BOREL CANTELLI LEMMA : For arbitrary sequence of events $\{A_n\}$, we have $$\sum_{n=1}^{\infty} P(A_n)<\infty \implies P(\limsup A_n)=0$$

CONVERSE BOREL CANTELLI LEMMA : For independent sequence of events $\{A_n\}$, we have $$\sum_{n=1}^{\infty} P(A_n)=\infty \implies P(\limsup A_n)=1$$

While I have no problem with the proof of Borel-Cantelli lemma, I'm stuck at constructing a proof of the converse. I have started with $P(\limsup A_n)^c$, breaking down the intersection-union definition of $\limsup$, applying de-morgan's law, taking product using independence and so on. But I'm not being able to finish it off.

Worse, I do not understand where to use the condition $\sum_{n=1}^{\infty} P(A_n)=\infty$ (In Borel-Cantelli, the condition $\sum_{n=1}^{\infty} P(A_n)<\infty$ was exploited with $\sum_{n=N}^{\infty} P(A_n)<\varepsilon$ for sufficiently large $N$). Any help (in constructing a proof of the converse Borel-Cantelli lemma) would be much appreciated.


1 Answer 1


Here's the proof I know. Surely this can be made more elegant. Let's show (equivalently) that

$$ 0 = \mathbb{P}((\limsup A_n)^c) = \mathbb{P}(\bigcup_{n\geq 1}(\bigcup_{k \geq n}A_k)^c) $$

by noting that each event $(\bigcup_{k \geq n}A_k)^c$ has probability zero for each $n \in \mathbb{N}$. In effect,

$$ \mathbb{P}((\bigcup_{k \geq n}A_k)^c) = \mathbb{P}(\bigcap_{k \geq n}A_k^c) = \lim_{N \to \infty}\mathbb{P}(\bigcap_{k = n}^NA_k^c) = \lim_{N \to \infty}\prod_{k=n}^N\mathbb{P}(A_k^c) = \\ = \prod_{k\geq n}1 - \mathbb{P}(A_k) \leq \prod_{k\geq n}e^{-\mathbb{P}(A_k)} = e^{-\sum_{k \geq n}\mathbb{P}(A_k)} = 0 $$

Here I use that $1-t \leq e^{-t}$ for $t \geq 0$, which can be proved by seeing that $g(t) = e^{-t}+t-1$ has non-negative derivative in $[0, \infty)$ and $g(0) = 0$. It's worth noting that in the last equality we can explicitly see that the divergence hypothesis really is necessary. Independence is used in the third equality.

  • $\begingroup$ This is sweet. I went along the same line but lacked the cleverness to implement $1-x \leq e^{-x}$. Very very sweet. Thank you very much. $\endgroup$
    – Adi
    Jun 27, 2018 at 4:31
  • $\begingroup$ No problem, glad I could help! $\endgroup$
    – qualcuno
    Jun 27, 2018 at 4:32
  • $\begingroup$ Just one thing : how can we justify the line : $\mathbb{P}(\bigcap_{k = n}^{\infty} A_k^c) = \lim_{N \to \infty}\mathbb{P}(\bigcap_{k = n}^NA_k^c)$ ? Is there an interversion of limit ? Thanks ! $\endgroup$
    – Maman
    Jan 8, 2019 at 1:14
  • $\begingroup$ The intersection for $k \geq n$ can be written as the intersection of finite intersections as a sequence (on $N$) that decreases, hence the probability of their intersection is the limit of the probabilities. $\endgroup$
    – qualcuno
    Jan 8, 2019 at 2:47
  • $\begingroup$ So you take $D_N = \bigcap_{k=n}^{N} A_k^C$ which is decreasing then you have $\bigcap_{N=0}^{+\infty}(\bigcap_{k=n}^{N}A_k^c)=\bigcap_{k=n}^{\infty} A_k^c$ and you apply the property of limit right ? $\endgroup$
    – Maman
    Jan 14, 2019 at 19:57

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