Shift Invariance of Backward Martingales for tail trivial probability measures 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. 
 A: As discussed in comments, the main question is not well-posed as it stands, since it depends on a choice of representative (mod $\mu$-a.e. equality) for the conditional expectation.
But as to your subsidiary question "Must $\mu$ be shift-invariant on tail events?", the answer is no.
Let $\nu$ be the probability measure on $\{0,1\}^2$ which assigns probability $1/2$ each to the outcomes $(0,0)$ and $(1,1)$.  Let $\mu = \prod_1^\infty \nu$ be the product of infinitely many $\nu$.  So under $\mu$, the random variables $X_1, X_3, X_5, \dots$ are iid coin flips, and a.s. we have $X_1 = X_2$, $X_3 = X_4$, etc.  (You can think of the "experiment" here as "flip infinitely many coins, but write down each flip twice".)  Then the tail $\sigma$-algebra is $\mu$-almost trivial, by the Kolmogorov 0-1 law.
Let $E_i$ be the event $\{X_{2i-1} = X_{2i}\}$ and $E = \{E_i \text{ a.a.}\}= \bigcup_{n=1}^\infty \bigcap_{i=n}^\infty E_i$ be the event that $X_{2i-1} = X_{2i}$ for all but finitely many $i$.  Clearly $E$ is a tail event and $\mu(E) = 1$.  But $\sigma^{-1}(E)$ is the event $F = \{F_i \text{ a.a.}\}$, where $F_i = \{X_{2i} = X_{2i+1}\}$.  The $F_i$ are independent with $\mu(F_i) = \frac{1}{2}$, so $\mu(F) = 0$.
To highlight the issue in your question, if we write $\mathcal{T}$ for the tail $\sigma$-algebra and take $f$ to be the constant function $1$, it is correct to say $\mu(f \mid \mathcal{T}) = 1_E$ a.s.  The statement "$1_E(x) = 1_E(\sigma x)$ a.s." is false.  But it is also correct to say $\mu(f \mid \mathcal{T}) = 1$ a.s.  The statement "$1(x) = 1(\sigma x)$ a.s." is true.  The definition of conditional expectation says nothing about whether $1_E$ or $1$ is a "preferred" representative.
