= Solution
Put $u(z)=\operatorname{Im}(z-g_A(z))$ on $D_A$. This is a <harmonic function>, tends to zero at infinity, and is bounded. To justify the <boundary> behavior without requiring a <locally connected space> as its <boundary>, let $f_A=g_A^{-1}$. Its expansion at infinity implies that bounded $z$ cannot have $|g_A(z)|\to\infty$. If $z_n$ approaches a finite point of $\partial D_A$ and $g_A(z_n)$ had a subsequential limit inside $\mathbb H$, <continuity> of $f_A$ would force that point to lie inside $D_A$, a contradiction. Thus <boundary degeneration under a mapping-out function> gives $\operatorname{Im}g_A(z_n)\to0$. Consequently $u$ extends continuously to the finite <boundary> with value $\operatorname{Im}z$.
Let $B$ be <planar Brownian motion> started at $z\in D_A$, and $\tau$ its <Brownian exit time> from $D_A$. This time is finite almost surely: it is at most the first time the imaginary coordinate, a one-dimensional <Brownian motion>, hits zero. By the <Itô formula>, $u(B_{t\wedge\tau})$ is a bounded <martingale>. The <optional stopping theorem> and <dominated convergence theorem> therefore give
$$
u(z)=\mathbb E_z[u(B_\tau)]=\mathbb E_z[\operatorname{Im}B_\tau].
$$
Boundedness is important here: directly stopping the unbounded imaginary-coordinate <martingale> would require an unjustified <uniform integrability> assertion.
At $z=iy$, the <Laurent series> gives $u(iy)=a/y+O(y^{-2})$. Hence
$$
\boxed{\operatorname{hcap}(A)=\lim_{y\to\infty}y\,\mathbb E_{iy}[\operatorname{Im}B_\tau]\geq0.}
$$
The sign follows because the exit point lies on $\partial D_A\subseteq\overline{\mathbb H}$. This proves the <Brownian representation of half-plane capacity>.
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