= Solution
Use the <intrinsic boundary of a simply connected domain>, so different approaches to the two sides of a slit remain distinct. The <mapping-out function> $g_K$ extends as a <homeomorphism> from this intrinsic compactification to that of the <complex upper half-plane>. Set $T=T(H)$. The imaginary coordinate of <planar Brownian motion> hits zero in finite time almost surely, and $T$ is no larger than that time. Thus $T<\infty$ and continuity gives a finite Euclidean exit point.
By <conformal invariance of planar Brownian motion>, $g_K(B_t)$ is a <planar Brownian motion> in the <complex upper half-plane>, run with clock
$$
A(t)=\int_0^t|g_K'(B_s)|^2\,ds.
$$
The terminal clock cannot be infinite: that would make the transformed Brownian motion stay in the upper half-plane forever. It cannot stop while the transformed path is in the interior either, since continuity of $g_K^{-1}$ would then put the original exit point inside $H$. Hence the terminal clock is precisely the transformed <Brownian exit time>. The transformed path converges to a real boundary point, and applying the extended inverse proves \b[almost sure convergence to a point of the intrinsic boundary].
Write $g_K(x+iy)=u+iv$, and let $E=g_K(S)\cap\mathbb R$. The point at infinity has zero <harmonic measure>. <Conformal invariance of planar Brownian motion> and the <Poisson kernel for the upper half-plane> give
$$
\mathbb P_{x+iy}(\widehat B_T\in S)
=\int_E\frac{v}{\pi((t-u)^2+v^2)}\,dt.
$$
The <hydrodynamic normalization at infinity> gives $g_K(z)=z+O(1/z)$, so along the specified approach $v/y\to1$ and $u/y\to0$. For each fixed real $t$,
$$
\frac{yv}{(t-u)^2+v^2}\longrightarrow1,
\qquad
0\le\frac{yv}{(t-u)^2+v^2}\le\frac yv.
$$
If $\operatorname{Leb}(E)<\infty$, <dominated convergence> applies because $y/v$ is eventually bounded. If $\operatorname{Leb}(E)=\infty$, <Fatou's lemma> makes the limit infinite. This proves the <harmonic-measure asymptotic at infinity>
$$
\boxed{\lim_{y\to\infty,\ x/y\to0}
\pi y\,\mathbb P_{x+iy}(\widehat B_T\in S)
=\operatorname{Leb}(g_K(S)),}
$$
with the equality understood in the extended nonnegative reals.
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