Characterization of tail $\sigma$-algebra and symmetries of $\{0,1\}^{\mathbb{N}}$ Recently the following question was asked:

If a measurable subset of $\{0,1\}^{\mathbb{N}}$ (with the Borel
  $\sigma$-algebra) is invariant under the action of
  $\mathbb{Z}_{2}^{\oplus \mathbb{N}}$ (via pointwise addition mod 2),
  is it in the tail $\sigma$-algebra?

This was answered in the affirmative.  However, it was the first time I have seen $\mathbb{Z}_{2}^{\oplus \mathbb{N}}$ acting on $\{0,1\}^{\mathbb{N}}$ in a probabilistic context so it begged the question: is every tail event invariant under this action?  
This leads to a similar question: what about the action of $\mathbb{N}$?  That is to say, what about the shift?  This action is a pretty important example in ergodic theory, and many of the tail events one finds in practice are, in fact, shift-invariant.  Thus, one might ask: is every tail event shift invariant?  A harder question is: is the tail $\sigma$-algebra generated by shift invariant set?  
 A: Below I will prove: 

Every tail event is $\mathbb{Z}_{2}^{\oplus \mathbb{N}}$-invariant so this
  characterizes the tail $\sigma$-algebra.  On the other hand, there are
  tail events that are not invariant under the shift.  

I haven't put much thought into the question of whether or not shift-invariant sets generate the tail $\sigma$-algebra.
Denote by $T : \{0,1\}^{\mathbb{N}} \to \{0,1\}^{\mathbb{N}}$ the shift, that is, $T(x_{1},x_{2},\dots) = (x_{2},x_{3},\dots)$.  To see there is a tail event that is not shift-invariant, fix a sequence $(\bar{x}_{1},\bar{x}_{2},\dots)$ and let $A$ be the set given by 
\begin{equation*}
A = \bigcup_{N = 1}^{\infty} \{x \in \{0,1\}^{\mathbb{N}} \, \mid \, \forall j \geq N \, \, x_{j} = \bar{x}_{j}\}.
\end{equation*}
Observe that, for each $K \in \mathbb{N}$, 
\begin{equation*}
A = \bigcup_{N = K}^{\infty} \{x \in \{0,1\}^{\mathbb{N}} \, \mid \, \forall j \geq N \, \, x_{j} = \bar{x}_{j}\}.
\end{equation*}
Thus, $A \in \sigma(X_{K},X_{K + 1},\dots)$.  Since this is true regardless of the choice of $K$, it follows that $A$ is a tail event.  However, $T^{-1}(A)$ is easily characterized:
\begin{equation*}
T^{-1}(A) = \bigcup_{N = 1}^{\infty} \{x \in \{0,1\}^{\mathbb{N}} \, \mid \, \forall j \geq N \, \, x_{j + 1} = \bar{x}_{j}\}.
\end{equation*}
Thus, $\bar{x} \in A \setminus T^{-1}(A)$ and $(0,\bar{x}_{1},\bar{x}_{2},\dots) \in T^{-1}(A) \setminus A$ so $A$ is far from $T$-invariant.  
On the other hand, suppose $A$ is any tail event.  Given $\nu \in \mathbb{Z}_{2}^{\oplus \mathbb{N}}$, I will abuse notation and also define $\nu : \{0,1\}^{\mathbb{N}} \to \{0,1\}^{\mathbb{N}}$ by 
\begin{equation*}
\nu(x_{1},x_{2},\dots) = (x_{1} + \nu_{1},x_{2} + \nu_{2},\dots),
\end{equation*}
where addition is modulo $2$.  Now since $A$ is a tail event, it follows that, if $K \in \mathbb{N}$, then $A \in \sigma(X_{K + 1},X_{K + 2},\dots)$.  Thus, one can show that if $\pi_{K} : \{0,1\}^{\mathbb{N}} \to \{0,1\}^{\{K + 1, K +2,\dots\}}$ is the projection given by $\pi_{K}(x_{1},x_{2},\dots) = (x_{K + 1},x_{K + 2},\dots)$, then there is a Borel measurable set $A_{K} \in \{0,1\}^{\{K + 1,K + 2,\dots\}}$ such that $A = \pi_{K}^{-1}(A_{K})$.  
Given $\nu \in \mathbb{Z}_{2}^{\oplus \mathbb{N}}$, let $L(\nu) = \max\{j \in \mathbb{N} \, \mid \, \nu_{j} \neq 0\}$.  Observe that $\pi_{L(\nu)} = \pi_{L(\nu)} \circ \nu$.  Thus, we immediately obtain
\begin{equation*}
\nu^{-1}(A) = \nu^{-1}(\pi_{L(\nu)}^{-1}(A_{L(\nu)})) = \pi_{L(\nu)}^{-1}(A_{L(\nu)}) = A.
\end{equation*}
In particular, $A$ is $\nu$-invariant.  Since our choice of $\nu$ was arbitrary, we conclude that $A$ is $\mathbb{Z}_{2}^{\oplus \mathbb{N}}$-invariant.  Therefore, the tail $\sigma$-algebra equals the collection of Borel measurable $\mathbb{Z}_{2}^{\oplus \mathbb{N}}$-invariant events.  
