The limit function of decreasing sequence of subharmonic is also subharmonic

Question

Let $u(z)$ be a continuous function on a domain $D \subset \mathbb{C}$ to $[−\infty, \infty)$. Suppose $u_n(z)$ is a decreasing sequence of subharmonic functions on $D$ such that $u_n(z) \to u(z)$ for all $z \in D$.

Show that $u(z)$ is subharmonic.

Answer 1

Take any $z \in D$ and $r$ small enough so that $D(z,r) \subset \Omega$. Since

$$ u_n(z) \le \int_{\partial D(z,\epsilon)} u_n(\zeta)\,d\sigma(\zeta) $$ (normalized arc length measure) for every $\epsilon < r$, monotone convergence shows that $$ u(z) \le \int_{\partial D(z,\epsilon)} u(\zeta)\,d\sigma(\zeta). $$

By the sub-mean value characterization of subharmonic functions, $u$ is also subharmonic. (Note that you can replace monotone convergence with locally uniform convergence if you like. Pointwise convergence is not enough, though.)