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This seems like a really basic question, and yet I haven't managed to find the answer!

Let $(X,\Sigma,\mu,T)$ be a measure-preserving dynamical system. Does there necessarily exist at least one sequence $(x_n)_{n \in \mathbb{N}}$ of points in $X$ such that $T(x_{n+1})=x_n$ for all $n \in \mathbb{N}$?

If not, what about in the particular case that $(X,\Sigma)$ is a standard measurable space?

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This fact is very well known and is used to define the so-called natural extension of a non-invertible dynamical system: its state space is precisely the space of bilateral $T$-orbits (i.e., of sequences $(x_n)_{n=-\infty}^\infty$ with $Tx_n=x_{n+1}$ for any $n$). For its construction one has to consider the measures $\mu_N (N\in\mathbb Z)$ on the space of $T$-orbits in $X$ starting at time $N$ which are the images of $\mu$ under the maps $x \mapsto (x_N=x,x_{N+1}=Tx, \dots)$ and to apply the Kolmogorov consistency theorem.

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    $\begingroup$ Thank you. When you say "this fact is very well known", I presume you just mean in the case where $(X,\Sigma,\mu)$ is a standard probability space? I've realised that the statement is trivially false in the completely general case: take $X=\mathbb{N}$, take $\Sigma$ to be the trivial $\sigma$-algebra $\{\emptyset,\mathbb{N}\}$, take $\mu$ to be the trivial probability measure, and take $T:x \mapsto x+1$. Do you know any references that prove the fact in the case where $(X,\Sigma)$ is standard? (I've looked at numerous references on natural extensions, but they don't prove this fact.) $\endgroup$
    – Julian Newman
    Commented Mar 12, 2015 at 19:22
  • $\begingroup$ Oh well - the argument I was referring to is actually based on the assumption that the set of points with empty preimages is measurable (which is obviously not the case in your example). Otherwise it's really just an application of the Kolmogorov theorem (equivalently, existence of projective limits in the category of "reasonably" probability spaces). For more details see, for instance, section 10.4 "Endomorphisms and their natural extensions" of the book by Cornfeld, Fomin, Sinai. $\endgroup$
    – R W
    Commented Mar 12, 2015 at 22:40
  • $\begingroup$ Thank you for this. (I don't think I have access to this book - but I can see now how the Kolmogorov argument works anyway.) Sorry, I'm a bit of a perfectionist: if you would like me to officially Accept your answer, could you please first edit it to make mention of the fact that your positive answer to the question is specific to "reasonable probability spaces"; otherwise, future readers may be misled into thinking that your statement "This fact is very well known" is true of my general question "Let ... be a measure-preserving dynamical system - does there necessarily exist ...?" $\endgroup$
    – Julian Newman
    Commented Mar 13, 2015 at 2:04
  • $\begingroup$ PS: In the first sentence of your comment, are you suggesting that the statement should be true in any case where $T(X)\in\Sigma$? (Surely this isn't the case: in my example, you can just change $\Sigma$ to be $\sigma(\{1\})$ and have $\mu(\{1\})=0$. In fact, even if we assume that $T^n(X)$ is measurable for all $n\in\mathbb{N}$, the statement still isn't true in full generality: take $X_1=\mathbb{N}$, $X_2=\{(m,n)\in\mathbb{N}^2:m\leq n\}$, $X=X_1\cup X_2$, $\Sigma=\sigma(\{x\}:x\in X_2)$, $\mu(X_1)=1$, $T(n)=n+1$, $T(1,n)=1$, $T(m+1,n)=(m,n)$.) $\endgroup$
    – Julian Newman
    Commented Mar 13, 2015 at 2:27

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