We saw the following theorem: If $f \in L^{+}$(Space of measurable functions from a set to $[0,\infty]$) and $\int f < \infty$, then $\{x: f(x) = \infty\}$ is a null set and $\{x: f(x) = \infty\}$ is $\sigma$-finite.
We proved it as follows: Let $E = \{x : f(x) = \infty \}$, then $E$ is measurable(1.Why?) and for all $n \ge 1$ define $\varphi_n = n\chi_E$ which is a simple function with $0 \leq \varphi_n \leq f$, so
Thus, $\mu E \leq \cfrac{1}{n}\int f $ for all $n \ge 1$. Since $0 \leq \int f < \infty$ it follows that $\mu E = 0$.(2.Why?)
Now let $F= \{x: f(x) > 0 $ and $F_n = \{x: f(x) > \cfrac{1}{n} \}$. Then $\bigcup_{n \ge 1}F_n $. For each $n$, define the simple function $\psi_n = (\cfrac{1}{n})\chi_{F_n}$ with $0 \leq \psi_n \leq f$ so $ \int f \ge \int \psi_n = \cfrac{1}{n}\mu F_n$ from which it follows that $\mu F_n \leq n \int f < \infty$ for all $n$, so $F$ is indeed a $\sigma-finite $ set.
I couldn't understand the proof, especially the $2$ points. Can you explain this proof or give an alternative proof which? Both would be great.