In complex analysis, a locally rectifiable curve $\Gamma$ is said to be Ahlfors regular when it satisfies the following geometric condition: There exists a constant $C>0$ such that $$\ell(\Gamma\cap B)\leq C\cdot r,$$ for every disk $B$ of radius $r>0$ centered at $\Gamma$.
Let $D$ be a simply connected domain and $\Gamma\subset\partial D$ be an Ahlfors regular arc. Consider the harmonic measure $\omega(z,\Gamma,D)$ of $\Gamma$ at $z\in D$. This is a harmonic function in the $z$ variable bounded by $1$ from above and by $0$ from below. Moreover, if $z_0\in\Gamma\setminus\lbrace\mathrm{endpoints}\rbrace$, $\lim_{z\to z_0}u(z)=u(z_0)=1$, when $z$ approaches $z_0$ inside $D$ and non-tangentially, and the same limit is $0$ if $z\in\Gamma\setminus\lbrace\mathrm{endpoints}\rbrace$.
Let $\tau\in(0,1)$ and consider the level set $A_{\tau}=\lbrace z\in D:\omega(z,\Gamma)=\tau\rbrace$. $A_\tau$ is an analytic curve lying in $D$ and its endpoints meet the endpoints of $\Gamma$.
Do you know, whether or not, such a level curve $A_{\tau}$ is Ahlfors regular or not?
This is what I have tried.
Using the Riemann mapping theorem we can map the interior of $A_{\tau}\cup\Gamma$ to the unit disk with $A_{\tau}$ being mapped to an arc $\gamma_1$ in a smooth manner. Further, we can map the unit disk to the upper half plane conformally with $\gamma_1$ being mapped to $[0,1]$ smoothly. If we denote the first Riemann map with $\Psi$ and the second Riemann map with $\psi$, then $A_{\tau}=\Psi^{-1}\circ\psi^{-1}([0,1])$ and I think this has nice geometric properties.