Plane subharmonicity is invariant under biholomorphic change of coordinate

Assume first that $u$ is subharmonic on ${\mathrm{\Omega }}_2$. The map $G$ is continuous and $u$ is upper semicontinuous, so $v=u\circ G$ is upper semicontinuous. If $C$ is a connected component of ${\mathrm{\Omega }}_1$, then $G(C)$ is a nonempty open subset of ${\mathrm{\Omega }}_2$, and [F5] provides a point of $G(C)$ at which $u$ is finite; hence $v$ is not identically $-\mathrm{\infty }$ on $C$.

Let $\overline{D(a,r)}⋐{\mathrm{\Omega }}_1$ be a closed disc and let $h$ be continuous on $\overline{D(a,r)}$, harmonic on $D(a,r)$, with $h\ge v$ on $\mathrm{\partial }D(a,r)$. Put $V:=G(D(a,r))$ and $H:=h\circ G^{-1}$. Since $G$ is a homeomorphism, $V$ is a nonempty connected open subset of ${\mathrm{\Omega }}_2$ with $\bar{V}=G(\overline{D(a,r)})⋐{\mathrm{\Omega }}_2$, so $\bar{V}$ is compact and $\mathrm{\partial }V=G(\mathrm{\partial }D(a,r))$; moreover $H$ is continuous on $\bar{V}$, harmonic on $V$ by [F4] applied to the holomorphic map $G^{-1}$, and $u\le H$ on $\mathrm{\partial }V$ because $h\ge v$ on $\mathrm{\partial }D(a,r)$.

Suppose $u\le H$ failed somewhere in $V$, so $w=u-H$ is positive at some point of $V$. The function $w$ is upper semicontinuous on the compact set $\bar{V}$, takes no $+\mathrm{\infty }$ values, and is positive somewhere. It is bounded above: the relatively open sets $\{w<n\}$ for positive integers $n$ cover $\bar{V}$, so a finite subcover gives a finite upper bound. Let $M:={\sup }_{\bar{V}}w>0$. For every real $b<M$, the closed set $\{w\ge b\}$ is nonempty, and these sets have the finite intersection property; compactness gives a point where $w=M$. Since $w\le 0$ on $\mathrm{\partial }V$ by step 1.2, such a maximum point lies in $V$. Now let $z\in V$ satisfy $w(z)=M$, and choose $\rho >0$ with $\overline{D(z,\rho )}\subseteq V$. The submean inequality for $u$ and the mean-value property for $H$ give $M=w(z)\le \frac{1}{2\pi }{\int }_0^{2\pi }w(z+\rho e^{it})dt\le M,$ where the integral is defined because $u$ is locally integrable [F5]. Thus $w=M$ almost everywhere on $\mathrm{\partial }D(z,\rho )$.

The set $Z:=\{z\in V:w(z)=M\}=\{z\in V:w(z)\ge M\}$ is nonempty and closed in $V$, since $w$ is upper semicontinuous and $w\le M$. It is open as well. If $z\in Z$ and $\overline{D(z,\rho )}\subseteq V$, then for each $0<t<\rho$ step 2.1 gives $w=M$ almost everywhere on $\mathrm{\partial }D(z,t)$; that full-measure subset is dense on the circle, so upper semicontinuity gives $w=M$ at every point of that circle. Every point of $D(z,\rho )$ lies on one of these circles or is $z$, hence $D(z,\rho )\subseteq Z$. Since $V$ is connected, $Z=V$ and $w\equiv M$ on $V$. Choose $b\in \mathrm{\partial }V$, which is nonempty because $V$ is a nonempty bounded domain. A sequence from $V$ tending to $b$ and upper semicontinuity on $\bar{V}$ give $w(b)\ge M>0$, contradicting $w\le 0$ on $\mathrm{\partial }V$ from step 1.2. Therefore $u\le H$ on $V$, that is, $v\le h$ on $D(a,r)$.

By step 1.1 the function $v$ is upper semicontinuous and is not identically $-\mathrm{\infty }$ on any connected component of ${\mathrm{\Omega }}_1$, and by steps 1.2 and 3.1 every harmonic majorant of $v$ on the boundary of a compactly contained closed disc majorizes $v$ on the disc. The equivalence [F2] therefore makes $v=u\circ G$ subharmonic on ${\mathrm{\Omega }}_1$.

Conversely, if $u\circ G$ is subharmonic on ${\mathrm{\Omega }}_1$, apply the implication of step 4.1 to the biholomorphism $G^{-1}:{\mathrm{\Omega }}_2\to {\mathrm{\Omega }}_1$ and the function $u\circ G$: it gives $(u\circ G)\circ G^{-1}=u$ subharmonic on ${\mathrm{\Omega }}_2$. The two implications prove the equivalence.