Let $(X_n)_{n \in \mathbb{N}}$ be a (strictly) stationary process and let $T$ denote the left-shift on $\mathbb{R}^\mathbb{N}$, i.e. $T((x_n)_{n \in \mathbb{N}}) = (x_{n + 1})_{n \in \mathbb{N}}$. Endow $\mathbb{R}^\mathbb{N}$ with the usual (product-)Borel-$\sigma$-algebra and the probability measure $P$ that is induced by the process $X$. Now $X$ is called ergodic, if the $\sigma$-algebra of $T$-invariant sets is $P$-trivial.
A set $A$ is $T$-invariant iff $T^{-1}(A) = A$. If I'm not mistaken, the identity $T^{-1}(A) = \mathbb{R} \times A$ holds. Intuitively I would have said that $A = \mathbb{R} \times A$ can only hold if $A = \emptyset$ or $A = \mathbb{R}^\mathbb{N}$ [but then ergodicity wouldn't be a special property, so this must be false].
- Can someone provide an example for a nonempty set $A \subsetneq \mathbb{R}^\mathbb{N}$ that satisfies $A = \mathbb{R} \times A$?
- Can someone provide a (simple) example for a stationary process that is not ergodic?
For a nontrivial $T$-invariant set you can take $$ A=\bigl\{(x_n)_{n\in\mathbb N}:x_n\not\in[0,1] \text{ for all sufficiently large $n$}\bigr\} $$ and many other variations. For an example of a nonergodic measure it is sufficient to consider a Markov measure that forbids transitions to $[0,1]$ in any number of steps. Then the former set $A$, although $T$-invariant, cannot have full measure and so the measure is not ergodic.