Let $(\Omega, \mathscr F, \mathbb P)$ be a probability space.
Conjecture: Suppose we have events $A_1, A_2, ...$ s.t. $\forall \ A \in \bigcap_n \sigma(A_n, A_{n+1}, ...)$, $P(A) = 0$ or $1$. There exists an independent sequence of events $B_1, B_2, ...$ s.t.
$$\tau_{A_n} := \bigcap_n \sigma(A_n, A_{n+1}, ...) = \bigcap_n \sigma(B_n, B_{n+1}, ...) := \tau_{B_n}$$
Is this true?
I think there exists a function $f: \mathbb N \to \mathbb N$ s.t. $A_{f(n)}$'s are independent so we can choose $B_n = A_{f(n)}$. Is that true? Why/Why not? If not, how else can I prove or disprove the conjecture above? If it is true, I think it can be proven by modifying the proof of the Kolmogorov 0-1 Law (for events).
Perhaps one of these subsequences of sets is independent:
$$A_n$$
$$A_{2n}, A_{2n+1}$$
$$A_{3n}, A_{3n+1}, A_{3n+2}$$
$$\vdots$$
$$A_{mn}, A_{mn+1}, A_{mn+2}, ..., A_{mn+(m-1)}$$
$$\vdots$$
I think we have that
$$\tau_{A_n} = \tau_{A_{mn+i}} := \bigcap_n \sigma(A_{mn+i}, A_{m(n+1)+i}, ...)$$
where $m \in \mathbb N$ and $i \in \{0, 1, 2, ..., m-1\}$.
It seems like we need any such $f(n)$, if it exists, to satisfy the following condition:
$$\sigma(A_{f(n)}, A_{f(n+1)}...) \subseteq \sigma(A_n, A_{n+1}, ...) \tag{**}$$
which I guess is true if (and only if?) $f(n) \ge n$.
Other possible candidates for $f(n)$: (assume the variables are s.t. $f: \mathbb N \to \mathbb N$ is satisfied. If need be, $(**)$ or $f(n) \ge n$ too.)
$\sum_{i=0}^{m} a_i n^i$
$2^n, 3^n, ...$
$\sum_{i=1}^{m} b_i c_i^n$
$\lfloor{t^n}\rfloor, \lceil{t^n}\rceil$ (I guess $t > e^{1/e}$)
$\lfloor{\sum_{i=1}^{m} b_i c_i^n}\rfloor, \lceil{\sum_{i=1}^{m} b_i c_i^n}\rceil$
$\lfloor{\text{linear combination of trigonometric functions}}\rfloor, \lceil{\text{linear combination of trigonometric functions}}\rceil$
$\lfloor{\text{Some linear combination of the above}}\rfloor, \lceil{\text{Some linear combination of the above}}\rceil$
Assuming the conjecture is true, I guess it's not necessary to find $f(n)$ that works for all possible sequences of events $A_1, A_2, ...$ because such $f(n)$ may not even exist.
To disprove the conjecture: I guess we must show that such a sequence $B_n$ being independent implies $B_n$ tail will never equal $A_n$ tail since $B_n$ tail will be $\mathbb P-$trivial by Kolmogorov 0-1 Law (for events).
Something that might help: we could show that $\forall \ A \in \bigcap_n \sigma(A_{f(n)}, A_{f(n+1)}, ...), P(A) = 0$ or $1$ and $\forall n \in \mathbb N, A_{f(n)}, A_{f(n+1)}, ...$ is not independent, but I'm not quite sure that the conjecture is disproved because we could construct some $B_n$'s that look like:
$$B_n = A_{n+1} \setminus A_n$$
$$B_n = A_{n} \setminus A_{n-1}, A_0 = \emptyset$$
$$B_n = \bigcap_m A_{mn}$$
$$B_n = \bigcup_m A_{mn}$$
$$B_{2n} = \bigcap_m A_{mn}, B_{2n+1} = \bigcup_m A_{mn}$$
$$B_n = \limsup_m A_{mn}$$
$$B_n = \liminf_m A_{mn}$$
$$B_{2n} = \limsup_m A_{mn}, B_{2n+1} = \liminf_m A_{mn}$$
Not to say of course that any of those $B_n$'s satisfy $\tau_{A_n} = \tau_{B_n}$ but that $B_n$ need not be in the form $A_{f(n)}$.
Borel-Cantelli:
If $\sum_n P(A_n) < \infty \to 0 = P(\limsup A_n) = P(\limsup A_{mn}) \ \forall m \in \mathbb N$. Hence $B_m = \limsup A_{mn}$ is independent.
If $\sum_n P(A_n) = \infty$, then maybe this extension of Borel-Cantelli? Not quite sure I understand it or how it would be helpful. I don't think we can conclude anything if we have $P(\limsup A_n)$.
Then there's the case of $\sum_n P(A_n) = \infty$ but the conditions earlier aren't satisfied.
Based on: Converse of Kolmogorov's Zero-One Law
