Proof with total variation and proving absolutely continuous? Show that if $F$ is of bounded variation in $[a, b]$, then
$\bullet \int_{a}^{b}|F'(x)|dx \leq T_{F}(a,b)$.
$\bullet \int_{a}^{b}|F'(x)|dx = T_{F}(a,b)$, iff $F$ is absolutely continuous
Use the above to further show that the formula $L=\int_{a}^{b}|z'(t)|dt$ for the length of a
rectifiable curve parameterized by $z(t)$ holds iff $z$ is absolutely continuous.
Note here that $T_{F}(a,b)$ stands for the total variation of $f$.
 A: First, let $T(x) = T_F(a,x)$, for simplicity. We know that if $T(a,b) = T(b) < +\infty$, then $T'(x) = |F'(x)|,$ for a.e. $x\in (a,b)$. Moreover, since $T$ is nondecreasing in $x$, it follows in general that for every $x\in(a,b)$
$$\int_{(a,x)} T'(s)\, ds \le T(x-) - T(a+) = T(a,x).$$
Now, let $x=b$ and replace $T'$ by $|F'|$. One has 
$$\int_{(a,x)} |F'(s)|\, ds \le T(a,b).$$
As for equality, provided that $F$ is of bounded variation, we know that $T$ is absolutely continuous if and only $F$ is. Of course, $T$ is absolutely continuous if and only if $T'$ exists a.e. and
$$\int_{(a,b)} T'(x)\,dx = T(b) - T(a) = T(a,b).$$
As before, since $T' = |F'|$ a.e., the result follows. 
Intuitively, absolute continuity prevents our function from having continuous change over infinitely small intervals. A perfect example of a continuous function that changes over infinitely small intervals is the Cantor-Lebesgue function. When we integrate $f'$, we get $0$ because $f' = 0$ a.e. yet the function still has positive total variation. If you can apply this intuition to your rectifiable curves, you should be able to get it no problem.
