have $$\hat{\nu}(a) = \lim_{n\rightarrow \infty}
\int \frac{1-f^{n}(z)}{z-a} d\nu = 0$$ Since the set $\{a: \int \frac{d |\nu|} {|z-a|} < \infty\}$ has full area measure in the plane, it follows by a well-known fact (see Theorem \[t:light\] or the comments after it) that $\nu$ is the zero measure off $\overline{K^{\circ}}$.

Finally, since $\mu|\partial \Omega_{j} \ll \omega_{\Omega_{j}}$ for each $j$ (by Lemma \[ABS1\]), the conclusion follows.

The proof of the next lemma can be found in [@b_apr].

\[l:key\] Let $\Omega$ be an open subset whose boundary does not contain $\infty$. Suppose the components of $\Omega$ are multi-nicely connected and the harmonic measures of the components of $\Omega$ are mutually singular. Let $\{\Omega_{j}\}_{j=0}^{\infty}$ be the collection of the components of $\Omega$. If all but finitely many components of $\Omega$ are simply connected, then $A(\Omega)$ is boundedly pointwise dense in $H^{\infty}(\Omega)$.

For a finite positive measure $\mu$, let $A^{\infty}(K,\mu)$ denote the weak-star closure of $A(K)$ in $L^{\infty}(\mu)$.

Let $\Omega$ be a multi-nicely connected domain. Let $W$ be a circular domain that is conformally equivalent to $\Omega$ and let $\phi$ be a conformal map of $W$ onto $\Omega$. Then the boundary value function $\tilde{\phi}$ has a well-defined inverse, $\tilde{\phi}^{-1}$, on $\partial \Omega$. Moreover, $\tilde{\phi}^{-1}\in A^{\infty}
