= A^{q}(\overline{U}_{n}, \mu | \Delta_{n})$;

2\) each $U_{n}$ is a finitely connected domain that is conformally equivalent to a circular domain $W_{n}$; the connectivity of $U_{n}$ does not exceed the connectivity of the component of $K^{\circ}$ that contains $U_{n}$;

3\) the map $e$, defined by $e(f)=\hat{f}$, is an isometrical isomorphism and a weak-star homeomorphism from $A^{q}(K_{n}, \mu|\Delta_{n}) \bigcap L^{\infty}(\mu|\Delta_{n})$ onto $H^{\infty}(U_{n})$;

4\) $\mu| \partial U_{n} \ll \omega_{U_{n}}$; and if $u_{n}$ is a conformal map from $W_{n}$ onto $U_{n}$, then for each $f\in H^{\infty}(U_{n})$ has nontangential limits on $\partial U_{n}$ [*a.e.*]{} $[\mu] $ and $$e^{-1}(f)(a) = \lim_{z\rightarrow u_{n}^{-1}(a)} f\circ u_{n}(z) \hspace{.05in}
\mbox{\em{a.e.} on} \hspace{.05in}\partial U_{n}
\hspace{.05in}\mbox{ with respect to }\hspace{.05in}  \mu|\partial
U_{n};$$

5\) for each $f\in H^{\infty}(U_{n})$, if let $f^{*}$ be equal to its nontangential limit values on $\partial U_{n}$ and let $f^{*} = \hat{f}$ on $U_{n}$, then the map $m$, defined by $m(f) =
f^{*}|\Delta_{n}$, is the inverse of the map $e$.

Thomson proved 1), 2) and 3) of Theorem \[t:main\] in the case when $A^{q}(K,\mu)=P^{q}(\mu)$. For the polynomial case, 4) is the main result in [@oy]. The author proved 4) for $A^{q}(K,\mu)$ with a different method in [@comm].

3\) clearly implies that each $A^{q}(K,\mu_{n})$ is irreducible.

The condition on 