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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 necessarysufficient. Most likely it is not, but I don't see have a counterexample nowright 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."

1

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 necessary. Most likely it is not, but I don't see a counterexample 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."