for $A^{q}(K,\mu)$, then $$A^{q}(K,\mu) = A^{q}(\overline{\Omega}_{1},\mu_{1})  \oplus
A^{q}(\overline{\Omega}_{2},\mu_{2}),$$ where $\mu_{i}$, $i=1, 2$, are as in Theorem \[t:main\]. Let $v$ be a conformal map of $\Omega_{1}$ onto a circular domain $W$. Theorem \[t:main\] implies that there exists $v_{e}\in
A^{q}(\overline{\Omega}_{1}, \mu_{1})$ such that $\hat{v}_{e} = v$. Set $\eta = \mu_{1} \circ v_{e}^{-1}$. According to Lemma 2 in [@comm], $\eta$ is a measure on $\partial W$ such that $A^{q}(\overline {W},\eta)$ is irreducible and $\nabla A^{q}(\overline{W},\eta) = W$. Moreover, $\eta \ll
\omega_{W}$ by Lemma 3 of [@comm]. On the other hand, since $W$ is circular and since $\nabla A^{q}(\overline{W},\eta) = W$, it is easy to see that $A^{\infty}(\overline{W}, \eta)$, the weak-star closure of $A(\overline{W})$ in $L^{\infty}(\eta)$, is equal to $\widetilde H^{\infty}(W)$, which is the image of the map $f\rightarrow \tilde f$ from $H^{\infty}(W)$ into $L^{q}(\omega_{W})$ (where $\tilde f$ is the boundary value function of $f$ on $\partial W$). Since the support of $\mu_{1}\subset \partial \Omega_{1}$, it follows by a classical result that \[$\omega_{W}] = [\eta]$. Now, applying Lemma 3 in [@comm], we conclude that $[\omega_{\Omega_{1}}]=[\mu_{1}]$. Similarly, we have $[\mu_{2}]=[\omega_{\Omega_{2}}]$. But $\mu_{1}$ and $\mu_{2}$ are mutually singular, therefore $ \omega_{\Omega_{1}}$ and $
\omega_{\Omega_{2}}$ must be mutually singular.

The proof of the main result
============================

For 