suppose $f$ is AC on $[a,b]$. I would like to establish the relation $V_a^b(f)=\int_a^b |f'(t)|\mathrm{d}t$, denoting $V_a^b(f)$ as the variation of $f$ in $[a,b]$.
Since this questions involves many ideas of (the intro to) real analysis - I would appeariciate a proof-check. The same question (with a different solution): One approach to showing the total variation of an absolutely continuous function f is the integral of |f'|
Knowing that every AC function is the difference of 2 monotonic AC functions - I wanted to first show this for such functions (and this way i'll get some intuition and maybe expand the result easily).
For a non-decreasing AC function $g$ we have:
$$V_a^b(g) = \sup\{\sum_{i=1}^ng(x_i)-g(x_{i-1}) \mid a=x_0<x_1<\dots<x_n=b\}.$$ But the right hand side is just a the integral $\int(\cdot)\mathrm{d}g$ over a subset of simple functions of the form $\sum_{i=1}^n\mathbb{1}_{[x_i,x_i-1]}$ and therefore, $V_a^b(g)\leq \int_a^b \mathrm{d}g$.
Since $g$ is AC we have the equality $\int_a^b \mathrm{d}g =\int_a^b g'(t)\mathrm{d}t$. $g'$ is the radon-nykodim derivative of the measure $\mathrm{d}g$ (and also the derivative of $g$). The next inequaly (for the non-decreasing function $g$) $\int_a^b g'(t)\mathrm{d}t \leq g(b)-g(a)= V_a^b(g)$ gives the wanted result. $\color{blue}{\textrm{Can we understand the equality } \int_a^b\mathrm{d}g = \int_a^bg' \text{ without provoking radon-nykodim derivatives?}}$
Next step is to show this for a signed AC function $f$.We assume $f$ is positive (the negative case is similar) and write $f=g-h$. since $f\geq0$ we have $g\geq h$. Since both $g,h$ are of bounded variation, we have $V_a^b(g)-V_a^b(h)\leq V_a^b(g-h)$. As for the other inequality (note that the summands in the definition of the variations are signed), we choose a partition of $[a,b]$ such that $$V_a^b(h) \leq \sum_{i=1}^k h(x_i)-h(x_{i-1}) +\varepsilon/2 $$ fining the partition (and reindexing) we can come up with a finer partition, for which we have $$V_a^b(g-h) \leq \sum_{i=1}^n g(x_i)-g(x_{i-1}) -\sum_{i=1}^n h(x_i)-h(x_{i-1}) +\varepsilon.$$
Combining these 2 inequalities we have:$$V_a^b(h-g) \leq V_a^b(g)-V_a^b(h)+\varepsilon/2$$ and since this is true for all $\varepsilon>0$ we are done. The (almost) same proof shows a similar results for negative $f$. $\color{blue}{\textrm{Are there more intuitive ways to show this?}}$
In the general case, an AC function $f=g-h$ ($g,h$ non-decreasing AC) may be written as $f=f^+-f^-=(g-h)^+ - (g-h)^-$ both are signed and AC. We have: $$V_a^b(f)=V_a^b(g-h)^+ + V_a^b(g-h)^-=\dots =\int_a^b(g'-h')^++(g'-h')^- = \int_a^b|f'|$$
$\color{blue}{\textrm{The equality } V_a^b(f)=V_a^b(f^+)+V_a^b(f^-) \text{ is not so obvious to me} }$
