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Let $V \subseteq \mathbb C$ be an open connected set , $\{f_n\}$ be a normal family on $V$ such that $\{z \in V : \lim _{n\to \infty} f_n(z)$ exists $\}$ has a limit point in $V$ . Then is it true that there is a function $f:V \to \mathbb C$ such that $\{f_n\}$ converges uniformly to $f$ on each compact subset of $V$ ?

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  • $\begingroup$ Yes. It's Vitali's theorem. $\endgroup$ – ts375_zk26 Apr 11 '17 at 23:58
  • $\begingroup$ @ts375_zk26 : can you please provide a proof of it , or some reference with a proof of it ? Thanks in advance $\endgroup$ – user228168 Apr 12 '17 at 3:47
  • $\begingroup$ Sorry. Typo in my answer is fixed. $\endgroup$ – ts375_zk26 Apr 12 '17 at 23:25
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Since $\{f_n\}$ is a normal family, every subsequence $\{f_{n_k}\}$ has a subsequence which converges uniformly to some analytic function $f$ on each compact subset of $V$.
Suppose that another subsequence converges to $g$. Then $f(z)=g(z)$ for $z \in E$, $E=\{z\in V:\lim_{n\to \infty} f_n(z) \text{ exists}\}$, implies $f=g$ by uniqueness principle. In other words, every converging subsequence converges to the same $f$.

Suppose that ${f_n}$ does not converge uniformly to $f$ on each compact subset. Then \begin{align} \exists \varepsilon >0, \exists \text{ compact }\Delta \subset V, \exists \{z_k\}, &z_k\in \Delta ,\exists \{f_{n_k}\} \\ &\text{ such that } |f_{n_k}(z_k)-f(z_k)|\ge \varepsilon \quad (k=1,2,...).\tag{1}\end{align} However, since the family is normal, $\{f_{n_k}\}$ has a subsequence $\{f_{n_k^\prime}\}$ which converges to $f$ uniformly on $\Delta $. This contradicts $(1)$.

[Typo fixed]

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