Let $D \subset \Bbb C$ be a bounded domain, $f,g:\bar D \to \Bbb R$ smooth subharmonic, $D_1, D_2,...$ domains with analytic boundary such that $D = \cup D_i$ and $\bar D_i \subset D_{i+1}$.
Let $u_m$ be the the harmonic function on $D_m$ with boundary values given by the function $h=f-g$, and let $\tilde h=\sup\{f \text{ subharmonic on }D_m, \forall \zeta \in \partial D: \limsup_{z\to\zeta} f(z) \leq h(\zeta) \}$ be the Perron solution of the Laplace equation.
Prove that $\tilde u_m \to \tilde h$ uniformly on every compact subset $E$ of $D$.
This is a theorem of Wiener from the 1920s in a more general setting, where $h$ is a general continuous function. Here, however, we assume $h$ is smooth, so the argument should be somehow simpler.
It is not clear to me what we gain from expressing $h$ as a difference of two smooth sub-harmonic functions. By compactness $\exists M: E \subset D_M$. We do not assume that $D$ has analytic boundary, so the Perron solution $\tilde h$ might not attain the correct boundary values $h$ on $\partial D$. How can I prove it then?