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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.

Michael Jesurum
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MNM
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    Post your proof please. We are just used to call it the converse while it's not the actually converse proposition. Far as I know, independence is necessary. So if you don't use it, maybe some details are missed. – Oolong Milktea May 07 '21 at 08:23
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    because of the added assumption it is only a "partial" converse. As limsup A_n is a tail event its probability is either 0 or 1. So P(limsup A_n) ≠ 1 iff its = 0 – Calvin Khor May 07 '21 at 08:27
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    Consider $[0,1]$ with Lebesgue measure. Let $A_n=(0,1/n]$. Then $\limsup A_n=\emptyset$... (so your proof is wrong - if you want opinions on where the error is you need to show us the argument). – David C. Ullrich May 07 '21 at 11:38
  • I posted my corrected proof using the independence – MNM May 07 '21 at 14:31
  • I don't use the "converse" language here. I say there are two directions for Borel-Cantelli: an easy direction (not requiring independence) and a hard direction (requiring independence). – GEdgar May 07 '21 at 14:33
  • There is a rather interesting version of a "converse" Borel-Cantelli theorem that has some applications in Harmonic Analysis. In fact, David C. Ulrich used this sort of result here in his solution to a problem solved by GuieppeNegro. The results states that if $(E_n:n\in\mathbb{N})$ are measurable subsets of $[0,1]$ with $\sum_n\lambda(E_n)=\infty$,, then there are translates $(F_n=E_n+r_n)$ such that $\mathbb{1}_{[0,1]}=\limsup_nF_n$ a.s. – Mittens May 07 '21 at 22:34

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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.

MNM
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