Suppose $f$ is a non-negative, Lebesgue-measurable function on $X$, such that $\int f d\mu < \infty$. Is it true then that $\mu A = \mu(\{x \in X : f(x) = \infty\}) = 0$?
My idea is to let $A_n = \{x \in X : f(x) \geq n\}$, and then take the intersection of the $A_n$ and show that it has measure zero, but I'm not sure exactly about the details.
Yes, you have the idea. For every $n$, $A\subset A_n$, and $$\mu(A_n) = \int 1_{A_n}\, d\mu \le \int \frac{f}{n}1_{A_n}\, d\mu \le \frac{1}{n}\int f\, d\mu$$
Thus $\mu(A) \le n^{-1}\int f\, d\mu$ for every $n$, forcing $\mu(A) = 0$.