A variation of Borel Cantelli Lemma

Question

If $P(A_n) \rightarrow 0$ and $\sum_{n=1}^{\infty}{P(A_n^c\cap A_{n+1}})<\infty$ then $P(A_n \text{ i.o.})=0$.

How to prove this? Thanks.

Answer 1

Hint: $\lim \sup A_n \subseteq A_N \cup \bigcup_{n=N}^\infty (A^c_n \cap A_{n+1})$. Estimate the probability of this.

Answer 2

$$P[A_n^c\ i.o.] = \lim_{N\rightarrow\infty} P[\bigcup_{n=N}^{\infty} A_n^c] \geq \lim_{N\rightarrow\infty} P(A_n^c) = 1\Longrightarrow P[A_{n+1}^c\ i.o.] = 1$$ But from $1st$ Borel-Cantelli Lemma, we know $P[A_n\cap A_{n+1}^c\ i.o.] = 0$, as $\sum_{n=1}^{\infty}P(A_n\cap A_{n+1}^c) < \infty$.
$$P[A_n\cap A_{n+1}^c\ i.o.] = 0\\\Rightarrow \lim_{N\rightarrow\infty} P[\bigcup_{n=N}^{\infty} A_n\cap A_{n+1}^c] = 0\\ \Rightarrow \lim_{N\rightarrow\infty} P[(\cup_{n=N}^{\infty} A_n)\cap (\cup_{n=N}^{\infty} A_{n+1}^c)] = 0\\ \Rightarrow \lim_{N\rightarrow\infty} P[(\cup_{n=N}^{\infty} A_n)] + \lim_{N\rightarrow\infty} P[(\cup_{n=N}^{\infty} A_{n+1}^c)] - \lim_{N\rightarrow\infty} P[(\cup_{n=N}^{\infty} A_n)\cup (\cup_{n=N}^{\infty} A_{n+1}^c)] = 0\\ \Rightarrow P[A_n\ i.o.] + 1 - 1 = 0$$ Hence, we have showed the required result.