Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 202 / 1

Let $V(t)$ be the total variation of $f$ on $[0,t]$. The Jordan decomposition of a function of bounded variation isFor $s<t$, the inequality $V(t)-V(s)\ge |f(t)-f(s)|$ shows that both increments are nonnegative, so $g,h$ are nondecreasing and $f=g-h$. Right-continuity of the finite variation function $f$ implies right-continuity of $V$, and hence of $g,h$.

It suffices by part a to treat a nondecreasing right-continuous integrator, whose increments define a finite Lebesgue–Stieltjes measure ${\mu }_f$ on $[0,1]$. Let ${\alpha }_n$ be the left-endpoint step approximation on the dyadic intervals. The displayed sum is exactlyThe continuous function $\alpha$ is uniformly continuous on the compact interval, so $\Vert {\alpha }_n-\alpha {\Vert }_{\infty }\to 0$. ThereforeApply this separately to $g$ and $h$ to obtain the asserted Lebesgue-Stieltjes integral limit.

For any partition of $[0,t]$,The final sum is at most the largest increment of $f$ times its total variation. It tends to zero because $f$ is uniformly continuous. Part b then gives the integration-by-parts identityThus the formula stated in the question holds when $f(0)=0$; for a general initial value the necessary endpoint correction is $-f(0{)}^2$.