Plane subharmonic functions are locally integrable

Fix a compact disc $\overline{D(a,R)}\subseteq \mathrm{\Omega }$ and a smaller concentric disc $\overline{D(a,\rho )}$ with $0<\rho <R$. The function $u$ is upper semicontinuous on the compact circle $\mathrm{\partial }D(a,R)$, so [L2] gives a finite upper bound there. By the harmonic-comparison theorem [L1], $u$ is therefore bounded above on $\overline{D(a,\rho )}$ by the harmonic majorant obtained from any continuous boundary majorant on $\mathrm{\partial }D(a,R)$.

The set $A=\{z\in \mathrm{\Omega }:u(z)=-\mathrm{\infty }\}$ has empty interior. Suppose instead that $D(c_0,R)\subseteq A$ for some $R>0$, and fix any point $x$ in the connected component of $\mathrm{\Omega }$ containing $c_0$. By [L3], choose a polygonal path $\gamma$ in that component from $c_0$ to $x$. Because $\gamma ([0,1])$ is compact and lies in the open set $\mathrm{\Omega }$, choose $r>0$ with $3r<R$ and $\overline{D(y,4r)}\subseteq \mathrm{\Omega }$ for every $y\in \gamma ([0,1])$. Subdivide the path by points $c_0,c_1,\dots ,c_N=x$ on $\gamma$ with $\mathrm{\mid }{c}_{j+1}-c_j\mathrm{\mid }<r\mathrm{/}2$ for every $j$. We claim inductively that $D(c_j,3r)\subseteq A$ for all $j$. The case $j=0$ holds because $3r<R$. If $D(c_j,3r)\subseteq A$ and $b\in D(c_j,7r\mathrm{/}2)$, choose $\rho$ with $\max (0,\mathrm{\mid }b-c_j\mathrm{\mid }-3r)<\rho <r\mathrm{/}2$. Then the circle $\mathrm{\mid }z-b\mathrm{\mid }=\rho$ lies in $D(c_j,4r)\subseteq \mathrm{\Omega }$ and meets $D(c_j,3r)$ in an open arc, so $u=-\mathrm{\infty }$ on a set of positive arc-length measure on that circle. By [L2], the circle values are Borel measurable and bounded above, hence the circle average is $-\mathrm{\infty }$. The submean inequality therefore gives $u(b)=-\mathrm{\infty }$, proving $D(c_j,7r\mathrm{/}2)\subseteq A$. Since $\mathrm{\mid }{c}_{j+1}-c_j\mathrm{\mid }<r\mathrm{/}2$, one has $D(c_{j+1},3r)\subseteq D(c_j,7r\mathrm{/}2)\subseteq A$, completing the induction. In particular $x\in A$. Because $x$ was arbitrary in the component, this contradicts the subharmonic convention that $u$ is not identically $-\mathrm{\infty }$ there. Thus $A$ has empty interior, so every open subdisc contains a point where $u$ is finite.

Fix $a\in \mathrm{\Omega }$ and choose $r>0$ with $\overline{D(a,3r)}\subseteq \mathrm{\Omega }$. Step 1.1, applied with outer radius $3r$ and inner radius $2r$, gives a finite upper bound $M$ for $u$ on $D(a,2r)$. By step 1.2 choose $c\in D(a,r\mathrm{/}2)$ with $u(c)>-\mathrm{\infty }$, and then choose $r+\mathrm{\mid }c-a\mathrm{\mid }<s<2r-\mathrm{\mid }c-a\mathrm{\mid }\mathrm{.}$ These inequalities give $D(a,r)\subseteq D(c,s)\subseteq D(a,2r)$.

For every $0<t<s$, the submean inequality at $c$ gives ${\int }_0^{2\pi }(M-u(c+t{e}^{i\theta }))d\theta \le 2\pi (M-u(c))\mathrm{.}$ The integrand is nonnegative and Borel by [L2]. Multiply by $t$, integrate from $0$ to $s$, and apply [L4] to obtain ${\int }_{D(c,s)}(M-u)dA\le \pi s^2(M-u(c))<\mathrm{\infty }\mathrm{.}$ Thus the negative part of $u$ is integrable on $D(c,s)$, while its positive part is bounded there by $M$. Hence $u\in L^1(D(c,s))$, and therefore $u\in L^1(D(a,r))$.

Every point $a\in \mathrm{\Omega }$ admits such a disc $D(a,r)$, so $u\in L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1(\mathrm{\Omega })$.