Problem to understand Borel-Cantelli Lemma

Question

I needed to prove the Borel-Cantelli lemma, which states that

If $$ \sum_{n=1}^{\infty} P(A_n) < \infty \implies P(\limsup A_n) = > 0,$$

$$\sum_{n=1}^{\infty} P(A_n) = \infty \implies P(\limsup A_n) = 1 $$

I could understannd the proof, but I am struggling with a detail:

If this probability is defined in a probability space, say $(\Omega, \mathbb{F}, P) $, then $ \bigcup_{n=1}^{\infty} A_n \in \mathbb{F} $. Then how is the following equality possible? $$ \sum_{n=1}^{\infty} P(A_n) = \infty $$

I believe it is due to the fact that $$ P\left(\bigcup_{n=1}^{\infty}A_n\right) \leq \sum_{n=1}^{\infty} P(A_n) $$

which makes possible for $ \bigcup_{n=1}^{\infty} A_n \in \mathbb{F} $ and still the sum of the probabilities not to be convergent. But it's still quite misty to me, I can't imagine an example where the sum of the probabilities would not be convergent. Could you give me one example?

Thank you!

Answer 1

let $\{A_n,n\geq 1\}$ is a sequence of tossing a coin. define $A_n$ that we observe $\{H \}$ in n-Th .

$ lim sup A_n$ ,That means the event that observe infinitely many of $H$,. so and $P(lim sup A_n)=1$ .

according to lemma $\sum p(A_n)=\infty$ and the results are same.

let $B_n$ be that not any of $A_1,\cdots ,A_n$ be "H". $P(lim sup B_n)=0$

also $\sum (\frac{1}{2^n}) < \infty$ so the results are same