Let $f:[a,b]\to\Bbb R$. Define $PV(f,[a,b])$ to be the positive variation of $f$ on $[a,b]$. I need to show that $PV(f,[a,b])=\int_{[a,b]}(f')^+$. I've already shown that $TV(f,[a,b])=\int_{[a,b]} |f'|$ where $TV$ denotes the total variation. We'll also denote $NV$ as the negative variation, and we know $TV=PV+NV$. Here's what I've tried so far:
$$ TV(f,[a,b])=\int_{[a,b]} |f'| = PV(f,[a,b])+NV(f,[a,b]) = \int_{[a,b]}(f')^+ + \int_{[a,b]}(f')^- $$
I also know that $f'=(f')^+-(f')^-$ and so $\int_{[a,b]} f' = \int_{[a,b]} (f')^+-\int_{[a,b]}(f')^-$. It seems like a relationship between $\int_{[a,b]} f'$ and the variation would be productive. But the only one that I know is $\int_{[a,b]} f' \le f(b)-f(a)$. If I had equality then I could use this together with $TV(f,[a,b])=f(b)-f(a)$ but I'm not certain that the reverse inequality holds. In fact I have a theorem saying that absolute continuity is equivalent to this equality only under the hypothesis that $f'$ is monotonic. Since that may not be true here, then I suspect the equality does not necessarily hold.