A plane subharmonic function with an interior maximum is constant on its component

Let [given] $S=\{z\in \mathrm{\Omega }:u(z)=M\}\mathrm{.}$ Because $u$ is upper semicontinuous, $S$ is closed in $\mathrm{\Omega }$, and it is nonempty because $a\in S$.

Choose $r>0$ with $\overline{D(a,r)}\subseteq \mathrm{\Omega }$. For every $0<\rho <r$, [L1] gives [L1, given] $M=u(a)\le \frac{1}{2\pi }{\int }_0^{2\pi }u(a+\rho e^{it})dt\le M,$ so the average equals $M$. Since the integrand never exceeds $M$, it equals $M$ almost everywhere on the circle $\mathrm{\mid }z-a\mathrm{\mid }=\rho$. If some point of that circle had value $<M$, upper semicontinuity would make the value $<M$ on a short arc, forcing the average below $M$. Hence $u\equiv M$ on every circle $\mathrm{\mid }z-a\mathrm{\mid }=\rho$ with $0<\rho <r$.

Step 1.2 shows that every point of $D(a,r)$ lies in $S$, so $S$ is open in $\mathrm{\Omega }$. Since $\mathrm{\Omega }$ is connected and $S$ is nonempty, closed, and open, one has $S=\mathrm{\Omega }$. Therefore $u\equiv M$ on $\mathrm{\Omega }$.