $[a,x]$, and $h_2 = h_1 - g$
