Let $\alpha$ be a functions of bounded variations on $[a, b]$ and $f$ a buonded function such that $f \in R(\alpha)$ (i.e. integrable with $\alpha$ as integrator).$V(x)$ is the total variation of $\alpha$ in $[a, x]$ with $V(a)=0$.
I need to prove that
\begin{gather*}
\left| \ \int_a^b f \ d \alpha \ \right| \leq \int_a^b \mid f \mid \ d V
\end{gather*}
What I've tried
$D=V-\alpha$, then $V$ and $D$ are increasing functions and we can write:
\begin{gather*}
\int_a^b f \ d \alpha = \int_a^b f \ d V – \int_a^b f \ d D \leq \int_a^b \mid f \mid \ d V – \int_a^b f \ d D
\end{gather*}
I think I need some considerations on the link between the signs of $V$ and $D$.
Best Answer
It is true that if $f \in \mathcal{R}(\alpha)$ where $\alpha$ is of bounded variation , then $f$ is Riemann-Stieltjes integrable with respect to both the increasing functions $V$ and $D = V- \alpha$ as you claim. The function $D$ is not needed for this proof.
Take $g = (V+\alpha)/2$ and $h = (V-\alpha)/2$. Both $g$ and $h$ are increasing functions and, as linear combinations of $V$ and $\alpha$, it follows that $f$ is Riemann-Stieltjes integrable with respect to these integrators. Thus,
$$\tag{1}\left|\int_a^b f \, d\alpha\right| = \left|\int_a^b f \, d(g-h)\right| = \left|\int_a^b f \, dg - \int_a^b f \, dh\right| \leqslant \left|\int_a^b f \, dg \right|+ \left| \int_a^b f \, dh\right|$$
Since $g$ and $h$ are increasing we have $\left|\int_a^b f \, dg \right|\leqslant \int_a^b |f| \, dg $ and $\left|\int_a^b f \, dh \right|\leqslant \int_a^b |f| \, dh $, and, substituting into (1), it follows that
$$\tag{2}\left|\int_a^b f \, d\alpha \right| \leqslant \int_a^b |f| \, dg + \int_a^b |f| \, dh = \int_a^b |f| \, d(g+h) $$
Since $V = g+h$, we get upon substitution into (2)
$$\left|\int_a^b f \, d\alpha \right| \leqslant \int_a^b |f| \, dV $$