TITLE: Bounded variation and $\int_a^b |F'(x)|dx=T_F([a,b])$ implies absolutely continuous
QUESTION [6 upvotes]: If $F$ is of bounded variation defined on $[a,b]$, and $F$ satisfies
$$\int_{a}^b |F'(x)|dx=T_F([a,b])$$
where $T_F([a,b])$ is the total variation. How to prove that $F$ is absolutely continuous?

My Attempt: I used the inequality, that $P_F([a,x])$ and $N_F([a,x])$ (positive and negative variation )are both monotonic non-decreasing function, and thus their derivative exists a.e. Thus 
$$\int_{a}^b |F'(x)|dx\le \int_a^b |P_F'([a,x])|dx+\int_a^b|N_F'([a,x])|dx\le P_f([a,b])+N_F([a,b])=T_F([a,b])$$
Then by the condition, the middle inequality should all be replaced by equality.
But I cannot derive any useful information from the equalities, since the annoying absolute value cannot be diminished. I tried to prove 
$$F(x)=F(a)+\int_{a}^x F'(t)dt$$
but this doesn't work. Applying the definition of absolute continuity also failed to give me a clearer view.
Thanks for your attention!

REPLY [7 votes]: First establish equality: for every $x\in [a,b]$
$$
\int_a^x |F'| = T_F(a,x).
$$
For this write
$$
\Big( T_F(a,x) - \int_a^x  |F'| \Big) + \Big( T_F(x,b) - \int_x^b  |F'| \Big) = 0
$$
and note that each parentheses above is $\ge 0$ (  $T_F' = P_F' + N_F' \ge |P_F' - N_F'| =|F'|$, then note that each term is an increasing fct, should be part (a) of that excersice). Consequently, 
$$
\int_{a_k}^{b_k} |F'| = T_F(a_k,b_k)
$$
for any subinterval $[a_k, b_k]$. Since $F'$ is integrable, given $\epsilon>0$, let $\delta>0$ such that $\int_E |F'|< \epsilon$ whenever $m(E)<\delta$. It follows that for any set of disjoint intervals $(a_k, b_k)$ with $k=1,\ldots,N $ and $\sum_1^N (b_k-a_k)<\delta$
$$
\sum_1^N |F(a_k)-F(b_k)| \le \sum_1^N T_F(a_k, b_k) = \int_{\cup(a_k,b_k)} |F'| < \epsilon
$$