“Converse” of Borel-Cantelli Lemma

borel-cantelli-lemmasmeasure-theoryprobability theorysolution-verification

I've stumbled upon this unusual "converse" of BC Lemma. The usual one being the one stated for example here: Proof of the converse Borel-Cantelli lemma

The unusual one being:

If $P(\lim \sup A_n) = P(A_n \ \ \text{i.o.})= 0\ $ (infinitely often) and $ \{A_n\}_{n \in \mathbb{N}}$ is a family of independent events then

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

My doubt here comes as why is this the converse and where does the independence have to be used on the result as I haven't used it in my proof, but as always, if it weren't necessary it wouldn't be stated in the first place

I've changed my proof to force myself to use independence as not doing so makes the statement false (David C. Ullrich's comment):

Let's suppose to reach a contradiction that $$
\sum_{n=1}^{\infty}P(A_n) = \infty
$$

Now, we know $P(\lim_n \sup A_n) = 0$ so $P(\lim_n \inf A_n^c) = 1$

But $P(\lim_n \inf A_n^c) = \lim_{N \to \infty} P(\cap_{n=N}^{\infty} A_n^c)$

Then, $P(\cap_{n=N}^{\infty} A_n^c) = \Pi_{n=N}^{\infty}P(A_n^c) = \Pi_{n=N}^{\infty}(1-P(A_n) \leq \Pi_{n=N}^{\infty}e^{-P(A_n)} = e^{-\sum_{n=N}^{\infty}P(A_n)} = 0$

Where the first equality holds since $A_n$ being independent also holds for $A_n^c$, and the inequality is true since $1-x \leq e^{-x} \ \forall x \in \mathbb{R}$

Implying that $P(\lim_n \sup A_n^c) = 1$ so $P(\lim_n \inf A_n^c) = 0$ contradicting the initial hypothesis.

Best Answer

Answer was given as correct by a professor of mine that checked it, so I'm posting it.

Let's suppose to reach a contradiction that $$ \sum_{n=1}^{\infty}P(A_n) = \infty $$

Now, we know $P(\lim_n \sup A_n) = 0$ so $P(\lim_n \inf A_n^c) = 1$

But $P(\lim_n \inf A_n^c) = \lim_{N \to \infty} P(\cap_{n=N}^{\infty} A_n^c)$

Then, $P(\cap_{n=N}^{\infty} A_n^c) = \Pi_{n=N}^{\infty}P(A_n^c) = \Pi_{n=N}^{\infty}(1-P(A_n)) \leq \Pi_{n=N}^{\infty}e^{-P(A_n)} = e^{-\sum_{n=N}^{\infty}P(A_n)} = 0$

Where the first equality holds since $A_n$ being independent also holds for $A_n^c$, and the inequality is true since $1-x \leq e^{-x} \ \forall x \in \mathbb{R}$

Implying that $P(\lim_n \sup A_n^c) = 1$ so $P(\lim_n \inf A_n^c) = 0$ contradicting the initial hypothesis.

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