$0 = \int \lim_{n\rightarrow \infty} f^{n} d
\eta_{m}
 = \eta_{m}(\{a\})$. Hence $ a_{m} \in G$ (otherwise, $\nu_{m}$ is the point mass at $a_{m}$ and hence $\nu_{m} (\overline{G} - \{a_{m}\}) =0$. So we conclude that $\eta_{m}(\overline{G}) = \eta_{m}(\overline{G} -
\{a_{m}\}) + \eta_{m}(\{a_{m}\}) = 0+0=0$, a contradiction).

Let $G_{i_{k_{m}}}$ be the component that contains $a_{m}$ and let $\tilde{v}_{m}$ be the sweep of $v_{m}$ on $\partial G_{i_{k_{m}}}$. Then for each $ g \in A(G)$

$$\begin{aligned}
\int_{\partial G} g(z) d \tilde{v}_{m}   =
 \int_{\partial G_{i_{k_{m}}}} g(z) d \tilde{v}_{m}
= \int_{\overline{G}_{i_{k_{m}}}} g(z) d v_{m}  = \int_{\partial G_{i_{k_{m}}}} g(z) d \omega_{a_{m}},\end{aligned}$$

where $\omega_{a_{m}}$ is the harmonic measure of $G_{i_{k_{m}}}$ evaluated at $a_{m}$. It follows by the uniqueness that $
\tilde{v}_{m}= \omega_{a_{m}}$. Hence, we have $$\begin{aligned}
 \eta|\partial G   \ll (\sum v_{a_{m}})|\partial G
\ll \sum \tilde{v}_{a_{m}}|\partial G
 = \sum \omega_{a_{m}}|\partial G\end{aligned}$$ In particular, we have that $\eta|E \ll \omega_{\Omega_{j}} |E $.

Finally, suppose that $g\in L^{p}(\mu)$ such that $\int f g d \mu =
0$, for $f \in A(K)$, where $\frac{1}{q} + \frac{1}{p} = 1$. Then $g
\perp A(G)$ as well. Hence, $g\mu |E \ll \omega_{\Omega_{j}}|E$. This implies that $(g\mu)_{s}|E = 0$, where $(g\mu)_{s}$ is the singular part of the [*Lebesgue decomposition*]{} of $g\mu$ with respect to $\omega_{\Omega_{j}}|E$. Consequently, 