Inequality for Riemann-Stieltjes integral Suppose $f$ is Riemann-Stieltjes integrable with respect to $g$. If $g$ is increasing then it is easy to show
$$
\newcommand{\d}[1]{\,\mathrm{d}#1}
\bigg| \int_a^bf(x)\d{g(x)} \bigg| \leq \int_a^b|f(x)| \d{g(x)}
$$
just using $-|f| \leq f \leq |f|$ and the monotonicity of the integral.

Question: If $g$ has bounded variation but not monotone then the inequality is
  $$\bigg|\int_a^b f(x)\d{g(x)}\bigg| \leq \int_a^b|f(x)| \d{v_a^x(g)}$$

Here $v_a^x(g)$ is the total variation function.  How can this be proved?
 A: \begin{align}
     \left|\int_{a}^{b}fdg\right| &= \lim_{\|\mathcal{P}\|\rightarrow 0}\left|\sum_{n=1}^{N}f(x_j^*)(g(x_j)-g(x_{j-1}))\right| \\
  & \le \lim_{\|\mathcal{P}\|\rightarrow 0}\sum_{n=1}^{N}|f(x_j^*)||g(x_j)-g(x_{j-1})| \\
  & \le \lim_{\|\mathcal{P}\|\rightarrow 0}\sum_{n=1}^{N}|f(x_j^*)|v_{x_{j-1}}^{x_j}(g) \\
  & = \lim_{\|\mathcal{P}\|\rightarrow 0}\sum_{n=1}^{N}|f(x_j^*)|\{v_a^{x_j}(g)-v_a^{x_{j-1}}(g)\} \\
  & = \int_{a}^{b}|f(x)|d_{x}v_{a}^{x}(g).
\end{align}
A: Indeed for an increasing function $f$ on $[a,b]$ one always has: $$ \color{blue}{V^x_a (f) = f(x)-f(a)}$$ 
Then Note the following Property:

Another characterization states that the functions of bounded variation on a compact interval are exactly those $g:[a,b]\to \Bbb R$ which can be decomposed as $g= u_g -v_g$ where  both $ u_g $and  $v_g$    are bounded and increasing functions.( that is $g$ is the difference of two increasing functions.)  read the introduction here or   here which more difficult

Moreover, indeed from definition we know that $$ \color{red}{V^x_a (g)= u_g(x)-u_g(a) +v_g(x)-v_g(a)}\implies \color{blue}{dV^x_a (g)= du_g(x)+dv_g(x)} $$ 
therefore: 
$$\bigg|\int_a^b f(x)\d{g(x)}\bigg|= \bigg|\int_a^b f(x)\d{u_g(x)}- \int_a^b f(x)\d{v_g(x)}\bigg| \\\leq \bigg|\int_a^b f(x)\d{u_g(x)}\bigg|+\bigg| \int_a^b f(x)\d{v_g(x)}\bigg|
\\\leq\int_a^b|f(x)| \d{u_g(x)}+\int_a^b|f(x)| \d{v_g(x)}
\\\color{blue}{=\int_a^b|f(x)| d \left({u_g(x)+v_g(x)}\right)=\int_a^b|f(x)| \d{v_a^x(g)}}$$

Then you can conclude from the first case since you already done the proof for increasing function

