[Math] How to apply the Borel-Cantelli lemma here

Question

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 $\sum_{n=1}^\infty \mathbb P( A_n)= \infty$?

Answer 1

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 $c$.)