 Solution

Fix $\epsilon >0$ and choose a boundary neighbourhood $U$ of $z$ on which $|\phi -\phi (z)|<\epsilon$. The positive continuous barrier $w$ has a positive minimum $m$ on the compact set $\partial \Omega \setminus U$. Choosing $A$ large enough makeson all of $\partial \Omega$. The left function is subharmonic and belongs to the Perron family; the right function is superharmonic and dominates every family member by part (ii). ThereforeAs $x\to z$, continuity gives $w(x)\to 0$. Letting $\epsilon \downarrow 0$ proves $\overline{u}(x)\to \phi (z)$. Such a $w$ is a barrier for the Dirichlet problem, and $z$ is a regular boundary point.