I want to see why holomorphic mappings send sets of measure zero to sets of measure zero.
I found this statement reading about quasiconformal mappings. In fact, there is a theorem (see for instance L. Ahlfors (2006) Lectures on Quasiconformal Mappings, American Mathematical Society) that state that a quasiconformal mapping $\phi$ maps sets of measure zero to sets of measure zero. I don't know if it could help. It also states that for every measurable set $E$ \begin{equation*} m(\phi(E)) = \int_E \text{Jac}\phi\;dm, \end{equation*} where $\text{Jac}$ denotes the Jacobian and $m$ is the Lebesgue measure.
What I thought is to use the fact that an holomorphic map $f$ is locally quasiconformal almost everywhere (a.e. because of the isolated zeros of $f'$ and the fact that conformality implies locally quasiconformality), but sincerely I'm really stuck.
Thanks in advance!
If $f \colon \mathbb{C} \rightarrow \mathbb{C}$ is non-constant holomorphic function, then $\{f'=0\}$ is countable. For any point $z \notin \{f'=0\}$ we can find neighborhhoods $U$ of $z$ and $V$ of $f(z)$ such that $f$ restricted to $U$ is a biholmorphic map von $U$ to $V$.
So, if $\lambda^2(E) =0$, then $\lambda(f(E \cap U) \cap V) =0$ by the change of variables theorem. This, implies with a covering argument (and using that $\{f'=0\}$ is countable) that $\lambda^2(f(E))=0$. Note that the above argument also shows that $f(E)$ is measurable.