# Does holomorphic a.e. and continuous imply holomorphic everywhere?

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 $K:=f^{-1}(0)$.

• 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?
– user2093
Commented Mar 17, 2013 at 6:30
• @WIlliam: No, your function has an essential singularity at zero, it is not continuous there. Remember, sine is not bounded in the plane. Commented Mar 17, 2013 at 6:42
• Ah, yes... I'm too tired.
– user2093
Commented Mar 17, 2013 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.
• 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. Commented Mar 17, 2013 at 14:43