Let $\{A_n\}_{n=1}^{\infty}$ be a independent sequence of events such that $\sum_{n=1}^{\infty}P(A_n) <\infty$, then $P(A_n i.o.) =0 $.
We have that $P(A_n i.o.) =\bigcap_{m=1}^{\infty} \bigcup_{n=m}^{\infty} P(A_n)$ which I understand, its the definition.
However part of most of the proofs I have read do additional step where $\bigcap_{m=1}^{\infty} \bigcup_{n=m}^{\infty} P(A_n)= lim_{m\rightarrow \infty} P(\bigcup_{n=m}^{\infty} A_n) $.
I don't understand why those two values are equal.
Let $B_m=\bigcup_{n=m}^{\infty}A_n$ and $$B:=\bigcap_{m=1}^{\infty}B_m\tag1$$
This with: $$B_1\supseteq B_2\supseteq B_3\supseteq\cdots\tag2$$
Based on $(1)$ and $(2)$ that in can be shown that $P(B_n)\downarrow P(B)$.
Can you figure out why yourself?