Consider $(E, \mathcal E)$ – measurable space, the space $\Omega = E^\mathbb N$ equipped with product σ-sigma algebra $\mathcal F = \otimes_{n} \mathcal E$. Next define projections $\pi_n(\omega) = \omega_n$, their respective σ-algebras $\mathcal F_n = \sigma (\pi _n)$, and their generated tail σ-algebra by
$$\mathcal T _{\infty} = \cap _{n\in \mathbb N}\sigma (\cup_{m \geq n} \mathcal F _m)$$
Define the left-shift operator $\tau : \Omega \rightarrow \Omega$ by $(\pi_n \circ \tau) (\omega)= \omega_{n+1}$ and left-shift invariant σ-algebra
$$\mathcal I = \{A\in \mathcal F : \tau ^{-1}(A) = A\}$$
We want to show that $\mathcal I \subset \mathcal T _\infty$.
My attempt: I tried to prove it by induction: we need to show that for a fixed $A\in \mathcal I$ it also has to be in $\sigma (\cup_{m \geq n} \mathcal F _m)$ for all $n \in \mathbb N$. I showed the inductive step $n \rightarrow n+1$ by proving the statement for the sets that generate $\sigma (\cup_{m \geq n} \mathcal F _m)$ of the form $\pi ^{-1}_k(B)$ for $k \geq n$ and $B \in \mathcal E$ (since $\tau^{-1}\circ \pi _n^{-1} = \pi _{n+1} ^{-1}$). However, I struggle to show the inductive basis $A \in \sigma (\cup_{m \geq 1} \mathcal F _m)$. So I am not sure if my approach is viable.
Help appreciated.
Best Answer
We have $$\mathcal{F}=\bigotimes_n\mathcal{E}=\sigma(\pi_1,\pi_2,\pi_3,\dots)=\sigma\left(\bigcup_n\sigma(\pi_n)\right)=\sigma\left(\bigcup_n\mathcal{F}_n\right)$$ So $A\in\mathcal{F}\iff A\in\sigma\left(\bigcup_{m\geq 1}\mathcal{F}_m\right)$.