I'm reading through Holomorphic Functions and Integral Representations in Several Complex Variables and have come across a proof I can't get through concerning thin sets. I'll include the definition of a thin set since it's not too common a notion:
Let $D\subseteq \mathbb{C}^n$ be open and $E\subseteq D$. $E$ is said to be thin if for every $p\in D$, there is $r>0$ such that there is a non-constant holomorphic function $f$ on $B(p,r)$ satisfying $f|_{E\cap B(p,r)} \equiv 0$.
Effectively, thin sets are sets that locally are zero sets of holomorphic functions. It is easy to see that if thin sets are Lebesgue measurable, they have to have measure zero however I have no idea how to show they are Lebesgue measurable. Marrying the complex analysis definition with measurability conditions is escaping me. Can anyone provide a hint? (Not a full proof, I'd like to figure that out on my own.)
They are (locally) contained in zero sets of non-constant holomorphic functions, not necessarily the full zero set.
To prove the Lebesgue-measurability of thin sets, show that the zero set of a non-constant holomorphic function is a null-set, and that you need only countably many to cover the thin set.