# Does every measure-preserving dynamical system admit a backward orbit?

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?

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.
• 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.) – Julian Newman Mar 12 '15 at 19:22
• 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 $X_1=\mathbb{N}$, $X_2=\{(m,n)\in\mathbb{N}^2:m\leq n\}$, $X=X_1\cup X_2$, $\Sigma=\sigma(\{x\}:x\in X_2)$, $\mu(X_1)=1$, $T(n)=n+1$, $T(1,n)=1$, $T(m+1,n)=(m,n)$.) – Julian Newman Mar 13 '15 at 2:27