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For a connected open set $D$, finiteness of the Brownian exit time expectation at one point implies finiteness everywhere. The Brownian exit-time ball averaging identity gives integrability on a ball centred at any finite point. Every point of that ball has a smaller centred ball contained in it; the same identity then gives a finite value there. The set of finite points is open. It is also relatively closed: near a limit point in $D$, choose a fixed ball radius smaller than a quarter of its distance to $D^c$ and centre it at a nearby finite point. The limit point lies in that ball and hence is finite. Connectedness completes the proof.

Allow $g_D$ to take the value $+\infty$. Fix a closed ball $\overline{B(z,r)}\subset D$, independently choose a radius $R$ with density $n{s}^{n-1}/r^n$ on $(0,r)$, and let $S$ be the first time the Brownian motion from $z$ reaches that radius. Conditional on $R=s$, the Strong Markov property at $S$ decomposes the remaining Brownian exit time from $D$. Since $S<T_D$ and the conditional exit time has mean $s^2/n$, Tonelli's theorem and part (a) give the Brownian exit-time ball averaging identityThis identity is valid in the extended nonnegative reals; no finiteness was assumed to derive it.

If $g_D(z)<\infty$, the integral over $B(z,r)$ is finite. For any $y\in B(z,r)$ choose $\rho >0$ with $\overline{B(y,\rho )}\subset B(z,r)$. The same identity at $y$ has a finite right side because its integral is over a subset of the integrable ball. Thus $g_D(y)<\infty$. The set $F=\{z\in D:g_D(z)<\infty \}$ is therefore open relative to $D$.

It is also relatively closed. If $z_j\in F$ converges to $y\in D$, choose $r>0$ with $\overline{B(y,3r)}\subset D$. For large $j$, $|z_j-y|<r$, and $\overline{B(z_j,2r)}\subset D$. The preceding argument shows that every point in $B(z_j,2r)$, including $y$, belongs to $F$. Hence $F$ is closed in $D$. Since $F$ is nonempty and $D$ is connected, $g_D(y)<\infty$ for every $y\in D$. This proves finiteness propagation of Brownian mean exit times without assuming beforehand that $g_D$ solves a smooth Poisson equation.