The Perron family is nonempty and uniformly bounded by the boundary data

The constant function $m$ is harmonic, hence subharmonic, and its boundary limsup equals $m\le \phi$. Therefore $m\in \mathcal{P}(\phi ,\mathrm{\Omega })$ by [L1], so the Perron family is nonempty.

Let $v\in \mathcal{P}(\phi ,\mathrm{\Omega })$ and fix $\epsilon >0$. By the boundary limsup condition in [L1], every boundary point $\zeta$ has a neighbourhood $U_{\zeta }$ such that $v\le M+\epsilon$ on $U_{\zeta }\cap \mathrm{\Omega }$. The boundary is compact because $\mathrm{\Omega }$ is bounded, so finitely many such neighbourhoods cover $\mathrm{\partial }\mathrm{\Omega }$; their union leaves a compact set $K⋐\mathrm{\Omega }$. If $v$ exceeded $M+\epsilon$ somewhere in $\mathrm{\Omega }$, then upper semicontinuity would make $v$ attain its maximum over $K$ at an interior point with value $>M+\epsilon$, contradicting [L2] because $v$ is not constant with that value near the boundary collar. Hence $v\le M+\epsilon$ on $\mathrm{\Omega }$.

Letting $\epsilon \downarrow 0$ in step 1.2 gives $v\le M$ on $\mathrm{\Omega }$ for every $v\in \mathcal{P}(\phi ,\mathrm{\Omega })$. Together with step 1.1, this yields $m\le U_{\phi }\le M$.