Subharmonic pieces glue across a boundary under the limsup inequality

On $D$, the function $w=\max \{u,v\}$ is subharmonic by [L2]. Away from $\mathrm{\partial }D$ its upper semicontinuity is therefore clear. At $\zeta \in \mathrm{\partial }D\cap \mathrm{\Omega }$, the inside limsup is at most $u(\zeta )$ by the hypothesis and upper semicontinuity of $u$, while the outside limsup is at most $u(\zeta )=w(\zeta )$. Thus $w$ is upper semicontinuous on $\mathrm{\Omega }$.

Let $\bar{B}\subseteq \mathrm{\Omega }$ be a closed disc and let $h$ be continuous on $\bar{B}$, harmonic on $B$, and satisfy $h\ge w$ on $\mathrm{\partial }B$. Because $w\ge u$ there, [L1] first gives $h\ge u$ throughout $B$.

Let $C$ be a connected component of $B\cap D$. On $\mathrm{\partial }C\cap \mathrm{\partial }B$ one has $h\ge w\ge v$. At a boundary point of $C$ inside $B$, the seam hypothesis and step 1.2 give ${\mathrm{lim\hspace{0.17em}sup}}_C(v-h)\le u-h\le 0$. Thus the subharmonic function $v-h$ has boundary limsup at most $0$ on the bounded domain $C$. If it were positive somewhere, upper semicontinuity and the boundary bound would make it attain a positive interior maximum, contradicting [L3]. Hence $h\ge v$ on every such component.

Steps 1.2 and 2.1 give $h\ge u$ on $B$ and $h\ge v$ on $B\cap D$, hence $h\ge w$ throughout $B$. Every harmonic boundary majorant therefore majorizes $w$, so [L1] and step 1.1 make $w$ subharmonic on $\mathrm{\Omega }$.