I have a question regarding to a specific step of the following proof. I am confused to how did $\sum$ for all n implies $\sum $ for countable cases?
My initial thought is math induction. However, since induction only applies to finite cases, not countable. There must be some other reason. My current guess is that it have to do with some type of monotonicity, but I struggle to reason it out completely.
Please help me.
What you need to use here is that if $\{ a_n \}$ and $\{ b_n \}$ are two convergent sequences of reals with limits $a$ and $b$, respectively, such that $a_n \geq b_n$ for all $n$, then the inequality is preserved under taking limits, that is: $$ a = \lim_{n \to \infty} a_n \geq \lim_{n \to \infty} b_n = b. $$
In this case, we have $$ \mu^*(A) \geq \sum_{i = 1}^n \mu^*(A \cap E_i) + \mu^*(A \cap E'). $$ Note that in this case, one of the sequences is just the constant sequence. Taking $\lim_{n \to \infty}$ on both sides, we get $$ \mu^*(A) \geq \sum_{i = 1}^\infty \mu^*(A \cap E_i) + \mu^*(A \cap E'), $$ since an infinite series is defined to be the limit of its partial sums.