Let $f \colon [0,1] \to \mathbb{R}$ be a càdlàg function, and let $V(f) \in [0,\infty]$ be the total variation of $f$ over $[0,1]$.
Is it the case that for any $y<V(f)$ there exists a partition $0 = t_0 < t_1 < \ldots < t_{N+1} = 1$ of $[0,1]$ such that:
- $f$ is continuous at $t_i$ for all $0 \leq i \leq N$;
- we have $$ \sum_{i=0}^N |f(t_{i+1})-f(t_i)| > y \ ? $$
(This question is motivated by discussion in comments under my answer to Why is this function of bounded variation?)
Yes. It follows from the definition of the supremum, right-continuity of $f$ and the fact that discontinuities of any càdlàg function are at most countable.
If you want more detail: if $y<V(f)$, then by definition of the supremum, you can find a subdivision $0=t_0<t_1<\cdots<t_{N+1}=1$ such that $\sum_{i=0}^N\vert f(t_{i+1})-f(t_i)\vert>y$.
Suppose that $f$ is not continuous at $t_i$ for some $i\in\{1,\cdots,N\}$. As the set of discontinuities of $f$ is at most countable, there exists $h>0$ arbitrarily small such that $f$ is continuous at $t_i+h$. As $f$ is right-continuous, you can choose $h$ small enough such that $\vert f(t_{i+1})-f(t_i+h)\vert+\vert f(t_i+h)-f(t_{i-1})\vert$ is arbitrarily close to $\vert f(t_{i+1})-f(t_i)\vert+\vert f(t_i)-f(t_{i-1})\vert$. This means you can replace $t_i$ in your subdivision with some $t_i+h$ for $h$ small enough such that $f$ is continuous at this new $t_i$ and $\sum_{i=0}^N\vert f(t_{i+1})-f(t_i)\vert>y$ is still satisfied. Repeat for potential other discontinuities.
