Since $ f \in L(\mu) $, we know that $ \int_X |f|d\mu < \infty $. Let $ F_{n} \equiv \{|f| > 1/n\} $. Then we have $ \{f \ne 0\} = \cup_{n=1}^\infty F_n. $
Clearly all of the $ F_n $ must have finite measure, else $ \int_X |f|d\mu $ would be infinite. Thus, $ \{f \ne 0\} $ is $ \sigma $-finite.
