Is it true $P(\cap_{n=1}^\infty \cup_{m=n}^\infty A_m)=\lim_{n \to \infty}P(\cup_{m=n}^\infty A_m)$?

Question

I just need some confirmation on the truth of this since I didn't see it in a textbook so far. I think it is true by continuity of probability. Am I right?

score 3 · Answer 1 · 2019-11-27T20:01:27.037

3

Let $B_n:=\bigcup_{m\ge n}A_m$. Then $B_{n+1}\subseteq B_n$ and by the continuity from above, $$ \mathsf{P}\left(\bigcap_{n\ge 1} B_n\right)=\lim_{n\to\infty} \mathsf{P}(B_n). $$

edited Nov 27 '19 at 20:01

answered Dec 06 '18 at 03:32

M A Pelto · Accepted Answer · 2018-12-06T04:05:01.003

For each $n \in \mathbb N$, let $B_n=\bigcup_{m=n}^\infty A_m $. Clearly for each $n \in \mathbb N$ we have $B_n \supseteq B_{n+1}$, and so $\{B_n\}_{n=1}^\infty$ is a decreasing sequence of sets. Moreover, $P(B_1)<\infty$ as a probability measure is finite. Therefore, $$P\left(\bigcap_{n=1}^\infty B_n\right)=\lim_{n\to \infty} P(B_n)$$ by the continuity from above property of a measure which is $(3')$ here.

Is it true $P(\cap_{n=1}^\infty \cup_{m=n}^\infty A_m)=\lim_{n \to \infty}P(\cup_{m=n}^\infty A_m)$?

2 Answers2