Let $f$ be a holomorphic function on $\mathbb{C}$ that is not identically zero, and consider $\log|f|$ and view this as a morphism to $\mathbb{R} \cup \infty \cup -\infty.$ If $f(z)=0$ we of course let $log|f(z)| = -\infty.$
It is easy to see that this gives an upper semicontinuous function. It seems to me that this function is actually lower semicontinuous, so that it is a continuous function. Is this true? For the definition of semicontinuity, see: https://en.wikipedia.org/wiki/Semi-continuity#Formal_definition
$g(z)=\log|f|$ is clearly continuous on $\{z\in\Bbb C:f(z)\ne0\}$. Since $f$ is holomorphic and not identically $0$, its zeroes are isolated. Then, if $f(z_0)=0$, we have $\lim_{z\to z_0}g(z)=-\infty$. Thus, $g\colon\Bbb C\to\Bbb R\cup\{\pm\infty\}$ is continuous. However, $\lim_{z\to\infty}g(z)$ exists if and only if $f$ is a polynomial.