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We know that a set may be of zero harmonic measure without its Lebesgue measure being zero (see Armitage and Gardiner, classical potential theory, pg 178).

Now, consider the following problem. Let $ V $ be a bounded open set in $ \mathbb{R}^{m} $, $ m\geq 2 $, and $W$ the interior of the closure of $V$. Let $E$ be a subset of $ \partial V\cap W $ ($\partial$ means boundary). We suppose $E$ has the following property: for every $x\in E$ and every $r>0$ such that the ball $B(x,r)$ is relatively compact in $W$, the boundary $\partial B(x,r)$ of $B(x,r)$ contains at least one point of $E$. My question is: can the harmonic measure $\omega(E,x,V)$ of $E$ be zero?

Intuitively (Brownian motion interpretation of harmonic measure) the answer seems negative. (?)

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  • $\begingroup$ Yes. In $R^3$, $V$ is a ball with removed radius, and $E$ is this radius. $\endgroup$ – Alexandre Eremenko Dec 10 '19 at 13:29
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    $\begingroup$ Your assumption does not imply that $E$ is connected. Neither it implies that Lebesgue measure of $E$ is positive. $\endgroup$ – Alexandre Eremenko Dec 10 '19 at 14:00
  • $\begingroup$ Yes you are right. I edited the post. Thanks. $\endgroup$ – M. Rahmat Dec 11 '19 at 4:26
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It is easy to construct such an example in $R^n$ for $n\geq 3$ (see my comment). In $R^2$ let $V$ be a rectangle with removed segments: $$\{ x+iy:|x|<1,0<y<2\}\backslash \left((0,i]\cup\cup_{n=2}^\infty ((1/n,1/n+i]\cup(-1/n,-1/n+i])\right),$$ and $E=(0,i)$.

Similar example will work in $R^n$ if you want the $n-1$ Lebesgue measure positive. In fact you can even have $n$-dimensional Hausdorff measure positive, if you take something widely curved for $E$ instead of a straight thing.

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