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Let $(X,\mathcal B,\mu, T)$ be a measure preserving dynamical system on a standard Borel space. I'm trying to follow the construction of the natural extension by Michael Hochman on page 53 of his notes. This is very similar to the construction mentioned in this post, which unfortunately fails to answer the one question I have about the construction.

For convenience let me repeat the essentials of the construction: We consider $X^\mathbb{Z}$, with the normal product $\sigma$ algebra $\tilde{\mathcal{B}}$, ensuring that each projection $\pi_n$ is measurable. The Kolmogorov extension theorem guarantees the existence of a probability measure $\tilde\mu$ on $\tilde{\mathcal{B}}$, such that the pushforward $\pi_{n*}\tilde\mu=\mu$ for all $n\in\mathbb Z$. The left shift $\tilde T$ is invertible, measurable, and measure preserving on $(X^\mathbb{Z},\tilde{\mathcal{B}},\tilde\mu)$. Now let us define $$\tilde X=\{(x_n)\in X^\mathbb{Z}:Tx_n=x_{n+1},~n\in\mathbb Z\},$$ and redefine $\tilde{\mathcal{B}},\tilde\mu,$ and $\tilde T$ to be the relevant restrictions to $\tilde X^~$. Finally let us define $\pi:\tilde X\to X$ by $\pi=\pi_0|_{\tilde X}$.

The goal is to show that $\pi$ is a factor map. I am happy with showing that $\pi$ is measurable, measure preserving, and that $\pi\circ \tilde T=T\circ \pi$, but what I have been unable to do is prove that $\mu(\operatorname{im}\pi)=1$, which we need if we want to call $\pi$ a factor map. My idea is to show that the set $A$ consisting of all $x\in X$ for which there is an infinite chain in the graph of all preimages of $x$ has full measure. To be more precise: $A$ consists of all $x\in X$ such there exists a sequence $(x_n)_{n\in\mathbb N_0}$ in $X$ such that $Tx_{n+1}=x_n$ for all $n$, and $x_0=x$. If $A$ does have full measure then we would be done, because for every $x\in A$ it would follow that $(\dots,x_2,x_1,x,Tx,T^2x,\dots)\in \tilde X$. Now I believe that we could construct $A$ as a clever countable intersection of sets of full measure, but I am currently not clever enough to see it. If anyone can see it I'd be much obliged.

From scouring the interwebs I know that the existence of a natural extension of a dynamical system is closely related to the notion of the categorical inverse limit. Unfortunately category theory is one of the many need to be familiar with area of mathematics that I am not familiar with. I would still appreciate an answer phrased in terms of inverse limits (if it was explained in a gentle manner), because that can only help my familiarity with category theory. Any help is appreciated.

K.Power
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1 Answers1

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$\operatorname{im}\pi$ won't be measurable in general - e.g. take $X=[0,1]^3$ with uniform measure, and $\pi$ to be the identity except on $\{0\}\times [0,1]^2$ where it acts as your favourite measurable function with non-Borel image.

However, we can pick a full-measure set $S_k\subseteq \operatorname{im}T^k,$ then take the intersection. This will have full measure and infinite preimages. Take for example $S_k=\cup_n T^k(A_{k,n})$ where $A_{k,n}$ is a compact set as given by Lusin's theorem such that $\mu(A_{k,n})>1-1/n$ and $T^k$ is continuous on $A_{k,n}$ - note that $\mu(T^k(A_{k,n}))\geq \mu(A_{k,n})$ so $\mu(S_k)=1.$

Dap
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  • Ahh yes just because $\pi$ is measurable doesn't mean it's image is. Should have caught that. Thanks though, I think this works well – K.Power Mar 15 '18 at 15:05
  • Do you mind please expanding on your last statement that $\mu(T^k(A_{k,n}))\geq\mu(A_{k,n})?$ I have a feeling it might have something to do with analytic sets, but can't figure it out – K.Power Mar 15 '18 at 15:25
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    @K.Power: by the measure-preserving property, $\mu(T^{-k}(T^k(A_{k,n})))=\mu(T^k(A_{k,n})).$ So the inequality is just monotonicity of measures applied to $A_{k,n}\subseteq T^{-k}(T^k(A_{k,n})).$ – Dap Mar 15 '18 at 16:34
  • Simplicity is key. Thanks! – K.Power Mar 15 '18 at 16:55
  • Sorry I thought I followed your argument, but I now don't see how we can say that each $T^kA_{k,n}$ is Borel measurable? Because the space is standard Borel we know each $T^kA_{k,n}$ will be Lebesgue measurable, but is that enough? – K.Power Mar 15 '18 at 21:06
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    @K.Power: the image of a compact set is compact hence closed hence Borel. There is a gap though - Lusin's theorem only applies to $\mathbb R$ so you need to invoke Kuratowski's theorem that an uncountable standard Borel space is Borel isomorphic to $\mathbb R.$ – Dap Mar 16 '18 at 08:44
  • Oh I was thinking about the version of Lusin's theorem that I know, which only guarantees that the $A_{k,n}$ can be chosen to be closed, not necessarily compact, and so I misread your definition of them. If $X$ is locally compact though we can choose the $A_{k,n}$ exactly as you have. I wasn't originally thinking that $X$ was necessarily locally compact , but I'm happy to accept that as a condition (unless you don't actually use it in your proof, in which case I'd be curious as to how you choose compact $A_{k,n}$ then.) – K.Power Mar 16 '18 at 17:17