Let $(X, \mathcal{A})$ be a measurable space, $f: X \rightarrow X$ be a measurable dynamical system and $\mu$ a probability on $(X, \mathcal{A})$ such that $(f,\mu)$ is an invariant system.
Suppose $\mu(A \cap f^{-n}(B)) \xrightarrow{n} \mu(A) \mu(B), \forall A,B \in \mathcal{A} $.
Is $(f,\mu)$ somehow isomorphic to a Bernoulli shift*? If not, do we have a counterexample?
Thanks in advance. Lucas A.
*PS - what I mean by a Bernoulli shift is: Let $(Y, \mathcal{B}, \nu)$ be a probability space, $(Y^\mathbb{N}, \mathcal{B}^\mathbb{N}, \nu^\mathbb{N})$ be the associated product space and $\sigma: Y^\mathbb{N} \rightarrow Y^\mathbb{N}$ be given by $(y_n)_{n \ge 0} \mapsto (y_{n+1})_{n \ge 0}$. Then $(\sigma, \nu^\mathbb{N})$ is an invariant system called a Bernoulli shift.
This is not true. Look into Kolmogorov automorphisms. These are always mixing, but Ornstein $[1]$ provides an example of a system that is Kolmogorov, but not Bernoulli. There's undoubtedly an easier answer since not all mixing systems are Kolmogorov (K).
$[1]$ Ornestein, D. "An Example of a Kolomogorov Automorphism that is not a Bernoulli Shift." ADVANCES IN MATHEMATICS 10, 49-62 (1973)