Is this sigma-additivity or union bound?

Question

In the proof of the second Lemma of Borel Cantelli, we arrived at the following point:

$$P[\bigcup_{n \geq 1} \bigcap_{m \geq n} A_m^c] = \sum_{n \geq 1} P[\bigcap_{m \geq n} A_m^c]$$

but I am very confused about the equality: The equality suggests that we use $\sigma$-additivity here, but are these sets $\bigcap_{m \geq n} A_m^c$ for $n \geq 1$ really disjoint? So isn't this rather just an inequality using an union bound?

Answer 1

You are right. The sets $B_n : = \bigcap_{m \ge n} A_m^C$ are not disjoint but rather increasing. Hence, we have

$$P(\bigcup_n B_n) \color{red}{\le} \sum_n P(B_n)$$

Iirc, the point of the proof is to show that $P(\bigcup_n B_n) = 0$ by showing that $\sum_n P(B_n) = 0$ so it will turn out that

$$P(\bigcup_n B_n) \color{red}{=} \sum_n P(B_n)$$

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We can however define $C_n := B_n \setminus B_{n-1}$ with $B_0 := \emptyset$

Then $C_n$'s are pairwise disjoint, and we have

$$\sum_n P(C_n) = P(\bigcup_n C_n) = P(\bigcup_n B_n) = \lim P(B_n)$$