A decreasing limit of plane subharmonic functions is subharmonic or identically -infinity

A decreasing limit of upper semicontinuous functions is upper semicontinuous, so $u$ is upper semicontinuous on $\mathrm{\Omega }$. If $u\equiv -\mathrm{\infty }$, the first alternative of the statement holds and there is nothing more to prove. Assume from now on that $u$ is finite at least at one point.

Fix a closed disc $\overline{D(a,r)}\subseteq \mathrm{\Omega }$. For every $n$, [L1] gives [L1, L2, L3, choose] $u_n(a)\le \frac{1}{2\pi }{\int }_0^{2\pi }{u}_n(a+r{e}^{it})dt\mathrm{.}$ By [L2], the boundary functions are measurable and bounded above. Choose a constant $M$ larger than $u_1$ on the circle. Then $M-u_n(a+r{e}^{it})$ is an increasing sequence of nonnegative measurable functions of $t$, so [L3] yields ${\lim }_{n\to \mathrm{\infty }}\frac{1}{2\pi }{\int }_0^{2\pi }{u}_n(a+r{e}^{it})dt=\frac{1}{2\pi }{\int }_0^{2\pi }u(a+r{e}^{it})dt\mathrm{.}$

Passing to the limit in the inequalities of step 1.2 gives [step 1.1, step 1.2, L1] $u(a)\le \frac{1}{2\pi }{\int }_0^{2\pi }u(a+r{e}^{it})dt\mathrm{.}$ Since the disc was arbitrary and step 1.1 supplied upper semicontinuity, [L1] makes $u$ subharmonic.