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Suppose an unbounded domain contains the upper strip $(-1,1)\times (1,\infty )$ with its two vertical sides on the boundary. A bounded harmonic function with constant values $a$ on the right side and $b$ on the left side, and with valid bounded harmonic representation with almost sure Brownian exit, satisfies $u(x,y)\to a(1+x)/2+b(1-x)/2$ as $y\to \infty$. The horizontal Brownian exit time is finite; on that finite random interval the vertical Brownian path has a finite minimum. Consequently reaching the bottom before a side has probability tending to zero. The remaining two exit probabilities are the one-dimensional interval probabilities.

The domain is the union of the horizontal and vertical strips. The four connected components of its boundary carry the constants $a,b,c,d$ in the northeast, northwest, southwest and southeast, respectively.

First justify finiteness of $T$. A small open ball around $(2,2)$ lies outside the domain. By part (b), planar Brownian motion visits that ball almost surely in finite time. A continuous path from $z\in D$ to that ball must meet $\partial D$ first. Hence $T<\infty$ almost surely, and $B_T\in \partial D$ by continuity.

The harmonic function $u$ is smooth inside $D$. The Itô formula, localized on relatively compact subdomains, makes $u(B_{t\wedge T})$ a local martingale. Its boundary definition uses the continuous extension of $u$. Since $u$ is bounded, this stopped process is a true martingale with uniform integrability: pass through the compact exhaustion using bounded convergence theorem. In particular ${\mathbb{E}}_{z}u(B_{t\wedge T})=u(z)$.

Now $B_{t\wedge T}\to B_T$ almost surely as $t\to \infty$. The dominated convergence theorem, boundedness of $u$, and its continuous boundary values give the bounded harmonic representation with almost sure Brownian exit:Almost sure exit and boundedness of the harmonic function are both used. The domain's unboundedness causes no loss of expectation here because the stopped function values are uniformly bounded.