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.

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    $\begingroup$ Why doesn't the example $f(z) = z\sin(1/z)$ if $z\neq 0$ and $f(0) = 0$ work? It is holomorphic everywhere except $z = 0$, but is continuous on $\mathbb{C}$, no? $\endgroup$ – user2093 Mar 17 '13 at 6:30
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    $\begingroup$ @WIlliam: No, your function has an essential singularity at zero, it is not continuous there. Remember, sine is not bounded in the plane. $\endgroup$ – J. Loreaux Mar 17 '13 at 6:42
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    $\begingroup$ Ah, yes... I'm too tired. $\endgroup$ – user2093 Mar 17 '13 at 6:46

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.

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  • $\begingroup$ I thought the proof to Rado's theorem in your reference is not correct, since the proof reduces the analytic function $f$ as two separate harmonic function $u$ and $v$, and his argument about the harmonic function along is not correct. For example, take $u(x,y)=2y$ for $y\geq 0$, $u(x,y)=y$ for $y<0$, then $u$ is harmonic outside its zero set, but $u$ is not harmonic in the whole plane. $\endgroup$ – Yuchen Liu Mar 17 '13 at 14:43

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