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I am given a sequence of events $A_1, A_2, \dots$ that satisfy

$$ \sum_{n \geq k} \mathbb{P}\left(A_n \ \middle|\ \bigcap_{i=k}^{n-1} A_i^c \right) = \infty, \forall k \in \mathbb{N} $$ and asked to conclude that $\mathbb{P}(\limsup_n A_n) = 1$. So far, I've tried:

$$ \mathbb{P}(\limsup_n A_n) = \lim_{k\to\infty} \mathbb{P}(\bigcup_{n \geq k}A_n) = \lim_{k\to\infty} \mathbb{P}\left( \bigcup_{n \geq k} A_n \setminus \bigcup_{i=k}^{n-1}A_i \right) \\ = \lim_{k\to\infty} \sum_{n \geq k} \mathbb{P}\left( A_n \cap \left( \bigcap_{i=k}^{n-1} A_i^c \right) \right) \qquad \text{(union over disjoint events)} $$

However, I'm not sure how to link the fact I'm given about the conditional probabilities with the above limit. Any ideas?

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    $\begingroup$ Here, rather than using the Borel-Cantelli lemma (which would require some form of independence), you can simply repeat the proof of the Borel-Cantelli lemma in this more general situation. For example, compute the probability that none of the events $A_i$ occur for $i \geq k$, that is, $P[\cap_{i \geq k} A_i^c]$. $\endgroup$ – Michael Mar 13 '18 at 4:32

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