I'm stuck on the following exercise
If $f: [a,b]\to \mathbb R$ is of bounded variation, $t(x) = Var(f;[a,x])$ for ($a\le x \le b$), then $t$ is $\lambda$-a.e. differentiable and t' = |f'| almost everywhere. (here $\lambda$ denotes Lebesgue measure).
$t$ is monotonically increasing, so is differentiable almost everywhere. Also, we have for $y
$ \begin{align} \frac{t(x)-t(y)}{x-y} &= \frac{Var(f,[y,x])}{x-y} \\ &\ge \frac{|f(x)-f(y)|}{x-y} \\ \end{align} $
Letting $y\to x$ we obtain - at each point, where both limits exist - that t'(x) \ge |f'(x)|. I don't know how to prove the other inequality, though.
There is also a hint in the exercise that one should use the following result:
If $f_n: [a,b]\to \mathbb R$ is a sequence of montonically increasing functions such that $F = \sum_n f_n$ converges, then for almost all $x\in [a,b]$ we have F'(x) = \sum_n f_n'(x)
But I really don't see how this could be used here...
I have also thought about writing $f(x) = f^+(x) - f^-(x)$ for monotonically increasing $f^+$ and $f^-$. And then $t(x) = f^+(x) + f^-(x)$, so I would need to prove (f^+)'(x) + (f^-)'(x) = |(f^+)'(x) - (f^-)'(x)| for almost all $x$. This is the same as saying that for almost all $x$ we either have (f^+)'(x) = 0 or (f^-)'(x) = 0. This seems like an Ansatz, which might lead to something, but I can't push it to its conclusion.
Any help would be greatly appreciated. Thanks! =)