The upper-semicontinuous regularization of a locally bounded-above subharmonic supremum is subharmonic

For every compact set $K\subseteq \mathrm{\Omega }$, the hypothesis gives a real number $M_K$ with $u\le M_K$ on $K$, so both $u$ and $U=u^{\ast }$ are locally bounded above and never take the value $+\mathrm{\infty }$. Because $\mathcal{F}$ is nonempty and every $v\in \mathcal{F}$ satisfies $v\le u\le U$, the function $U$ is not identically $-\mathrm{\infty }$ on any connected component. By [L3], $U$ is upper semicontinuous and satisfies $u\le U$.

Let $\overline{D(a,r)}\subseteq \mathrm{\Omega }$ and let $h$ be continuous on the closure, harmonic on the disc, and satisfy $h\ge U$ on $\mathrm{\partial }D(a,r)$. Because $u\le U$, one also has $h\ge u$ on the boundary.

Fix any $v\in \mathcal{F}$. Since $h\ge v$ on $\mathrm{\partial }D(a,r)$, [L2] gives $h\ge v$ on $D(a,r)$. The same is therefore true for every finite maximum of members of $\mathcal{F}$, and [L1] keeps those maxima subharmonic. Taking the supremum over all $v\in \mathcal{F}$ yields $h\ge u$ on $D(a,r)$.

Since $h$ is continuous and dominates $u$, it also dominates the least upper-semicontinuous majorant $U$ by [L3]. Thus $h\ge U$ on $D(a,r)$. Another use of [L2] shows that $U$ is subharmonic on $\mathrm{\Omega }$.