Since $ \int_X|f|\mbox{d}\mu = \sum_{n = 1}^{\infty}\int_{E_n}|f|\mbox{d}\mu $, and $ E_n=\{x \in X : n-1 \leq |f| \leq n\} $, then $ \sum_{n = 1}^{\infty}(n-1)\mu(E_n) \leq \sum_{n = 1}^{\infty}\int_{E_n}|f|\mbox{d}\mu \leq\sum_{n = 1}^{\infty}n\mu(E_n) $.
$ (\Leftarrow) $ If $ \sum_{n = 1}^{\infty}n\mu(E_n) < \infty $, then $ \int_X|f|\mbox{d}\mu < \infty, $, i.e. $ f \in L^1 $.
$ (\Rightarrow) $ If $ f \in L^1 $, then $ \sum_{n = 1}^{\infty}(n-1)\mu(E_n) < \infty $. $ \sum_{n = 1}^{\infty}n\mu(E_n) = \sum_{n = 1}^{\infty}(n-1)\mu(E_n) + \sum_{n = 1}^{\infty}\mu(E_n) = \sum_{n = 1}^{\infty}(n-1)\mu(E_n) + \mu(X) < \infty $, since $ X $ is a finite measure space.
