Let $(X, \mathcal X, \mu)$ be a probability space and let $T:X\to X$ be a measure preserving map (note that $T$ may not be surjective).
Define $\tilde X=\{(x_k)_{k\in \mathbb Z}\in X^{\mathbb Z}:\ T({x_k})=x_{k+1} \text{ for all } k\in \mathbb Z\}$.
Question. Is $\tilde X$ non-empty?
A friend of mine suggested an idea which requires an additional assumption.
Assume that each $T^n(X)$ is a measurable set in $X$. It is clear that each $T^n(X)$ is of full measure. Define $B_n=T^{n}(X)\setminus T^{n+1}(X)$. Thus $B_n$ has measure $0$. Therefore $B=\bigcup_{n\geq 0} B_n$ is of measure $0$. Now if $x_0\in X\setminus B$ we have members $x_{-1}, x_{-2}, x_{-3}, \ldots\in X$ such that $T({x_{i}})=T({x_{i+1}})$ for each $i< 0$. Define $x_k=T^k(x_0)$ for each $k>0$. The sequence $(x_k)_{k\in \mathbb Z}$ is in $\tilde X$.
Can someone see how to get rid of the extra assumption. Thanks.
Often, the first place to look for ill-behaved examples is to take the silly sigma-algebra, $\mathcal{X} = \{ \emptyset, X \}$. Then every map from $X$ to itself is measurable and giving a probability measure is just specifying some number $\mu(X)$. In that case every map $X \to X$ is measure-preserving (with respect to any measure on $\mathcal{X}$), so there are actually no restrictions on what $T$ can be.
In that case it is easy to find examples where $\tilde{X}$ is empty. For example, take $X = \mathbb{N} \cup \{ 0 \}$ and $T(n) = n+1$. Then every sequence $(x_k)$ in $\tilde{X}$ would need to take the value $0$ at some point, but since $0$ is not in the image you can go no further back in time.