each $f \in A^{q}(K,\mu),\hspace{.03in} \hat{f}=f$ on $\nabla
A^{q}(K,\mu)$ a.e. $[\mu]$.

Let $ a$ be an $abpe$ and choose a sequence $\{f_{n}\} $ in $A(K)$ such that $f_{n}\rightarrow f $ in $L^{q}(\mu)$. Since $f_{n}\rightarrow \hat{f}$ uniformly in a neighborhood of $a$, it follows (by passing a sequence if necessary) that $f_{n}\rightarrow
f$ [*a.e.*]{} $[\mu]$ and consequently $$\hat{f}(a) = \lim_{n\rightarrow \infty} \hat{f}_{n}
= \lim_{n\rightarrow \infty} f_{n} = f(a) \hspace{.07in}a.e.
\hspace{.07in} [\mu].$$

The following lemma is elementary too.

\[l:elem1\] If $f\in L^{\infty}(\mu) \cap A^{q}(K,\mu)$ and $g\in A^{q}(K,\mu)$, then $fg \in A^{q}(K,\mu)$ and $\widehat{fg} = \hat{f}\hat{g}$.

The next lemma is proved in [@comm].

\[l:ISO\] Let $\Omega = \nabla A^{q}(K,\mu)$. If $\Omega$ is finitely connected, then every component of ([**C**]{} $\setminus \Omega$) has nonempty interior.

[*Representing measures*]{}. Let $\cal B$ be a closed subalgebra of $C(K)$. A complex representing measure of $\cal B$ for $a\in K$ is a finite measure $\nu$ on $K$ such that $$f(a) = \int f d\nu, \hspace{.05in} f\in \cal B.$$ A [*representing measure*]{} for $a$ is a probability measure that satisfies the above condition. Note, if $a$ is a peak point, then the only representing measure for $a$ is the point mass $\delta_{a}$.

[*The sweep of a measure*]{}. Let $G$ be 