Is the absolute value of the derivative of a BV function equal to the derivative of its total variation? For a given function $f\in BV([0,1])$, its function of total variation $V_f(x)$ is defined by $V_f(x)=sup_{0=t_0<t_1<...<t_n=x,\ n\in\mathbb N}\{\Sigma_{i=1}^n|f(t_i)-f(t_{i-1})|\}$, then  $f$ and $V_f$ is differentiable almost everywhere, and by Jordan's decomposition theorem, we have $|f'(x)|\leq V_f'(x)$
Furthermore， I wonder if $|f'(x)|= V_f'(x)$ almost everywhere or not, since this statement holds for all monotone functions, and every BV function could be decomposited to two monotone functions.
Any suggestion is welcomed, thanks a lot!
 A: Here is a reference to a proof.
Cater, F. S. A new elementary proof of a theorem of de la Vallée Poussin.
Real Anal. Exchange 27 (2001/02), no. 1, 393–396.

Summary: "We give a new elementary proof of the classical theorem: Let $f$ be of bounded variation on $[a,b]$ and let $V$ be its total
variation function. Then there is a set $N $  such that
$m(V(N))=m(f(N))=m(N)=0$, and for each $x$ not in $N$, $f$ and $V$
have derivatives, finite or infinite, and $V′(x)=|f′(x)|$.''

Here $m$ denotes Lebesgue measure and $f(N)$ and $V(N)$ are the images under $f$ and $V$ of the set $N$.   Note this doesn't say that at every point where one of  $f'(x)$ and $V'(x)$ exist they are equal.  Consider $f(x)=|x|$ which has no derivative at $x=0$ but its total variation function does.
In the Monthly article cited below, the author gives some examples where $f'(x)$ and $V'(x)$ both exist and have different values.  Thus the de la Vallée Poussin theorem just describes the set on which they exist and agree, not individual points.
G. A. Heuer.  The Derivative of the Total Variation Function. The American Mathematical Monthly, Dec., 1971, Vol. 78, No. 10 (Dec., 1971), pp. 1110-1112.
