Consider the infinite cartesian product $\Omega=\{0,1\}^{\mathbb{N}}$ as a measurable space endowed with the $\sigma$-algebra $\mathscr{F}$ generated by the cylinder sets and $\sigma:\Omega\to\Omega$ the left shift map. Denote by $\sigma^n(\mathscr{F})$ the $\sigma$-algebra generated by the family of r.v. $\{X_i(\omega)=\omega_i: i\geq n\}$. Suppose that $\mu$ is a Borel probability measure such $\mu(E)\in \{0,1\}$ for any $E\in \cap_{i\in\mathbb{N}}\sigma^i(\mathscr{F})$. I would like to know if $\mu(f|\cap_{i\in\mathbb{N}}\sigma^i(\mathscr{F}))(x)=\mu(f|\cap_{i\in\mathbb{N}}\sigma^i(\mathscr{F}))(\sigma x), \mu$ a.s. for any continuous function $f$.

In such generality I suspect that the answer is no, but I was not able to find a counter example. Of course, the answer is positive if $\mu(E)=\mu(\sigma^{-1}(E))$ for all $E\in \cap_{i\in\mathbb{N}}\sigma^i(\mathscr{F})$ (which is the case of $\sigma$-invariant measures). I spent some time trying to prove that the triviality hypothesis of $\mu$ implies the shift invariance (for events on the tail $\sigma$-algebra). This would be true if either $E=\sigma^{-1}(E)$ or both set have zero $\mu$ measure. For all the tail events I know this is true, but this does not sound reasonable statement to me in such generality.

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