Let $(A_n)$ be a sequence of independent events with $\mathbb P(A_n)<1$ and $\mathbb P(\cup_{n=1}^\infty A_n)=1$. Show that $\mathbb P(\limsup A_n)=1$.

It looks like the problem is practically asking to apply the Borel-Cantelli. Yet the suggested solution went differently: via $\prod_{n=1}^\infty \mathbb P( A_n^c)=0$.

How can we apply the Borel-Cantelli lemma here? I.e. how to show that $\sum_{n=1}^\infty \mathbb P( A_n)= \infty$?


2 Answers 2


What you are asked to show:

If $\mathbb P(A_n)\lt1$ for every $n$ and $\prod\limits_{n=1}^\infty \mathbb P( A_n^c)=0$ then $\sum\limits_n\mathbb P( A_n)$ diverges.

Thus, Borel-Cantelli lemma is not involved in the proof that the series $\sum\limits_n\mathbb P( A_n)$ diverges, which is purely a problem of real analysis. In full generality:

Consider some nonnegative sequence $(x_n)$ such that $x_n\lt1$ for every $n$ and $\prod\limits_{n=1}^\infty (1-x_n)=0$ then the series $\sum\limits_nx_n$ diverges.

Can you think of a simple approach to show this?

If $x_n\geqslant\frac12$ infinitely often then $\sum\limits_nx_n$ diverges. Otherwise, $x_n\leqslant\frac12$ for every $n$ large enough, say, for every $n\geqslant N$, and $\prod\limits_{n=N}^\infty (1-x_n)=0$ (this is where we use that $x_n\ne1$ for every $n$).
For every $x$ in $[0,\frac12]$, $1-x\geqslant\mathrm e^{-cx}$ for some suitable $c$ hence $\prod\limits_{n=N}^\infty (1-x_n)\geqslant\exp\left(-c\sum\limits_{n=N}^\infty x_n\right)$, which shows that $\sum\limits_{n=N}^\infty x_n$ diverges, QED. (Exercise: Find $c$.)


You want to show $P(\cap_{n=1}^{+\infty}\cup_{k=n}^{+\infty}A_k) = 1$, which is equivalent to $P(\cup_{n=1}^{+\infty}\cap_{k=n}^{+\infty}A_k^c) = 0$.

At the same time, we have $$P(\cup_{n=1}^{+\infty}\cap_{k=n}^{+\infty}A_k^c) = \lim_{n\to +\infty } P(\cap_{k=n}^{+\infty}A_k^c) = \lim_{n\to +\infty } \prod_{k=n}^{+\infty}P(A_k^c)$$

Remark that $\prod_{n=1}^{+\infty}P(A_n^c) = 0$ implies $\prod_{k=n}^{+\infty}P(A_k^c) = 0$ for any $n$ (since all $P(A_n^c) > 0$), so the above limit is $0$.

We've proven $P(\cap_{n=1}^{+\infty}\cup_{k=n}^{+\infty}A_k) = 1$, in consequence we have $\sum_{n=1}^{+\infty}P(A_n) = +\infty$

For a direct proof, see the related question

  • $\begingroup$ Thank you for the answer but I rather meant how to show $\sum_{n=1}^\infty \mathbb P( A_n)= \infty$ which then would imply (due to the second Borel-Cantelli lemma) that $\mathbb P(\limsup A_n)=1$. And as far as I understand, you showed $\mathbb P(\limsup A_n)=1$ directly (this was exactly what I meant by the 'suggested solution' which I had) and then showed $\sum_{n=1}^\infty \mathbb P( A_n)= \infty$ with the first Borel-Cantelli lemma. $\endgroup$
    – Leo
    Sep 16, 2014 at 10:48
  • $\begingroup$ @Leo read the proof and the first comment in this post. Then you can get a direct proof $\endgroup$ Sep 16, 2014 at 11:08

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