Let $\nu$ be a finite signed Borel measure on the interval $[a,b]$, and let $F_\nu$ be the function given by $F_\nu(x) = \nu([a,x])$. I'd like to show that the total variation $V(F_\nu, [a,b])$ is the same as $|\nu|([a,b])$. Let's require that $F_\nu(a) = 0$ because the result doesn't seem to hold in general.
I was able to prove the inequality $V(F_\nu,[a,b]) \leq |\nu|([a,b])$; it just amounted to showing that for every partition of $[a,b]$, the corresponding sum of absolute differences of $F_\nu$ is less than $|\nu|([a,b])$. But I'm struggling with the other inequality. I feel like the idea should be to construct a partition such that the associated sum of absolute differences is arbitrarily close to $|\nu|([a,b])$, but how do we do this?
Here is my inital idea: We break $[a,b]$ into a Hahn decomposition $P \sqcup N$ for $\nu$, and then find subsets $P' \subseteq P$ and $N' \subseteq N$ such that $P'$ and $N'$ are finite unions of closed intervals and $|\nu|(P - P')$ and $|\nu|(N - N')$ are arbitrarily small. Then we can use the sets $P'$ and $N'$ to build a suitable partition for $[a,b]$. I'm having trouble finding the sets $P'$ and $N'$, though. If $|\nu|$ were just the normal Lebesgue measure $\mu$, then this would be easy, but for an arbitrary Borel measure $\nu$ I don't know if it's possible.