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.

  • $\begingroup$ Are there some $i$ and $n$ mixed up in your notation? $\endgroup$ – Nate Eldredge Jan 19 '16 at 19:39
  • $\begingroup$ I think there's some ambiguity in this question. The conditional expectation $\mu(f|\cap_{i\in\mathbb{N}}\sigma^i(\mathscr{F}))$ is only well-defined up to $\mu$-null sets; it doesn't identify a function on $\Omega$ but only an equivalence class in $L^0(\mu)$. Each representative $g$ of this equivalence class potentially leads to a different set $\{x : g(x) = g(\sigma x)\}$; for some choices of $g$ it might have full measure and for others it might not. $\endgroup$ – Nate Eldredge Jan 19 '16 at 23:59
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    $\begingroup$ In fact, since the tail $\sigma$-algebra is assumed almost trivial, we can always choose a representative $g$ which is a constant, in which case $\{x : g(x) = g(\sigma x)\} = \Omega$. $\endgroup$ – Nate Eldredge Jan 20 '16 at 0:00
  • $\begingroup$ @NateEldredge I guess your last comment answer the question. But I need to digest it because it implies that any probability measure trivial on "this" tail $\sigma$-algebra is shift invariant (its restricition to the tail $\sigma$-algebra) which seems very strong claim. Anyway thanks a lot for this clarification. $\endgroup$ – Leandro Jan 20 '16 at 1:02
  • $\begingroup$ No, it doesn't imply that; in fact the latter is false. I'll post an answer. $\endgroup$ – Nate Eldredge Jan 20 '16 at 1:57

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.

  • $\begingroup$ This construction is nice, but I did not understand why we can apply the zero-one law since the r.v. $(X_i)_{i\in\mathbb{N}}$ are not independent. Of course, the projections $Y_i\in \{0,1\}^2$ are independent and the theorem applies, but its tail $\sigma$-algebra is different from the tail $\sigma$-algebra generated by the $X_i$ variables (which seems richer). $\endgroup$ – Leandro Jan 20 '16 at 4:51
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    $\begingroup$ @Leandro: Actually, the tail $\sigma$-field of the $Y_i$ is identical to that of the $X_i$. Note $\sigma(Y_i : i \ge n) = \sigma^{2n-1}(\mathcal{F})$. Since the $\sigma$-algebras $\sigma^n(\mathcal{F})$ are decreasing, we have $\bigcap_n \sigma^n(\mathcal{F}) = \bigcap_n \sigma^{2n-1}(\mathcal{F})$. $\endgroup$ – Nate Eldredge Jan 20 '16 at 4:57
  • $\begingroup$ @Leandro: Alternatively, just note that the proof of the Kolmogorov 0-1 law goes through for the $X_i$ almost without change, if you use a little care on the cases of odd and even indices. $\endgroup$ – Nate Eldredge Jan 20 '16 at 5:03
  • $\begingroup$ I was just verifying that :) $\endgroup$ – Leandro Jan 20 '16 at 5:03

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