TITLE: Differentiability almost everywhere of the variation of a function
QUESTION [5 upvotes]: 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 $y<x$:
$$
\begin{align}
\frac{t(x)-t(y)}{x-y} &= \frac{Var(f,[y,x])}{x-y} \\
&\ge \frac{|f(x)-f(y)|}{x-y} \\
\end{align}
$$
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! =)

REPLY [2 votes]: We prove this step by step:

Consider a "bump function" that bridges $f$ to $(-\infty,+\infty)$ such that $f$ is defined everywhere in $\mathbb{R}$, $f(x)\rightarrow 0$ as $x\rightarrow -\infty$. This bump function should not influence the conclusion except we notice that the total variation may be changed by a constant from the original total variation. 
We have the modified $f\in NBV$. So there corresponds a unique Borel measure $\mu$ such that $f(x)=\mu(-\infty,x)$. Moreover for this $\mu$ we have $$T_{f}(x)=|\mu|(-\infty,x)$$
By Lesbegue decomposition theorem we have $$|\mu|=\mu_{a}+\mu_{b},\mu_{a}\le \lambda, \mu_{b}\perp \lambda$$ (therefore $|\mu|'=\mu_{a}'$ almost everywhere)
By Lesbegue decomposition theorem we have $$\mu_{a}=\int_{E}hd\lambda$$for some unique $h\in L^{1}(\lambda)$. Now we verify that $h=|f'|$ almost everywhere.