 Solution

Interpret $f$ as a conformal bijection onto $D^{\prime }$. If it is merely an injective map into a larger target, the correct exit domain is $f(D)$ instead. Let $\zeta$ be the exit time from $D$, write $f=u+iv$, and identify planar Brownian motion with $B^1+i{B}^2$. The Cauchy-Riemann equations give $u_x=v_y$, $u_y=-v_x$, and both $u,v$ are harmonic functions. The Itô formula, localized on compact subsets of $D$, yieldsThus both coordinates are continuous local martingales, withSince $f^{\prime }$ never vanishes, $A$ is strictly increasing before $\zeta$. Let ${\eta }_s=\inf \{t:A_t>s\}$, defined for $s<A_{\zeta -}$. The optional time-change theorem gives continuous local martingale coordinates for ${\beta }_s=f(B_{{\eta }_s})$, with bracket matrix $sI$. The Lévy characterization of multidimensional Brownian motion makes $\beta$ a standard planar Brownian motion started at $f(z)$, up to its lifetime. Therefore conformal invariance of planar Brownian motion takes the formTo identify the lifetime, exhaust $D$ by relatively compact open sets $D_n$ with nested closures. The images $f(D_n)$ exhaust $D^{\prime }$ because $f$ is a homeomorphism onto $D^{\prime }$. Each stopped transformed path exits $f(D_n)$ at precisely the clock value of its exit from $D_n$. Increasing these exit times gives the exit lifetime of Brownian motion from $D^{\prime }$, including a possibly infinite lifetime. This identifies $A_{\zeta -}$ with that lifetime and needs no global extension of $f$ to the boundary.