Riemann-stieltjes integral and the supremum of f Prove that if $f$ is continuous on $[a,b]$ and $g$ is bounded variation on $[a,b],$ then $$\vert\int_a^bfdg\vert\le [sup_{a\le t \le b} \vert  f(t) \vert] V_{[a,b]}g$$
Proof:
As f is continuous on [a,b] and g is BV([a,b]) then f is riemann-stieltjes integrable, i.e. $f\in R(g)$.
But I don't know how to prove $\vert\int_a^bfdg\vert\le [sup_{a\le t \le b} \vert  f(t) \vert]$ sup{$\sum_{k=1}^n|g_k(x)-g_{k-1}(x)|:  \{x_0,x_1,...,x_n\}\in P[a,b] \}$.
How can I continue the proof?
Can someone give  me a hint/solution?
Note: $V_{[a,b]}g$ means the total variation of g on [a,b]
 A: The "straightforward" proof relies on many other results pertaining to the Riemann-Stieltjes integral with respect to a general integrator function of bounded variation.
Let the function $h$ be defined as the total variation of $g$ on the interval $[a,x]$:
$$h(x) = V_{[a,x]}g$$
It can be shown that $f$ is Riemann-Stieltjes integrable with respect to $h$ and $h - g$. Also if $f$ is integrable then so is $|f|$.
Now take $\alpha = (g + h)/2$ and $\beta = (h-g)/2$. Both are increasing functions on $[a,b]$ and using basic properties of Riemann-Stieltjes integration we have
$$\begin{align}\left|\int_a^b f \, dg \right| &= \left|\int_a^b f \, d(\alpha - \beta)  \right|  \\ &= \left|\int_a^b f \, d\alpha -  \int_a^b f \, d\beta \right| \\ &\leqslant   \left|\int_a^b f \, d\alpha \right| +  \left|\int_a^b f \, d\beta \right| \\ &\leqslant \int_a^b |f| \, d\alpha  +  \int_a^b |f| \, d\beta  \\ & = \int_a^b |f| \, d(\alpha + \beta) \\ &= \int_a^b |f| \, dh \\ &\leqslant \sup_{t \in [a,b]}|f(t)|(h(b) - h(a)) \\ &= \sup_{t \in [a,b]}|f(t)|V_{[a,b]}g\end{align}$$
