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I was studying the construction of the modular lambda function and I started thinking about the following question. Suppose that $\Omega\subset \mathbb{C}$ is an open connected set and $f:\Omega\to \mathbb{C}$ is continuous on $\Omega$ and holomorphic except on a set of measure zero (with respect to 2-dimensional Lebesgue measure). Is $f$ actually holomorphic on all of $\Omega$ ?

If this statement is false, how much does one have to weaken the measure zero condition? For example if the set of measure zero is a $C^1$ curve the above follows from Morera's theorem.

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(Comment, since I am in a rush and have not thought through the details correctly.) This form of the Cauchy integral formula suggests the answer should be yes, if I have not made a mistake. – Yemon Choi Jan 28 '13 at 16:11
Oops - the above result is only for C^1 functions. I will leave the previous comment in case it suggests something more useful – Yemon Choi Jan 28 '13 at 16:13
up vote 11 down vote accepted

The following is too long for a comment. I will suppose here that your set of measure zero is compact.

In this case, Your question is closely related to so-called continuous analytic capacity. Let $K$ be a compact set in the plane, and let $\Omega$ be the complement of $K$ with respect to $\mathbb{C}_\infty$. The continuous analytic capacity $\alpha(K)$ is defined as

$$\alpha(K):= \sup \{ |f'(\infty)| : f \in A(\Omega), \|f\|_{\infty} \leq 1 \},$$

where $A(\Omega)$ is the set of all functions holomorphic in $\Omega$ that extend continuously to $\mathbb{C}_\infty$.

It is not difficult to prove that $$\alpha(K)=0$$ if and only if for every open set $U$ with $K \subseteq U$, every $f$ holomorphic on $U \setminus K$, continuous on $U$, extend analytically to the whole of $U$.

So what you're looking for is a characterization of the compact sets $K$ with $\alpha(K)=0$. This is certainly very difficult, since it took more than a hundred years to solve a similar problem (but with analytic capacity $\gamma$ instead of $\alpha$), the so-called Painlevé's problem.

It is easy to prove however that if $\alpha(K)=0$, then the area of $K$ must be zero. You're asking if this condition is sufficient. Most likely it is not, but I don't have a counterexample right now.. Usually, for problems related to removable sets for holomorphic functions, it is more useful to consider hausdorff measure instead.

Anyway, I suggest you look in Garnett's book "Analytic Capacity and measure". See also the recent book by Dudziak, "Vitushkin's conjecture for removable sets."

EDIT (Answer to the question) Every compact set $K$ with Hausdorff dimension strictly between $1$ and $2$ is a counterexample. An easy way to see this is to use Frostman's lemma. Indeed, using this lemma, there exists a nonzero Borel measure $\mu$ supported on $K$ with growth $\mu(B(z,r)) \leq C r^{1+\epsilon}$. The growth of $\mu$ implies that the Cauchy Transform $$f(z):=\int_{K} \frac{1}{z-\zeta}d\mu(\zeta)$$ is continuous on $\mathbb{C}$ and holomorphic on $\mathbb{C} \setminus K$. Furthermore, $f(z)\rightarrow 0$ as $z \rightarrow \infty$ and so in particular, $f$ is bounded on $\mathbb{C}$. However, $f$ is non-constant because $zf(z) \rightarrow \mu(K) \neq 0$. Therefore $f$ does not extend analytically to all of $\mathbb{C}$, for otherwise it would be constant by Liouville's Theorem.

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I know nothing about this subject. However, if I understand correctly the MSE post linked by Claudio Gorodski above: if $K$ is removable for continuous functions, it is also removable for Hölder $C^\alpha$ functions for every $\alpha>0$, hence its $(1+\alpha)$-dimensional Hausdorff measure is $0$. In other words, a necessary condition is that $K$ has Hausdorff dimension at most $1$. Thus, any compact set of Hausdorff dimension strictly between $1$ and $2$ is a counterexample. – Emil Jeřábek Jan 28 '13 at 17:04
Thanks, I'll check those out! – Matt Jacobs Jan 28 '13 at 17:39
@EmilJeřábek : I didn't know about the removability criterion for the Hölder functions. This is very interesting. – Kalim Jan 28 '13 at 17:44

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