= Solution
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 $ns^{n-1}/r^n$ 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 identity>
$$
g_D(z)=\mathbb E_zS+\mathbb E_zg_D(W_S)
=\frac{r^2}{n+2}+\frac1{|B(z,r)|}\int_{B(z,r)}g_D(w)\,dw.
$$
This 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>, \b[$g_D(y)<\infty$ 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>.
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