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''?

Let $E_n = \{x \in \Omega | n \le f(x) < n + 1\}$ and let $m_n = min_{x \in E_n} { f(x)}$. 
Now, if $\mu(E_n) = \infty$ for any n, then $\int_{\Omega} |f| \ge \sum_{n = -\infty}^{\infty} \mu (E_n) m_n = \infty$, , contradicting that f is integrable.  
