= Solution
A finite original exit time alone does not imply that the integral defining the <conformal Brownian clock> is finite. To establish this, let $\psi=\phi^{-1}$. On the event $\{\widetilde T=\infty\}$, the Brownian path of part (a) lives forever in $D'$, while its inverse clock satisfies
$$
\tau(t)=\int_0^t|\psi'(\widetilde B_r)|^2\,dr<T.
$$
Choose a closed disc $C$ compactly contained in $D'$. Its radius may be decreased so that $|\psi'|^2\geq c>0$ on $C$. <Recurrence of planar Brownian motion>, together with the <Strong Markov property>, gives infinite total <Brownian occupation time> in $C$: return repeatedly to a smaller concentric disc, and use the fixed positive probability of remaining in $C$ for a fixed positive duration. The successive trials imply infinitely many such durations. This is the same mechanism as the <divergence of a positive planar Brownian occupation integral>.
Hence, on any infinite-lifetime Brownian path,
$$
\int_0^\infty|\psi'(\widetilde B_r)|^2\,dr
\geq c\int_0^\infty\mathbf1_C(\widetilde B_r)\,dr=\infty.
$$
This contradicts $\tau(t)<T<\infty$. Thus $\widetilde T$ is finite <almost surely>, and equality in law from part (a) gives \b[<finite exit from a conformal image of a bounded planar domain>]:
$$
\boxed{\mathbb P_{z'}(T'<\infty)=1.}
$$
This asserts almost-sure finiteness, not finiteness of the expected exit time.
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