Problem 4.11(d) from Billingsley's Probability and Measure book states: Show that $P\left(\limsup A_n\right)=1$ if and only if $\sum_n P(A\cap A_n) $ diverges for each $A$ of positive probability.

Notably, there is no assumption that the $A_i's$ are independent. I'm not sure how to proceed here! I assume there is a nice trick. Any hints are most welcome.


1 Answer 1


It seems to be simpler to show the equivalence of the negation of the assertions, that is, we will prove that $$ \mathbb P\left(\limsup_{n\to\infty}A_n\right)<1\Leftrightarrow \mbox{ there exists }B \mbox{ of positive probability such that }\sum_n\mathbb P\left(B\cap A_n\right)<\infty. $$ Let us show $\Leftarrow$. By the classical Borel-Cantelli lemma applied to $(B\cap A_n)_n$, we derive that $\mathbb P\left(\limsup_n (B\cap A_n)\right)=0$. By elementary operations on sets, we derive that $\mathbb P(A\cap B)=0$, where $A=\limsup_nA_n$. As a consequence, $\mathbb P(A)=\mathbb P(A\cap B^c)\leqslant \mathbb P(B^c)$. Since $\mathbb P(B)>0$, $\mathbb P(B^c)<1$ hence $\mathbb P(A)<1$.

It remains to show $\Rightarrow$. By assumption, $\mathbb P(\liminf_n A_n^c)=\mathbb P\left(\left(\limsup_n A_n\right)^c\right)>0$. By definition of $\liminf$, there exists a $k_0$ such that $\mathbb P\left(\bigcap_{k\geqslant k_0}A_k^c\right)>0$. Let $B=\bigcap_{k\geqslant k_0}A_k^c$. For $n\geqslant k_0$, $B\cap A_n$ is empty hence the series $\sum_n\mathbb P\left(B\cap A_n\right)$ converges.

  • $\begingroup$ Amazing! Thanks so much @Davide Giraudo $\endgroup$ Commented Aug 10, 2022 at 16:48
  • $\begingroup$ You are welcome! $\endgroup$ Commented Aug 10, 2022 at 17:37

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