Hi everyone: Let $ \omega $ be a bounded open set in $ \mathbb{R}^{q} $, $ q\geq 2 $, and $ E $ a subset of the boundary $ \partial\omega $ that has harmonic measure zero in $ \omega $. Let $ V $ be the interior of the closure of $ \omega $. We know that some points of $ E $ can be inside $ V $. If I take an open ball $ B $, with the closure of $ B $ inside V, can I say that $ \partial B\cap E $ has harmonic measure zero in $ B $?



Harmonic measure
No. Let $V$ be the disk $|z|<2$ in the complex plane, $E$ the arc $\{ e^{it}/2:|t|\leq1\}\subset V$.
Now draw a simple arc $\gamma(t),\; 0\leq <1$, $\gamma(0)=1$, $\gamma\backslash\gamma(0)\subset V\backslash E$, and $\gamma$ is spiraling around $E$,
so that the limit set of $\gamma(t)$ as $t\to 1$ equals $E$.
And let $\omega=V\backslash\{\gamma\cup E\}=V\backslash\overline{\gamma}$. Then $V$ is the interior
of the closure of $\omega$. Take $B=\{ z:|z|=1/2\}$. So $E\cap\partial B$
has non zero harmonic measure with respect to $B$. 


On the other hand $E$ has zero harmonic measure with respect to $\omega$,
simply because no point of $E$ is accessible from $\omega$, and the set
of non-accessible boundary points has zero harmonic measure. (This is completely evident if one recall the Brownian motion interpretation of the harmonic measure).


