TITLE: Does holomorphic a.e. and continuous imply holomorphic everywhere?
QUESTION [9 upvotes]: Suppose $D$ is a domain in $\mathbb{C}$, $f:D\rightarrow \mathbb{C}$ is a continuous function.
Suppose $f$ is holomorphic outside the zero set $f^{-1}(0)$, and $f^{-1}(0)$ has Lebesgue measure zero.
Question: Is $f$ holomorphic on the whole domain $D$ or not?
The point that I'm confused with is that it seems that $f$ is a weakly holomorphic function, but I cannot prove it. Weakly holomorphic corresponds to $\int_D f\cdot\partial_{\bar{z}}\phi=0$ for every $\phi\in C_c^\infty(D)$, but I can only prove that $\int_D f\cdot\partial_{\bar{z}}\phi=0$ for every $\phi\in C_c^\infty(D-K)$, where $K:=f^{-1}(0)$.
Any answer or comment is welcome. I really appreciate your help.

REPLY [6 votes]: You don't even have to assume that the zero set of $f$ has zero measure. The result is known as Radó's theorem: A continuous function which is holomorphic outside its zero set is holomorphic.
It was surprisingly difficult to find an online reference, but here is one: A simple proof of Radó's theorem.