converges to $r_{n}$ in $L^{\infty}(\omega)$. This proves the claim.

Next we show that $r_{n}$ belongs to the weak-star closure of $A(\overline{K^{\circ}})$ in $L^{\infty}(\mu)$. By Proposition \[p:ABS2\], $\mu|\partial K \ll \omega$. Thus $$\lim_{m\rightarrow} \int_{\partial K} f q_{m} \hspace{.05in}d\mu = \int_{\partial K}
f r_{n} d \mu, \hspace{.05in}f \in L^{1}(\mu).$$ Since $\{q_{m}\}$ is bounded and pointwise converges to $r_{n}$ on $K^{\circ}$, it follows by the bounded convergence theorem that $$\lim_{n\rightarrow } \int fq_{m} \hspace{.05in}d\mu = \int f r_{n} \hspace{.05in}d\mu, \hspace{.05in}
f \in L^{1}(\mu).$$ Therefore, $r_{n}$ belongs the weak-star closure of $A(\overline{K^{\circ}})$ in $L^{\infty}(\mu)$. But this closure is contained in the weak closure of $A(\overline{K^{\circ}})$ in $L^{q}(\mu)$. Because a convex set is norm-closed if and only if it is weakly closed in $L^{q}(\mu)$, we have $r_{n}$ is in $A^{q}(\overline{K^{\circ}},\mu)$.

Now for each $n \geq 1$, choose $x_{n} $ in $A(\overline{K^{\circ}})$ such that $$\|r_{n}- x_{n} \|_{A^{q}(\overline{K^{\circ}},\mu)}| \leq \frac{1}{n}.
$$ Then $$\begin{aligned}
\|x_{n}-h\|_{L^{q}(\mu)} & = \|x_{n}-r_{n}\|_{L^{q}(\mu)} +
 \|r_{n}-h\|_{L^{q}(\mu)}  \\ &
\leq \frac{1}{n} + \|r_{n}-h\|_{L^{q}(\tau)} \rightarrow 0,
\hspace{.07in} \mbox{as}\hspace{.05in}\rightarrow \infty.\end{aligned}$$ Thus we have that $h\in A^{q}(\overline{K^{\circ}},\mu)$. Because $h$ is an arbitrary element in $A(\overline{\Omega},\tau)$, we conclude that $A^{q}(\overline{\Omega},\tau) \subset
A^{q}(\overline{K^{\circ}},\mu).$ Since $A^{q}(K,\mu)$ is pure and since $\mu$ is supported on $\overline{K^{\circ}}$ (for $\mu|\partial K \ll \omega$ and 