{wiki=Bounded_variation}

A process $A$ has finite variation on compact intervals when every sample path has finite <total variation of a function> there. Such a process can be integrated pathwise by the Lebesgue-Stieltjes integral.

Finite-variation process

Every dyadic partition is among the finite partitions on the right-hand side, so the displayed supremum is at least $\lVert f\rVert$. For the converse, the claim is immediate if $\lVert f\rVert=\infty$. If it is finite, apply part (a) to every interval of an arbitrary partition $0\leq t_0<\cdots<t_n=1$:
$$
|f(t_k)-f(t_{k-1})|
\leq\lVert f_{t_k}\rVert-\lVert f_{t_{k-1}}\rVert.
$$
Summing telescopes and gives
$$
\sum_{k=1}^n|f(t_k)-f(t_{k-1})|
\leq\lVert f_1\rVert-\lVert f_{t_0}\rVert
\leq\lVert f\rVert.
$$
Taking the supremum proves that the dyadic definition equals the usual <total variation of a function>.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 202 / 3 / b / Solution

{title2=$V_a^b(f)$}
{wiki=Total_variation}

The total variation of $f:[a,b]\to\mathbb R$ is
$$
V_a^b(f)=\sup_{a=t_0<\cdots<t_n=b}
\sum_{k=1}^n|f(t_k)-f(t_{k-1})|.
$$
The function has bounded variation when this quantity is finite.

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 Incoming links: Total variation of a function