If there exists an integrable function that is not zero a.e., then the measure is $\sigma$-finite Suppose $f\in L^1(\Omega,\mathcal{A},\mu)$ and $f(x)\neq 0$ for almost every $x\in \Omega$. How to prove $\mu$ is $\sigma-$finite?
I only got that $\Omega=\cup_{n=1}^\infty \{x\in \Omega:|f(x)|\geq \frac{1}{n}\}\cup \{{ x\in \Omega: f(x)=0\}} $, because $\mu\{x\in \Omega:|f(x)|\geq \frac{1}{n}\}\leq n||f||_1<\infty,\ \ \mu\{x\in \Omega:f=0\}=0$, so $\Omega$ is $\sigma-$finite. 
But how it works to deal with ''$\mu$ is $\sigma-$finite''?

As you observed, in the decomposition 
$$\Omega=\bigcup_{n=1}^\infty \left\{x\in \Omega:|f(x)|\geq \frac{1}{n}\right\}\cup \{{ x\in \Omega: f(x)=0\}}$$
every set on the right  has finite measure. Therefore, the measure is $\sigma$-finite. 
If needed, one can also be more precise and say: "$\mu$ is $\sigma$-finite on $\Omega$", or (not as often) "$\Omega$ is a  $\sigma$-finite set for $\mu$". 
