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Let $(E,\mathcal E,\lambda)$ be a probability space and $\lambda$ be a measurable map on $(E,\mathcal E)$ with $\lambda\circ\tau^{-1}=\lambda$.

I would like to show that $\tau$ is $\lambda$-ergodic by reducing this claim to the following well-known result: If $(E,\mathcal E)$ is a measurable space and $(Y_i)_{i\in\mathbb N}$ is an independent stationary $(E,\mathcal E)$-valued process on any probability space (we denote the law of $Y_i$ by $\mathcal L(Y_i)$ for $i\in\mathbb N$), then the coordinate process $(X_i)_{i\in\mathbb N_0}$ on $(\Omega,\mathcal A,\operatorname P):=(E^{\mathbb N},\mathcal B(E)^{\otimes\mathbb N},\bigotimes_{i\in\mathbb N}\mathcal L(Y_i))$ is independent and stationary. Thus, the shift $$\theta:\Omega\to\Omega\;,\;\;\;(x_i)_{i\in\mathbb N}\mapsto(x_{i+1})_{i\in\mathbb N}$$ is $\operatorname P$-ergodic.

Now assume $$\tau^i=\varphi\circ\theta^iY\;\;\;\text{for all }i\in\mathbb N.\tag1$$ for some $(\mathcal A,\mathcal E)$-measurable $\varphi:\Omega\to E$ and an $(E,\mathcal E)$-valued independent stationary process $(Y_i)_{i\in\mathbb N}$ on $(E,\mathcal E,\lambda)$.

Can we somehow use the identity $(1)$ to derive that $\tau$ is $\lambda$-ergodic from knowing that $\theta$ is $\operatorname P$-ergodic? I guess we only need to show that $\varphi$ is sufficiently nice for this conclusion to hold.

Feel free to assume that $(Y_i)_{i\in\mathbb N}$ is identically distributed, if necessary.

Remark: I've asked for the special case, where $(Y_i)_{i\in\mathbb N}$ is the dynamical system generated by the Bernoulli shift, back over on MSE: https://math.stackexchange.com/q/3657914/47771.

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1 Answer 1

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I think you’re asking whether a factor a Bernoulli process is ergodic. In fact it’s a simple fact that any factor of any ergodic process is ergodic.

You should think of $\phi$ as mapping the sequence of $Y$’s to the sequence of $\tau$’s. Now take any measurable shift-invariant subset of $\tau$ sequences. Its preimage is a measurable shift-invariant subset of $Y$-sequences. This therefore has measure 0 or 1. It follows that the subset of $\tau$ sequences has measure 0 or 1 as required.

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  • $\begingroup$ I'm not sure whether I can follow. Let $\mathcal I_\tau:=\{B\in\mathcal E:\tau^{-1}(B)=B\}$ and $\mathcal I_\theta:=\{A\in\Omega:\theta^{-1}(A)=A\}$. Are you claiming that if $B\in\mathcal I_\tau$, then $\varphi^{-1}(B)\in\mathcal I_\theta$? This would require $\varphi$ to be $(\mathcal I_\theta,\mathcal I_\tau)$-measurable. It is clearly $(\mathcal A,\mathcal E)$-measurable and since $\mathcal I_\theta\subseteq\mathcal A$, we get $\varphi^{-1}(B)\in\mathcal A$. What am I missing? $\endgroup$
    – 0xbadf00d
    Commented May 9, 2020 at 17:19
  • $\begingroup$ I've asked for that separately here: math.stackexchange.com/q/3667845/47771. Maybe I'm missing something. $\endgroup$
    – 0xbadf00d
    Commented May 10, 2020 at 10:31
  • $\begingroup$ I think the additional ingredient is the fact that $\theta\circ\phi=\phi\circ\theta$. Here the first $\theta$ is the shift on $\tau$ sequences; and the second $\theta$ is the shift on $Y$ sequences. The fact that the inverse image of an invariant set is invariant follows from this. $\endgroup$ Commented May 10, 2020 at 16:34
  • $\begingroup$ I'm sorry, we've obviously got different backgrounds, which is why I really struggle to even understand your terminology (and maybe you've misunderstood mine at some point). $\endgroup$
    – 0xbadf00d
    Commented May 10, 2020 at 18:52
  • $\begingroup$ Could you please write down explicitly what your left and right $\theta$'s are (and please use different symbols for them). Moreover, what do you mean by a "$\tau$-sequence" or a "$Y$-sequence"? I've no idea what you could mean with these terms, since $\tau$ is a measurable map on the space $(E,\mathcal E)$ which is not a product space. Having said that, $\varphi$ maps into this space $(E,\mathcal E)$ and hence I've no idea what could be "shifted" (which is why $\theta\circ\varphi$ doesn't make sense to me). $\endgroup$
    – 0xbadf00d
    Commented May 10, 2020 at 18:52

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