If $A$ is a normal family, why does $\bar A$ have to be?
I am not sure how to prove this. Formally, given a sequence $\{ f_n \}$ of functions in $\bar A$ we have to show that this has a subsequence which converges uniformly on evwry compact set in the domain. If $\{f_n\}$ has only finite number of functions from $\bar A - A$ the ln this is true, but how can prove the general case?
A basic theorem about normal families says that a family of analytic functions is normal iff the the family is uniformly bounded on each compact subset of the domain. If $K$ is compact and $|f(z)| \leq M$ for all $f \in A$ for all $z \in K$ then $|g(z)| \leq M$ for all $z \in K$ for any function $g$ in the closure of $A$, hence this closure is normal. [Note that closure here is w.r.t. uniform convergence on compact sets. So $g \in \overline {A}$ implies that $f_n \to g$ uniformly on $K$ for some seqeunce $\{f_n\}$ in $A$].