Suppose $W$ is a bounded open subset of $\mathbb{R}^{n}$ and $n\geq2$. Let $V$ be the interior of the closure of $W$ and $E$ a subset of the boundary of $V$. If $\omega(x,W)(E)=0$ ($\omega(x,W)$ is the harmonic measure of $W$ at $x\in W$), can we conclude that $\omega(x,V)(E)=0?$



A question about harmonic measure 2
Not really. Take one of the usual counter-examples in potential theory: $$W = (0,1) \times (0,1) \setminus \biggl(\bigcup\_{n=2}^\infty\{\tfrac{1}{n}\} \times (\tfrac{1}{n}, 1-\tfrac{1}{n})\biggr).$$ Then $V = (0,1) \times (0,1)$. If $E = \{0\} \times (0, 1)$, then $\omega(x, W)(E) = 0$, but of course $\omega(x, V)(E) > 0$.


