Let $(\Omega, \mathcal B, P)$ be a probability space and $\mathcal A = \bigcap_n \mathcal{A}_n$, where $\mathcal{B} \supset \mathcal{A}_1 \supset ... \supset \mathcal{A}_n \supset ...$ is a decreasing sequence of sub-sigma-fields.
I recently came across the following statement:
$P$ is 0-1 valued on $\mathcal{A}$ if and only if $\sup_{A \in \mathcal{A}_n}|P(A \cap B) - P(A)P(B)| \to 0$ for all $B \in \mathcal{A}_1$.
In other words, "asymptotic independence on the tail field is equivalent to $P$ being trivial."
The result seems pretty intuitive to me as a generalization of the Kolmogorov 0-1 law, but I can't find a proof of it.
One direction seems easy to me (is it correct?):
Assume $\sup_{A \in \mathcal{A}_n}|P(A \cap B) - P(A)P(B)| \to 0$ for all $B \in \mathcal{A}_1$, and let $B \in \mathcal A$. Then, $B \in \mathcal{A}_1$ and for all $n$
\begin{align} |P(B) - P(B)P(B)| &= |P(B\cap B) - P(B)P(B)|\\ & \leq \sup_{A \in \mathcal{A}_n} |P(A \cap B) - P(A)P(B)| \to 0. \end{align}
So, $P(B) = P(B)^2$, which implies that $P$ is 0-1 valued on $\mathcal A$.
I don't see how to prove the converse though, so any hints, help, or references would be appreciated.
An attempt:
If $\lim_n\sup_{A \in \mathcal{A}_n}|P(A \cap B) - P(A)P(B)| \neq 0$, then for some $\epsilon>0$, we can find a subsequence $A_{n_1}, A_{n_2},...$ such that $A_{n_j} \in \mathcal{A}_{n_j}$ and $$|P(A_{n_j} \cap B) - P(A_{n_j})P(B)| \geq \epsilon$$ for all $j$. Furthermore, we can assume without loss of generality that all of the $A_{n_j}$ are identical. Indeed, if $A_{n_j} \neq A_{n_k}$ for some $j<k$, then since the sequence of sub-sigma-fields is decreasing, $A_{n_k} \in \mathcal{A}_{n_j}$, and we can replace $A_{n_j}$ with $A_{n_k}$ and still have a sequence of events with the stated property. Using the decreasing property of the sequence of sub-sigma-fields again, we can also assume without loss of generality that the subsequence $n_1,n_2,...$ is in fact all of $1,2,...$. Indeed, if there were some $n \in \mathbb{N}$ not occurring among the $n_1,n_2,...$, then, for $n_j > n$, we have $A_{n_j} \in \mathcal{A}_n$, and so we can simply add $A_n$ to our subsequence.
So, we have a set $A$ in every $\mathcal{A}_n$---which is to say $A \in \mathcal A$---with the property that $|P(A \cap B) - P(A)P(B)| \geq \epsilon$. And this implies that $P(A) \notin \{0,1\}$.
A bonus question is
Is there an easy proof of the Kolmogorov 0-1 law using the result above?
Your approach appears correct to me. Perhaphs another way to see the "non-trivial" part: \begin{align} & \lim_{n\to\infty} \sup_{A \in \mathcal{A}_n}|P(A \cap B) - P(A)P(B)|\\ & \le \lim_{n\to\infty} \sup_{A \in \mathcal{A}_n\setminus \mathcal{A}} |P(A \cap B) - P(A)P(B)| + \underbrace{\sup_{A \in \mathcal{A}} |P(A \cap B) - P(A)P(B)|}_{=\alpha} \end{align} Since $P$ is 0-1 valued on $\mathcal{A}$, then $\alpha=0$. Then, \begin{align} & \lim_{n\to\infty} \sup_{A \in \mathcal{A}_n\setminus \mathcal{A}} |P(A \cap B) - P(A)P(B)|\\ % & \le \lim_{n\to\infty} \sup_{A \in \mathcal{A}_n\setminus \mathcal{A}} P(A) + P(A)P(B) \\ & \le (1+P(B)) P(\lim_{n\to\infty} \sup_{A \in \mathcal{A}_n\setminus \mathcal{A}} P(A)) \\ & = (1+P(B)) P(\emptyset) =0. \end{align}