Why is an i.i.d. sequence of random variables indexed in discrete time, $\{X_t\}_{t \in \mathbb{Z}}$, ergodic? Here, we define ergodicity as follows:
Let $T$ be the left shift operator. i.e. $T( (..., X_{-1}, X_0, X_1, ...) ) = (..., X_0, X_1, X_2, ...)$. A set $A$ is called invariant if and only if $$A = \{ \{X_t\}_{t\in \mathbb{Z}} \in B\} = \{ T^n(\{X_t\}_{t\in \mathbb{Z}}) \in B\} \quad \quad \forall n \in \mathbb{Z}$$ for some $B$ = Borel set in $\mathbb{R}^\mathbb{Z}$. It is easy to show that the set of invariant sets is a sigma algebra, and we denote it by $\mathscr{I}_X$. A process $\{X_t\}_{t \in \mathbb{Z}}$ is ergodic if and only if its invariant sigma algebra is trivial; that is, $P(A) \in \{0,1\} \quad \forall A \in \mathscr{I}_X$
It is clear to me that when $X$ is indexed solely with $\mathbb{N}$ and we define shift operator $$T((x_0, x_1, x_2, ...)) = (x_1, x_2, x_3, ...) $$ then any event $A$ such that $$A = \{(X_0, X_1,...) \in B \} = \{T^n((X_0, X_1, ...)) \in B \} = \{(X_n, X_{n+1}, ...) \in B \} \in \sigma(X_n, X_{n+1}, ...)$$ for each $n$ and is thus in the tail sigma field by definition. Hence by Kolmogorov's 0-1 law we have $P(A) \in \{0,1\}$ and then the invariant sigma algebra for $X$ is clearly trivial so $X$ is ergodic by definition. However, for the $\mathbb{Z}$ indexed case, I cannot reconcile it in this way and I'm utterly and completely confused, so any help would be massively appreciated.