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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!

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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.

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  • $\begingroup$ Okay! Only to check if I understood the argument: You take into account that $f$ is biholomorphic almost everywhere (because it is locally biholomorphic for all $z$ such that $f'(z)\neq 0$ and $Z = \{z\;f'(z) = 0\}$ has measure zero). Then, $\lambda(E) \propto \lambda(f(E))$ by the change of variables theorem and we finish it. Am I right (obviously writen rigorously with neighbourhoods and coverings)?? $\endgroup$
    – dudas
    Mar 15, 2018 at 14:28
  • $\begingroup$ And, about the change of variables theorem: I read that $\lambda(E) = |J_f|\lambda(f(E))$, but what does $|J_f|$ mean? I know that it is the Jacobian, but it doesn't make sense that $\lambda(E)$ doesn't depend no any point and $J_f$ does...what I don't understand of the change of variables theorem? (I am really new in measure theory) $\endgroup$
    – dudas
    Mar 15, 2018 at 14:38
  • $\begingroup$ There is a multi-dimensional variant of the 'substituation formula' - called sometimes ' transformation theorem', see here. In your case, we have $\lambda^2(f(E \cap U)) = \int_{E \cap U} | \det [(D f^{-1}(x,y)] | \, \rm{d} x \rm{d}y$. If $E$ is a nullset, this implies $\lambda^2(f(E \cap U)) =0$. $\endgroup$
    – p4sch
    Mar 15, 2018 at 20:16

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