Let $\Omega$ and $\Omega'$ be two bounded open sets of ${R}^{n}$ with $n\geq2$. Suppose $E$ is a common subset of the boundaries of both $\Omega$ and $\Omega'$.
My question is: if the measure of $E$ with respect to the harmonic measure of $\Omega$ is zero, can we also conclude that the measure of $E$ with respect to the harmonic measure of $\Omega'$ is also zero? ( notice that harmonic measure depends on the open sets).
The answer is no. For instance, let $n=2$, $\hspace{3pt}E=C\times{\{0\}}$ where $C$ is the standard middle- thirds Cantor set [1], $\Omega=[0,1]\times{[0,1]}$ and $\Omega^{'}=B(0,2)\setminus{(C\times{\{0\}})}$.
Since harmonic measure on $\Omega$ is absolutely continuous with respect to arclength and $C$ is a set of measure $0$ with respect to Lebesgue measure on $[0,1]$, $hm(z,E,\Omega)=0$ for all $z\in{\Omega}$.
On the other hand, $C\times{\{0\}}$ is a nonpolar subset of $\mathbb{R}^{2}$ [2] which means that it is eventually hit by a Brownian motion in the plane. In particular, it follows that with positive probability, we will hit $C\times{\{0\}}$ before we hit the boundary of the ball $B(0,2)$. That is, $hm(z,E,\Omega^{'})>0$ for all $z\in{\Omega^{'}}$.
It's not hard to see that an analogous construction exists for all $n>2$. The one last thing I'll point out is that if $\Omega\supseteq{\Omega^{'}}$ (regardless of whether or not $\Omega$ and $\Omega^{'}$ are bounded) the desired result DOES indeed hold, simply because:
$$hm(z,E,\Omega)\geq{hm(z,E,\Omega^{'})} \hspace{10pt}\text{for all}\hspace{10pt}z\in{\Omega^{'}}$$
[1] https://en.wikipedia.org/wiki/Cantor_set#Construction_and_formula_of_the_ternary_set
[2] See example 4.7 in chapter 3 of "Harmonic Measure" by Garnett and Marshall