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.
Best Answer
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.