In An Introduction to Teichmüller Spaces, by Imayoshi and Taniguchi, we have the following definition for an absolutely continuous function:
where, I suppose, a function $g:I\to \mathbb{C}$, $I \subset \mathbb{R}$ interval, is said to be absolutely continuous if, $\forall \epsilon>0$, $\exists \delta>0$ such that,
$$[x_k,y_k]\subset I, k\in \{1,\dots,n\}\text{ pairwise disjoint intervals with }\sum_{k=1}^n|y_k-x_k|<\delta$$ $$\implies\sum_{k=1}^n|f(y_k)-f(x_k)|<\epsilon.$$
The authors want to define quasiconformal mappings which are, in particular, ACL. The very first example given for quasiconformal mappings is, of course, the conformal ones. But while giving this example in the book, it is not shown that the conformal map is ACL (it is only discussed the other requirements, which are indeed easier to see).
So my question is: $f:D\to \mathbb{C}$ conformal $\implies$ $f$ ACL? Is this due to some classical result? Thank you very much!
A conformal map is differentiable on every line and the derivative is continuous, so in particular is in $L^1$ locally. This implies that the fundamental theorem of calculus is true on lines, i.e. $f(y)-f(x)= \int_{[x,y]} f' ds$ and this in turn implies absolute continuity.
The problem with the quasiconformal maps is that they are not necessarily $C^1$, but they still have weak partial derivatives in local $L^2$. This is enough to yield the ACL property.
If $f$ is in local $L^2$ and has partial derivatives of first order in local $L^2$, then we say that $f$ is in the Sobolev space $ W^{1,2}_{\textrm{loc}}$. It turns out that this is equivalent to saying that $f$ is absolutely continuous on almost every line (and thus has partial derivatives in "almost every" direction), and the partial derivatives are in local $L^2$.