Kolmogorov's zero-one law and triviality of tail sigma-algebra

Question

I study ergodic theory and K-automorphisms in particular. One of the equivalent definitions of $K$-system demands trivial tail sigma-algebra. So I came up with the following question:

$\textbf{Kolmogorov's zero-one law}$

Consider probability space $(\Omega,\mathcal A, \mathbb P)$ and let $\mathcal F_n$ be a sequence of independent sub-$\sigma$-algebras of $\mathcal A$. Consider $\sigma$-algebra $$\mathcal F= \bigwedge\limits_{n=1}^{\infty} \bigvee\limits_{k=n}^\infty \mathcal F_k$$

Then $\forall E\in \mathcal F$ either $\mathbb P(E)=0$ or $\mathbb P(E)=1.$

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So my question is: does this theorem imply triviality of $\mathcal F$, i. e. $\mathcal F=\{\varnothing, \Omega\}$?

For me it's almost obvious that no - because triviality is much more stronger property. But I don't know how to see it - may be there's an example in which I can see this sigma-algebra is not trivial, but still probabilities of all sets are either $0$ 0r $1$?

Thanks

Answer 1

Example: $\mathcal F_k=\mathcal F=\{\emptyset, A, A^{c},\Omega\}$ where $P(A)=0$. Indepedence hypothesis is satisfied in this example.

Answer 2

In ergodic theory, the trivial $\sigma$-algebra is not the algebra $\{\emptyset, \Omega\}$, it is the $\sigma$-algebra containing all negligible sets and their complements. The Kolmogorov 0-1 law asserts the triviality of the tail $\sigma$-algebra in that sense.

Two $\sigma$-algebras are said to be equal mod 0 if for any set $A$ in one of the algebras, there is a set $A'$ in the other such that $\mu(A \Delta A') = 0$. All equalities should be considered mod 0.