# What do these alternative formulations of the second Borel-Cantelli lemma (Durrett theorem 4.3.4 and 4.5.5) say?

I am reading Durrett and I don't understand what do these formulations of the second Borel-Cantelli lemma say:

Specifically, what does $$P(B_n|F_{n-1})$$ mean?

• $P(B_n\mid \mathcal F_{n-1})(\omega)$ is the $\mathcal{F}_{n-1}$ measurable random variable $E[\mathsf 1_{B_n}\mid \mathcal F_{n-1}](\omega)$. Apr 4, 2021 at 20:44

It's been awhile since I've done probability, but...

this appears to say

for 4.3.4:

1. $$\limsup B_n = \{\sum_{n=1}^{\infty} E(1_{B_n}|\mathcal F_{n-1}) = \infty\}$$

2. P.S. I thought $$\mathcal F_0$$ was always trivial...sounds like something a book should say in earlier chapters.

for 4.5.3:

1. the adapted thing is weird. adapted sounds like something you use to describe random variables not events. I think the events $$\{B_n\}$$ are adapted to the filtration $$\{\mathcal F_n\}$$ if and only if the random variables$$\{1_{B_n}\}$$ are adapted to the filtration $$\{\mathcal F_n\}$$. So basically the same as 4.3.4 where you had $$B_n \in \mathcal F_n$$

2. without the 'on' clause this means

• 4.1. $$\liminf \frac{\sum_{m=1}^n 1_{B_m}}{\sum_{m=1}^n p_m} = \limsup \frac{\sum_{m=1}^n 1_{B_m}}{\sum_{m=1}^n p_m}$$ (I guess surely...or at least almost surely) s.t. we can define $$\lim \frac{\sum_{m=1}^n 1_{B_m}}{\sum_{m=1}^n p_m}$$ as their common value like $$[\lim \frac{\sum_{m=1}^n 1_{B_m}}{\sum_{m=1}^n p_m}](\omega):=[\liminf \frac{\sum_{m=1}^n 1_{B_m}}{\sum_{m=1}^n p_m}](\omega)=[\limsup \frac{\sum_{m=1}^n 1_{B_m}}{\sum_{m=1}^n p_m}](\omega)$$, where $$[\liminf \frac{\sum_{m=1}^n 1_{B_m}}{\sum_{m=1}^n p_m}](\omega):=\liminf [\frac{\sum_{m=1}^n 1_{B_m}}{\sum_{m=1}^n p_m}(\omega)]$$ and similar for $$\limsup$$

• 4.2. and $$\lim \frac{\sum_{m=1}^n 1_{B_m}}{\sum_{m=1}^n p_m} = 1$$ almost surely, i.e. $$P(\omega \in \Omega | [\lim \frac{\sum_{m=1}^n 1_{B_m}}{\sum_{m=1}^n p_m}](\omega) = 1) = 1$$

• 4.3. I think $$\lim \frac{A_n}{B_n} = 1$$ almost surely if and only if $$\lim A_n = \lim B_n$$ almost surely aaaand I guess $$B_n(\omega) \ne 0$$ for all $$n$$ and for all $$\omega$$ so...

• 4.4. ...I guess that $$p_m(\omega) \ne 0$$ for all $$m$$ and for all $$\omega$$...or at least $$\sum_{m=1}^n p_m \ne 0$$ for all $$n$$ and for all $$\omega$$.

3. Now including the 'on' clause...hmmm...this appears to be basically an 'if' clause, sooo..... I guess $$\{\omega \in \Omega | \sum_{m=1}^{\infty} p_m(\omega) = \infty \} \subseteq \{\omega \in \Omega | [\lim \frac{\sum_{m=1}^n 1_{B_m}}{\sum_{m=1}^n p_m}](\omega) = 1\}$$

4. Sounds pretty much like saying if $$\sum_{m=1}^{\infty} p_m = \infty$$ a.s., then $$\sum_{m=1}^\infty 1_{B_m} = \infty$$ a.s. I didn't learn this in probability class, but maybe this can be understood in some examples of $$\{B_n\}$$ and $$\{\mathcal F_n\}$$ s.t.

• 6.1. $$\sum_{m=1}^{\infty} p_m$$ that is not almost surely infinite but still $$\sum_{m=1}^\infty 1_{B_m} = \infty$$ a.s.

• 6.2. $$\sum_{m=1}^{\infty} p_m$$ that is not almost surely infinite and $$\sum_{m=1}^\infty 1_{B_m} = \infty$$ is not almost surely infinite