BV functions and absolute continuity short inquiry This comes as a complement to: Relation between total variation and absolute continuity; I was wondering if the following holds:
Let $F$ be a function of bounded variation on $[a,b]$, then $\int_{a}^{b}{|F'(x)|dx} = T_{F}(a,b)$ implies $F$ is absolutely continuous (same notations).
Any help is welcomed.
I guess that we actually have that if $G$ is an increasing continuous function for which $G'(x) < \infty$ a.e, then $G$ is absolutely continuous. (?)
 A: 
Let $F$ be a function of bounded variation on $[a,b]$, then $\int_{a}^{b}{|F'(x)|dx} = T_{F}(a,b)$ implies $F$ is absolutely continuous (same notations).

Yes.  $T_F(a,b)$ is the norm of the measure $|dF|$.  In general we have the decomposition 
$F(x) = F_s(x) + \int_a^x F'(t)\ dt$ where $F_s$ is singular with respect to Lebesgue measure, and $T_F(a,b) = \||dF_s|\| + \int_a^b |F'(x)|\ dx$.   $T_F(a,b) = \int_a^b |F'(x)|\ dx$ iff $dF_s = 0$ iff $F(x) = \int_a^x F'(t)\ dt$ iff $F$ is absolutely continuous.

I guess that we actually have that if $G$ is an increasing continuous function for which $G'(x) < \infty$ a.e, then $G$ is absolutely continuous. (?)

No, that's wrong.  $G'(x) < \infty$ a.e. for any increasing function.
A: Another approach to prove this:
Since, $\int_{a}^{b}{|F'(x)|dx} = T_{F}(a,b)$, and we know that $\int_{c}^{d}{|F'(x)|dx}\le T_{F}(c,d)$ for any $[c,d]\subset[a,b]$ we can show that $\int_{c}^{d}{|F'(x)|dx} = T_{F}(c,d)$ for all subintervals $[c,d]$ actually.
Then for $\{(a_k,b_k)\}_{k=1}^n$ if $\sum_{k=1}^n(b_k-a_k)<\delta$ consider $\sum_{k=1}^n|F(b_k)-F(a_k)|\le\sum_{k=1}^nT_{F}(a_k,b_k)=\sum_{k=1}^n\int_{a_k}^{b_k}{|F'(x)|dx}=\int_{\cup_{k=1}^n(a_k,b_k)}|F'(x)|dx$. 
Then since $F^{\prime}\in L^1$, we have that for every $\epsilon>0$ there exists $\delta>0$ such that whenever $m(A)<\delta$ then $\int_A|F^{\prime}|<\epsilon$. By applying this to above we can conclude that $F$ is absolutely continuous.
