I have read that this theorem would be untrue if subharmonic were continuous instead of just upper semicontinuous. Why is that?
Theorem Let $(u_n)_{n \geq 1}$ be subharmonic functions on an open set $U$ in $\mathbb{C}$, and suppose $u_1 \geq u_2 \geq \cdots$ on $U$. Then $u:=\lim_{n \to \infty}u_n$ is subharmonic on U.
Thanks!
Because this limit is non necessarily continuous, even when $u_n$ are continuous subharmonic functions. For example, let $U$ be the unit disk, and $$u_n(z)=\sum_{k=2}^n\frac{1}{k^2}\log|z-1/k|.$$ The limit $u$ exists and is finite at $0$. But it equals $-\infty$ at all points $1/k$.
If you do not want to call $\log|z|$ continuous, replace it by by $\max\{\log|z|,-M\}$, where $M>0$ is a large number (larger than $|u(0)|$)