I am trying to prove that when $f$ and $g$ are of bounded variation on $[a, b]$, $f/g$ is of bounded variation on [a, b] if there exists an $\varepsilon\gt 0$ such that $|g(x)|\ge \varepsilon$ for $x\in [a, b]$.
Let $\Gamma$ be a partition of $[a, b]$ by $\Gamma=\{x_0, x_1, \cdots, x_m: x_0=a, x_m=b\}$.
Let $\displaystyle S_\Gamma[f; a, b] = \sum_{i=0}^m \left[f(x_i)-f(x_{i-1})\right]$.
Let $\displaystyle V[f; a, b] = \sup_{\Gamma} S_\Gamma[f; a, b]$.
\begin{align} S_\Gamma[f/g; a, b] &= \sum_{i=1}^m \left|\frac{f(x_i)}{g(x_i)}-\frac{f(x_{i-1})}{g(x_{i-1})}\right|\\ &=\sum_{i=1}^m \left| \frac{g(x_{i-1})\left(f(x_i)-f(x_{i-1})\right)-f(x_{i-1})\left(g(x_i)-g(x_{i-1})\right)}{g(x_i)g(x_{i-1})} \right|\\ &\le\sum_{i=1}^m \left|\max(g)\right|\left|\frac{f(x_i)-f(x_{i-1})}{g(x_i)g(x_{i-1})}\right| - \sum_{i=1}^m \left|\max(f)\right|\left|\frac{g(x_i)-g(x_{i-1})}{g(x_i)g(x_{i-1})}\right|\\ &= \left|\max(g)\right| \left|\frac{S_\Gamma[f; a, b]}{\sum_{i=1}^m g(x_i)g(x_{i-1})}\right| - \left|\max(f)\right| \left|\frac{S_\Gamma[g; a, b]}{\sum_{i=1}^m g(x_i)g(x_{i-1})}\right|\\ &\le \left|\max(g)\right| \left|\frac{V_\Gamma[f; a, b]}{\sum_{i=1}^m g(x_i)g(x_{i-1})}\right| - \left|\max(f)\right| \left|\frac{V_\Gamma[g; a, b]}{\sum_{i=1}^m g(x_i)g(x_{i-1})}\right| \end{align}
I proved until here but I can't proceed more.
I don't know whether $\left|\frac{A}{B}\right|$ is less than or equal to, or greater than or equal to $\frac{\left|A\right|}{\left|B\right|}$
I have the following conditions from the proposition: $$V[f; a, b] < \infty \\ V[g; a, b] < \infty$$
Moreover, the other problem is when $\varepsilon$ become smaller and smaller, the value of $\frac{1}{g}$ become bigger and bigger.
Can someone give me the way to prove this proposition?
