I was reading the Durrett's book and encounter some questions about the proof of generalized version of the second Borel-Cantelli lemma: Here is the setting:

Second Borel-Cantelli lemma, II. Let $\mathcal F_n, n \ge 0$ be a filtration with $F_0 = \{\emptyset, \Omega\}$ and $A_n , n \ge 1$ a sequence of events with $A_n ∈ \mathcal F_n$ . Then $$ \{A_n \,i.o.\} = \left\{\sum_{n \ge 1} P (A_n |\mathcal F_{n−1}) =\infty \right\}. $$

The proof use the following fact: a bounded increments martingale either converge of oscillate between $\pm \infty$; i.e., Let $X_1,X_2...,$ be a martingale with $|X_{n+1} - X_n| \le M < \infty$. Let \begin{array}{l} C: = \left\{ {\mathop {\lim }\limits_n {X_n}\begin{array}{*{20}{c}} {} \end{array}{\rm{exists}}\begin{array}{*{20}{c}} {} \end{array}{\rm{and}}\begin{array}{*{20}{c}} {} \end{array}{\rm{is}}\begin{array}{*{20}{c}} {} \end{array}{\rm{finite}}} \right\}\\ D: = \left\{ {\mathop {\lim \sup }\limits_n {X_n} = \infty \begin{array}{*{20}{c}} {} \end{array}{\rm{and}}\begin{array}{*{20}{c}} {} \end{array}\mathop {\lim \inf }\limits_n {X_n} = - \infty } \right\} \end{array} Then, $P(C \cup D) =1$.

Hence, from the Durrett's proof of second Borel-Cantelli lemma, we first define $${X_n}: = \sum\limits_{m = 1}^n {{1_{{A_m}}}} - \sum\limits_{m = 1}^n {P({A_m}|{F_{m - 1}})} $$ Then this $X_n$ is a bounded increments martingale with $|X_n - X_{n-1}| \le 2$. Hence, apply the fact, we can split to two cases:

case 1: for the event $C$, $$ \left\{ {{A_n}\begin{array}{*{20}{c}} {} \end{array}i.o.} \right\} = \sum\limits_{n = 1}^\infty {{1_{{A_n}}}} = \infty \Leftrightarrow \sum\limits_{m = 1}^n {P({A_m}|{F_{m - 1}})} = \infty $$

case 2: for the event $D$, $$\left\{ {{A_n}\begin{array}{*{20}{c}} {} \end{array}i.o.} \right\} = \sum\limits_{n = 1}^\infty {{1_{{A_n}}}} = \infty \; \text{ and } \; \sum\limits_{m = 1}^\infty {P({A_m}|{F_{m - 1}})} = \infty$$.

Since we have $P(C \cup D) =1$, the desired result follows. $\square$

Here is my question:

  1. for the event $C$ (case 1): is this because that if $\{A_n i.o.\} = \sum_{n=1}^\infty 1_{A_n}=\infty$, but by the definition of $X_n$, we know $${\lim X_n} = \sum\limits_{m = 1}^\infty {{1_{{A_m}}}} - \sum\limits_{m = 1}^\infty {P({A_m}|{F_{m - 1}})} $$ and since $\lim X_n$ is bounded in this case, we must have $\sum\limits_{m = 1}^\infty {P({A_m}|{F_{m - 1}})} = \infty$ ? Is this thinking process correct?

  2. for the event $D$ (case 2): in this case, we have both $\limsup X_n = \infty$ and $\liminf X_n = -\infty$. So, for $\{A_n i.o\}$, I don't see why $\sum\limits_{m = 1}^\infty {P({A_m}|{F_{m - 1}})} = \infty$ also hold because I was thinking that $\infty- \infty$ happens.

Thank you


1 Answer 1


Regarding your first question: Yes, this is correct.

Regarding your second question: For brevity, set $$A := \{A_n \, \text{i.o.}\} \qquad \quad B := \{\sum_n \mathbb{P}(A_n \mid \mathcal{F}_{n-1})=\infty\}.$$ We want to show that $A \cap D = B \cap D$.

  • Let $\omega \in A \cap D$. Suppose that $\sum_{n \geq 1} \mathbb{P}(A_n \mid \mathcal{F}_{n-1})(\omega)<\infty$. Then it follows that $$X_n(\omega) = \sum_{m=1}^n 1_{A_m}(\omega) - \sum_{m=1}^n \mathbb{P}(A_m \mid \mathcal{F}_{m-1})(\omega) \geq \sum_{m=1}^n 1_{A_m}(\omega) - \sum_{n \geq 1} \mathbb{P}(A_n \mid \mathcal{F}_{n-1})(\omega).$$ Since, by assumption $\sum_{m=1}^n 1_{A_m}(\omega) \uparrow \infty$, this means $$\liminf_{n \to \infty} X_n(\omega) > - \infty.$$ Obviously, this contradicts $\omega \in D$; hence, $A \cap D \subseteq B \cap D$.
  • Let $\omega \in B \cap D$ and suppose that $\sum_{n \geq 1} 1_{A_n}(\omega) < \infty$. Then $$X_n(\omega) \leq \sum_{n \geq 1} 1_{A_n}(\omega)- \sum_{m=1}^n \mathbb{P}(A_m \mid \mathcal{F}_{m-1})(\omega).$$ Since the second term on the right-hand side is non-decreasing, we get $$\limsup_{n \to \infty} X_n(\omega) < \infty.$$ Again, this contradicts $\omega \in D$. Consequently, $B \cap D \subseteq A \cap D$.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.