Given a sequence of events $A_n$, construct a sequence of independent events $B_n$ such that $\mathbb{P}(\limsup A_n) = \mathbb{P}(\limsup B_n$)

Question

Let $(\Omega,\mathcal{A},\mathbb{P})$ be a probability space. Let $(A_n)$ be a sequence of events in $\mathcal{A}$. Does there exist a sequence of (pairwise) independent events $(B_n)$ such that $\mathbb{P}(\limsup A_n) = \mathbb{P}(\limsup B_n)$? If yes, how to construct this out of $(A_n)$?

Note that this can also be done if the limsups of both the sequences are same. But this is possibly harder to construct.

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I am working on a problem that involves showing that a sequence of events satisfying some conditions has probability of limsup equal to 1. I figured that if I can construct a sequence of independent events such that the probabilities of limsups are equal, then by Borel-Cantelli lemma, I can say that the probability of limsup is either 0 or 1. And using the conditions, I can possibly show that the probability cannot be 0.

Answer 1

I think your comment shows that the answer is false.

Let $A_n = A$ be an event with $P(A)=\frac{1}{2}$, then clearly $P(\limsup A_n)=\frac{1}{2}$. Let $B_n$ be independent events.

By Borel-Cantelli, if $\sum P(B_n) < \infty$, then $P(\limsup B_n) = 0$. Furthermore, if $\sum P(B_n)=\infty$, then since $B_n$ are indepdent $P(\limsup B_n)=1$.