A boundary point is regular exactly when it admits a barrier

Assume first that $b$ is a barrier at $\zeta$, and let $\phi \in C(\mathrm{\partial }\mathrm{\Omega })$. Put $H=H_{\phi }$, which is harmonic by [L1]. Fix $\epsilon >0$. Choose a boundary neighbourhood $V$ of $\zeta$ with $\mathrm{\mid }\phi (\eta )-\phi (\zeta )\mathrm{\mid }<\epsilon$ on $V\cap \mathrm{\partial }\mathrm{\Omega }$, and choose $A>0$ so large that the negative boundary bound from [L3] forces both [L3, given, choose] $\phi (\eta )-\phi (\zeta )-\epsilon +A\mathrm{lim\hspace{0.17em}sup}b(\eta )\le 0,-\phi (\eta )+\phi (\zeta )-\epsilon +A\mathrm{lim\hspace{0.17em}sup}b(\eta )\le 0$ for $\eta \in \mathrm{\partial }\mathrm{\Omega }\setminus V$.

Assume conversely that $\zeta$ is regular, and define a continuous boundary datum on $\mathrm{\partial }\mathrm{\Omega }$ by $\psi (\eta )=-\mathrm{\mid }\eta -\zeta {\mathrm{\mid }}^2$. Let $B=H_{\psi }$, which is harmonic on $\mathrm{\Omega }$ by [L1]. Regularity gives $B(z)\to \psi (\zeta )=0$ as $z\to \zeta$. Now let $v$ be any member of the Perron family for $\psi$. By [L4], the function $q(z)=\mathrm{\mid }z-\zeta {\mathrm{\mid }}^2$ is subharmonic on $\mathrm{\Omega }$, so $v+q$ is subharmonic there. For every boundary point $\eta \in \mathrm{\partial }\mathrm{\Omega }$, the defining Perron inequality gives ${\mathrm{lim\hspace{0.17em}sup}}_{\begin{array}{c}z\to \eta \\ z\in \mathrm{\Omega }\end{array}}(v(z)+q(z))\le \psi (\eta )+\mathrm{\mid }\eta -\zeta {\mathrm{\mid }}^2=0.$ If $v+q$ were positive somewhere in $\mathrm{\Omega }$, then upper semicontinuity and boundedness of $\mathrm{\Omega }$ would produce a positive interior maximum, contradicting [L2]. Hence $v(z)\le -\mathrm{\mid }z-\zeta {\mathrm{\mid }}^2$ on $\mathrm{\Omega }$ for every lower function $v$. Taking the supremum over the Perron family and then upper-semicontinuous regularizing yields $B(z)\le -\mathrm{\mid }z-\zeta {\mathrm{\mid }}^2<0(z\in \mathrm{\Omega })\mathrm{.}$ Now let $V$ be any neighbourhood of $\zeta$. The compact set $\mathrm{\partial }\mathrm{\Omega }\setminus V$ has ${\delta }_V:={\min }_{\eta \in \mathrm{\partial }\mathrm{\Omega }\setminus V}\mathrm{\mid }\eta -\zeta {\mathrm{\mid }}^2>0,$ so the displayed inequality gives ${\mathrm{lim\hspace{0.17em}sup}}_{\begin{array}{c}z\to \eta \\ z\in \mathrm{\Omega }\end{array}}B(z)\le -{\delta }_V(\eta \in \mathrm{\partial }\mathrm{\Omega }\setminus V)\mathrm{.}$ Thus $B$ is negative on $\mathrm{\Omega }$, tends to $0$ at $\zeta$, and stays uniformly below a negative constant away from $\zeta$. Hence $B$ is a barrier at $\zeta$.

The functions [L2, step 1.1] $s_{+}(z)=H(z)-\phi (\zeta )-\epsilon +Ab(z),s_{-}(z)=-H(z)+\phi (\zeta )-\epsilon +Ab(z)$ are subharmonic on $\mathrm{\Omega }$ because $H$ and $-H$ are harmonic and $b$ is subharmonic. Step 1.1 shows that both have boundary limsup at most $0$. If either had a positive value in the interior, upper semicontinuity would produce a positive interior maximum, contradicting [L2]. Hence $s_{\pm }\le 0$ on $\mathrm{\Omega }$.

Step 2.1 gives [step 2.1, L3] $\phi (\zeta )-\epsilon +Ab(z)\le H(z)\le \phi (\zeta )+\epsilon -Ab(z)\mathrm{.}$ Letting $z\to \zeta$ inside $\mathrm{\Omega }$ and using $b(z)\to 0$ gives $\phi (\zeta )-\epsilon \le {\mathrm{lim\hspace{0.17em}inf}}_{\mathrm{\Omega }\ni z\to \zeta }H(z)\le {\mathrm{lim\hspace{0.17em}sup}}_{\mathrm{\Omega }\ni z\to \zeta }H(z)\le \phi (\zeta )+\epsilon \mathrm{.}$ Since $\epsilon$ is arbitrary, $H(z)\to \phi (\zeta )$. Thus $\zeta$ is regular.

Steps 1.1 through 4.1 prove both directions, so $\zeta$ is regular exactly when it admits a barrier.