Bounded variation and $\int_a^b |F'(x)|dx=T_F([a,b])$ implies absolutely continuous

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!

Solution 1:

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 $\sum_1^N (b_k-a_k)<\delta$ $$ \sum_1^N |F(a_k)-F(b_k)| \le \sum_1^N T_F(a_k, b_k) = \int_{\cup(a_k,b_k)} |F'| < \epsilon $$