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Hello everyone,

I would like some help on proving the following statement:

Let $f\in\mathrm{BV}[a,b]$, i.e. $f$ is of bounded variation and let $T(x)$ be the total variation of $f$ on $[a,x]$ for $x\in[a,b]$, then the derivative of $T$ is equal to the derivative of $|f|$ a.e. that is, $$T'= |f|' \quad\mathrm{almost\;everywhere}.$$

Any help is appreciated.

First I've simplified the problem. If $f$ is continuously differentiable, then $f$ is of bounded variation. So I showed that the right derivative of $T$ at $x=c$ equals $|f'(c)|$. This can be done by taking a partition and after a careful estimation with the Mean-Value theorem on each subinterval, this case follows.

This argument clearly fails in the general case, since the Mean-Value theorem might not work. I have no idea how to prove the general case.

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  • $\begingroup$ Fundamental Theorem of Calculus? $\endgroup$
    – Aaron Hoffman
    Commented Jan 12, 2012 at 2:09
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    $\begingroup$ Recall that the derivative of a function of bounded variation exists a.e. Then show that $T'\ge |f'|$ at all points where both parts make sense (note that ${}'$ has to be inside $|\,|$, not outside). Finally, show that $T'\le |f'|$ at every point that is a Lebesgue point of $f'$ and the zero-density point of the singular part of $dT$ simultaneously. It isn't FTC, rather Radon-Nicodim. $\endgroup$
    – fedja
    Commented Jan 12, 2012 at 7:23

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