Theorem 7.1.3. Let $g:\Bbb R^{0,1,...} → \Bbb R$ be measurable. If $X_0,X_1,...$ is an ergodic stationary sequence, then $Y_k = g(X_k, X_{k+1}, . . .)$ is ergodic.
Proof. Suppose $X_0,X_1,...$ is defined on sequence space with $X_n(\omega) = \omega_n$. If $B$ has $\left\{\omega : \left(Y_0,Y_1,...\right) \in B\right\} = \left\{\omega : (Y_1,Y_2,...) \in B\right\}$ then $A = \left\{\omega : (Y_0, Y_1,...) \in B\right\}$ is shift invariant.
My question: I really cannot understand the proof. I understand every word in the proof however I fail to see how the $A$ is shift invariant implies $Y_k$ is ergodic. Could anyone help explain this? Thanks so much!
Since $A$ is shift invariant (for the shift on the sequence space) and the sequence $\left(X_n\right)_{n\geqslant 0}$ is ergodic, its probability is $0$ or $1$. This proves that an invariant event for the shift on the sequence $\left(Y_n\right)_{n\geqslant 0}$ and measurable with respect to $\sigma\left( Y_n,n\geqslant 0\right)$ has probability is $0$ or $1$, from which ergodicity of the sequence $\left(Y_n\right)_{n\geqslant 0}$ follows. Indeed, all the measurable events with respect to $\sigma\left( Y_n,n\geqslant 0\right)$ have the form $\left\{\omega\mid \left( Y_n\right)_{n\geqslant 0}\in B \right\}$ where $B\in\mathbb R^{\mathbb N}$.