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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

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1 Answer 1

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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, $B \cap D \subseteq A \cap D$.
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