$K$ is the best possible one. What we mean here is that in order to have Theorem \[t:main\] holds for any positive finite measure supported on $K$, it is necessary and sufficient that each component of $K^{\circ}$ is multi-nicely connected and the harmonic measures of the components of $K^{\circ}$ are mutually singular.

Now we outline a proof for this fact. Let $\Omega$ be a component of $K^{\circ}$ and let $\mu$ be a harmonic measure for $\Omega$. By Theorem 3 of [@comm], the map $f\rightarrow \hat{f}$, from $A^{q}(\overline{\Omega}, \mu) \bigcap L^{\infty}(\mu)$ onto $H^{\infty}(\Omega)$ is surjective if and only if $\Omega$ is a multi-nicely connected domain. Hence we know that $\Omega$ must be multi-nicely connected.

Now let $\Omega_{1}$ and $\Omega_{2}$ be two components of $K^{\circ}$. We want to show that $\omega_{\Omega_{1}}$ and $\omega_{\Omega_{2}}$ are mutually singular. Set $\mu=\omega_{\Omega_{1}} +\omega_{\Omega_{2}}$. Then it is easy to see that $A^{q}(K,\mu)$ is pure. Since the harmonic measure at a given point is a representing measure, it follows by the definition of $abpe$ and the Harnack’s inequality that $ \nabla A^{q}(K,\mu)
\supset \Omega_{1} \cup \Omega_{2}. $ Since each $\Omega_{i}$ has no boundary slit, it follows clearly that $\nabla A^{q}(K,\mu) =
\Omega_{1} \cup \Omega_{2}.$ If Theorem \[t:main\] holds 