Sort-of converse of Kolmogorov zero-one theorem

Question

Let $(\Omega, \mathscr F, \mathbb P)$ be a probability space. The Kolmogorov zero-one theorem states that

Suppose we have independent random variables $X_1, X_2, ...$. Then $\forall \ A \in \bigcap_n \sigma(X_n, X_{n+1}, ...)$, $P(A) = 0$ or $1$.

If we choose $X_k = 1_{A_k}$ for events $A_1, A_2, ...$, then we have:

Suppose we have independent events $A_1, A_2, ...$. Then $\forall \ A \in \bigcap_n \sigma(A_n, A_{n+1}, ...)$, $P(A) = 0$ or $1$.

Now, is this following conjecture true? If not, can it be modified slightly to be true?

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.

$$\bigcap_n \sigma(A_n, A_{n+1}, ...) = \bigcap_n \sigma(B_n, B_{n+1}, ...) \tag{*}$$

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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 Theorem (for events).

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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

$$\bigcap_n \sigma(A_n, A_{n+1}, ...) = \bigcap_n \sigma(A_{mn+i}, A_{m(n+1)+i}, ...)$$

where $m \in \mathbb N$ and $i \in \{0, 1, 2, ..., m-1\}$.

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Based on what @FedorPetrov pointed out, it seems like we need $f(n)$'s s.t.

$$\sigma(A_{f(n)}, A_{f(n+1)}...) \subseteq \sigma(A_n, A_{n+1}, ...) \tag{**}$$

which I guess is true if (and only if?) $f(n) \ge n$.

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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.)

1. $\sum_{i=0}^{m} a_i n^i$

2. $2^n, 3^n, ...$

3. $\sum_{i=1}^{m} b_i c_i^n$

4. $\lfloor{t^n}\rfloor, \lceil{t^n}\rceil$ (I guess $t > e^{1/e}$)

5. $\lfloor{\sum_{i=1}^{m} b_i c_i^n}\rfloor, \lceil{\sum_{i=1}^{m} b_i c_i^n}\rceil$

6. $\lfloor{\text{linear combination of trigonometric functions}}\rfloor, \lceil{\text{linear combination of trigonometric functions}}\rceil$

7. $\lfloor{\text{Some linear combination of the above}}\rfloor, \lceil{\text{Some linear combination of the above}}\rceil$

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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.

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To disprove the conjecture: There's of course showing that any sequence that satisfies $(*)$ will not be independent, but I have a feeling it's more of showing that any independent sequence will never satisfy $(*)$.

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:

1. $$B_n = A_{n+1} \setminus A_n$$

2. $$B_n = A_{n} \setminus A_{n-1}, A_0 = \emptyset$$

3. $$B_n = \bigcap_m A_{mn}$$

4. $$B_n = \bigcup_m A_{mn}$$

5. $$B_{2n} = \bigcap_m A_{mn}, B_{2n+1} = \bigcup_m A_{mn}$$

6. $$B_n = \limsup_m A_{mn}$$

7. $$B_n = \liminf_m A_{mn}$$

8. $$B_{2n} = \limsup_m A_{mn}, B_{2n+1} = \liminf_m A_{mn}$$

Not to say of course that any of those $B_n$'s satisfy $(*)$ but that $B_n$ need not be in the form $A_{f(n)}$.

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Borel-Cantelli:

1. 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.

2. 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 $P(\limsup A_n)$.

3. Then there's the case of $\sum_n P(A_n) = \infty$ but the conditions earlier aren't satisfied.

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Based on: https://math.stackexchange.com/questions/605301

Answer 1

I don't think that there can be such a selection function $f$.

Define $A_i$ through independent events $C_i$ and $D_i$, each of which occurs with $50\%$ chance, as follows. $A_1=D_1$ and for $i\ge 2$, if $C_i$, then $A_i=D_i$, otherwise $A_i=A_{i-1}$. If $A \in \bigcap_n \sigma(A_n, A_{n+1}, ...)$, then $P(A) = 0$ or $1$ should hold, with a similar proof as for the Kolmogorov Zero-One Theorem. On the other hand, no $A_i$ and $A_j$ are independent.