Locality of subharmonicity in the plane and on Riemann surfaces

Under the hypothesis of part 1, $u$ is not identically $-\mathrm{\infty }$ on any connected component $C$ of $\mathrm{\Omega }$. Otherwise, for a point $x\in C$, take a connected open neighbourhood $V\subseteq \mathrm{\Omega }$ of $x$ on which $u$ is subharmonic. Since $V$ is connected and meets $C$, it lies in $C$, so $u\equiv -\mathrm{\infty }$ on $V$, contrary to [F1].

Harmonic comparison. Let $\overline{D(a,r)}\subseteq \mathrm{\Omega }$ be a closed disc and let $h$ be continuous on $\overline{D(a,r)}$, harmonic on $D(a,r)$, with $h\ge u$ on $\mathrm{\partial }D(a,r)$; then $h\ge u$ on $D(a,r)$. Suppose not and put $s:=u-h$ on $\overline{D(a,r)}$; then $s$ is upper semicontinuous (difference of an upper semicontinuous function and a continuous one), $s\le 0$ at every point of $\mathrm{\partial }D(a,r)$, and $s$ is positive somewhere on $D(a,r)$. Put $M:={\sup }_{\overline{D(a,r)}}s>0$. Compactness supplies boundedness and attainment without selecting a sequence: the open sets $\{s<m\}$, for positive integers $m$, cover the closed disc since $s$ has no $+\mathrm{\infty }$ values; a finite subcover bounds $s$ above. For each real $b<M$, the closed set $K_b=\{s\ge b\}$ in the closed disc is nonempty, and these sets have the finite intersection property because a finite list has a largest $b<M$. Their intersection is nonempty: otherwise their open complements cover the compact disc and a finite subcover contradicts that finite intersection property. At an intersection point $z$ one has $s(z)\ge b$ for every $b<M$, hence $s(z)=M$. Since $s\le 0$ on the boundary and $M>0$, this point is interior. The maximum set $Z:=\{z\in D(a,r):s(z)=M\}$ is nonempty and closed in $D(a,r)$ [F4]. It is also open: given $z\in Z$, choose a small open disc $V$ about $z$, contained in $D(a,r)$ and in a neighbourhood on which $u$ is subharmonic; then $s=u+(-h)$ is subharmonic on the domain $V$ [F1, F2] and attains the finite maximum $M$ at the interior point $z$, so $s\equiv M$ on $V$ [F3], a neighbourhood of $z$ contained in $Z$. Hence $Z$ is a nonempty clopen subset of the connected disc $D(a,r)$, so $Z=D(a,r)$ and $s\equiv M>0$ on $D(a,r)$. But for $\zeta \in \mathrm{\partial }D(a,r)$ and any sequence $z_j\in D(a,r)$ with $z_j\to \zeta$ one has ${\mathrm{lim\hspace{0.17em}sup}}_{z\to \zeta }s(z)\ge {\lim }_{j}s(z_j)=M>0$, while $s(\zeta )\le 0$ and $s$ is upper semicontinuous, a contradiction. Therefore $M\le 0$ and $h\ge u$ on $D(a,r)$.

Part 1 follows: $u$ is upper semicontinuous by hypothesis, is not identically $-\mathrm{\infty }$ on any component by step 1.1, and satisfies the harmonic comparison of step 1.2, so the harmonic-majorant characterisation makes $u$ subharmonic on $\mathrm{\Omega }$.

Surface case. Let $W\subseteq X$ and $u$ be as in part 2. First $u$ is upper semicontinuous on $W$: for $x\in W$ choose an open neighbourhood $V$ of $x$ on which $u$ is subharmonic, hence upper semicontinuous, and ${\mathrm{lim\hspace{0.17em}sup}}_{y\to x}u(y)\le u(x)$ computed with $y\in V$ is the limsup over $W$. Next, for every chart $\phi :U\to U^{\mathrm{\prime }}$ of the atlas with $U\cap W\mathrm{\ne }\mathrm{\varnothing }$, the chart expression $u_{\phi }$ on $\phi (U\cap W)$ is upper semicontinuous, being the composition of the upper semicontinuous $u$ with the continuous inverse chart, and every point of $\phi (U\cap W)$ has a connected open neighbourhood on which $u_{\phi }$ is subharmonic: if $V\subseteq W$ is an open neighbourhood of the corresponding point of $U\cap W$ on which $u$ is subharmonic, take an inverse-chart image of a small plane disc contained in $\phi (U\cap V)$, so the resulting neighbourhood is connected and contained in $U\cap V$, so by [F6] the chart expression of $u$ is plane subharmonic on its image. Apply step 2.1 (part 1) separately to each connected component of the open plane set $\phi (U\cap W)$; it makes $u_{\phi }$ subharmonic on each component; as $\phi$ was an arbitrary chart of the atlas, [F6] makes $u$ subharmonic on $W$.

Part 1 is step 2.1 and part 2 is step 3.1, so the lemma holds: every assertion above used only the displayed hypotheses, and no choice principle was invoked.