The proposition and a part of its proof are given below:
My questions are:
In the forward direction: 1- How is the countable additivity of Lebesgue measure is used Should not the given set be described as a countable union of disjoint subsets?
2- How this proves that $f$ equals 0 a.e.?
Could anyone help me in answering these questions, please?
In (1) is used the subaddittivity of the Lebesgue measure and the fact that $$\{f>0\}=\bigcup_{n=1}^{\infty}\{f \geq \frac{1}{n}\}$$
For (2) $m(\{f>0\}) \leq \sum_{n=1}^{\infty}m(\{f \geq \frac{1}{n}\})=0$
Since $f$ is non-negative anf $m(\{f>0\})=0$ then $f=0$ a.e