we have $g\perp
\chi_{\Delta\cap E}$, where $\Delta$ is the carrier of $\mu_{s}|\partial \Omega_{j}$ and $\mu_{s}$ is the singular part of the [*Lebesgue decomposition* ]{} of $\mu$ with respect to $\omega_{\Omega_{j}}$. Now an application of [*Hahn-Banach theorem*]{} yields $\chi_{\Delta\cap E} \in A^{q}(K,\mu)$. By purity, $\chi_{\Delta\cap E} = 0$ a.e. $[\mu]$ and therefore $\mu|E \ll
\omega_{\Omega_{j}} |E $. Since $E$ is an arbitrary component of $\partial \Omega_{j}$, it follows that $\mu|\partial \Omega_{j} \ll
\omega_{\Omega_{j} }$.

\[p:ABS2\] Let $K$ be a compact subset such that the components of $K^{\circ}$ are multi-nicely connected and the harmonic measures of $K^{\circ}$ are mutually singular. Let $\{\Omega_{j}\}_{j=0}^{\infty}$ denote the collection of the components of $K^{\circ}$. Set $\omega =
\sum_{j=0}^{\infty} \frac{1}{2^{j}} \omega_{\Omega_{j}}$. If $A^{q}(K,\mu)$ is pure, then $\mu|\partial K \ll\omega$.

By Lemma 17.10 in [@conw p. 246], there exists a function $g\in A^{q}(K,\mu)^{\perp}$ such that $|f|\mu \ll |g|\mu$ for each $f\in A^{q}(K,\mu)^{\perp}$. Since $A^{q}(K,\mu)$ is pure, we see that $g \neq 0$ on $supp(\mu)$ [*a.e.*]{} $[\mu]$. Thus $[|g|\mu] = [\mu]$ [*a.e.*]{} $[\mu]$. Set $\nu = |g|\mu$.

Let $a\in K -\overline{K^{\circ}}$ be such that $\int \frac{d |\nu|} {|z-a|} < \infty$ and let $f$ be a peak function for $a$. For each integer $n\geq 1$, $\frac{1-f^{n}(z)}{z-a} \in A(K)$, so we 