$g$ is Lebesgue Integrable Implies $g < \infty$ a.e. (Papa Rudin 1.41)

Question

In the last sentence of the proof of 1.41, Rudin RCA, he says that if $g \in L^1(\mu)$ then $g < \infty$ a.e. (in this case $g$ is a function into $[0, \infty]$). I understand how this works intuitively, the sets at which $g$ is equal to $\infty$ are not contributing to the integral and thus the integral remains finite. But I'm having trouble showing this.