Ergodicity and Kolmogorov 0-1 law Consider the balanced coin tossing measure in $2=\{0,1\}$ and let $\mu$ be the product measure on the Cantor space $2^\mathbb{N}\cong (\mathbb{Z}_2)^\mathbb{N}$, defined on the Borel subsets of $2^\mathbb{N}$. I'd like to show that this measure is ergodic with respect to the action of the group $\bigotimes_{n\in\mathbb{N}} \mathbb{Z}_2$. In other words, given a Borel subset $A$ with the following property:

For all $x,y\in 2^\mathbb{N}$ which differ on only finitely many coordinates, i.e. $\exists m\forall n\ge m (x(n)=y(n))$, 
  $$x\in A\iff y\in A$$

$\mu(A)=0$ or $\mu(A)=1$. 
This has already been asked here, but that question received no answers. 
Fix $n\in\mathbb{N}$ and consider the $\sigma$-algebra
$$
\mathcal{F}_n:=\left\{\prod_i A_i\in \mathscr{B}(2^\mathbb{N}): \forall i\neq n (A_i=2)\right\}
$$
where $\mathscr{B}(2^\mathbb{N})$ is the Borel $\sigma$-algebra on the Cantor set. To apply the Kolmogorov 0-1 law as stated here, it would suffice to see that
$$A\in \sigma\left(\bigcup_{k\ge n} \mathcal{F}_k\right)$$
for all $n$. This is intuitively clear, since such an $A$ should not depend on its first $n$ coordinates for any $n$, but this is too hand-wavy for my tastes.
 A: It turns out this is an interesting question.  Henceforth let $\{R_{j}\}_{j \in \mathbb{Z}_{2}^{\oplus \mathbb{N}}}$ denote the action of $\mathbb{Z}_{2}^{\oplus \mathbb{N}}$ on $Z_{2}^{\mathbb{N}}$ given by 
\begin{equation*}
R_{j}(x_{1},x_{2},\dots) = (x_{1} + j_{1},x_{2} + j_{2},\dots),
\end{equation*}
where addition is understood modulo $1$.  I prove that if $A \subseteq \mathbb{Z}_{2}^{\mathbb{N}}$ is Borel measurable and $R_{j}^{-1}(A) = A$ independently of $j$, then 
\begin{equation*}
A \in \bigcap_{n = 1}^{\infty} \sigma \left(\bigcup_{k = n}^{\infty} \mathcal{F}_{k} \right).
\end{equation*}
That is, $A$ is a tail event.  Note that "being a tail event" does not depend on the measure we put on $\mathbb{Z}_{2}^{\mathbb{N}}$.  Therefore, if $\mu$ is any product measure on $\mathbb{Z}_{2}^{\mathbb{N}}$, the 0-1 law implies $\mu(A) \in\{0,1\}$.
However, to show that $A$ is a tail event, it is most convenient to pick the product measure $\mu_{2}$ satisfying $\mu_{2}(\{X_{j} = 1\}) = \frac{1}{2}$ for each $j \in \mathbb{N}$.  Note that $\{R_{j}\}_{j \in \mathbb{Z}_{2}^{\oplus \mathbb{N}}}$ preserves $\mu$.  (Henceforth I will let $\mathbb{P} = \mu_{2}$.)  Thus, if $a_{1},\dots,a_{n} \in \mathbb{Z}_{2}$, $j \in \mathbb{Z}_{2}^{\oplus \mathbb{N}}$, and $n \in \mathbb{N}$, then
\begin{equation*}
\mathbb{P}(A \cap \{X_{1} = a_{1},\dots,X_{n} = a_{n}\}) = \mathbb{P}(A \cap \{X_{1} = a_{1} + j_{1},\dots,X_{n} = a_{n}\}
\end{equation*}
since $A$ is an $R_{j}$-invariant set.  By fixing $n$ and varying $j$, we conclude that 
\begin{equation*}
\mathbb{P}(A \cap \{X_{1} = a_{1},\dots,X_{n} = a_{n}\}) = 2^{-n} \mathbb{P}(A).
\end{equation*}
In particular, $A$ is independent of $\sigma(\mathcal{F}_{1} \cup \dots \cup \mathcal{F}_{n})$.  
I claim that it follows that $A$ is $\sigma\left(\bigcup_{k \geq n + 1} \mathcal{F}_{k}\right)$-measurable.  For convenience, let $\mathcal{G}_{n} = \sigma\left(\bigcup_{k \geq n + 1} \mathcal{F}_{k}\right)$.  If $a_{1},\dots,a_{m} \in \mathbb{Z}_{2}$ and $m \geq n$, then
\begin{align*}
\mathbb{P}(A \cap \{X_{1} = a_{1},\dots,X_{m} = a_{n}) &= \mathbb{P}(A \cap \{X_{n + 1} = a_{n + 1},\dots,X_{m} = a_{m}\}) \mathbb{P}\{X_{1} = a_{1},\dots,X_{n} = a_{n}\} \\
&= \mathbb{E}(\mathbb{P}(A \mid \mathcal{G}_{n})) : X_{n + 1} = a_{n + 1},\dots, X_{m} = a_{m}) \mathbb{P}\{X_{1} = a_{1},\dots,X_{n} = a_{n}\} \\
&= \mathbb{E}(\mathbb{P}(A \mid \mathcal{G}_{n}) : X_{1} = a_{1},\dots,X_{m} = a_{m}).
\end{align*}
Since events of the form $\{X_{1} = a_{1},\dots,X_{m} = a_{m}\}$ for $m \geq n$ generate the Borel $\sigma$-algebra and $A$ is Borel measurable, it follows that $1_{A} = \mathbb{P}(A \mid \mathcal{G}_{n})$.  Therefore, $A \in \mathcal{G}_{n}$.  
Since the $n$ above was arbitrary, we conclude that $A \in \bigcap_{n = 1}^{\infty} \mathcal{G}_{n}$, which is the tail $\sigma$-algebra.  Thus, by the 0-1 law, if $\mu$ is any product measure on the Borel $\sigma$-algebra, then $\mu(A) \in \{0,1\}$.  We have used $\mu_{2}$ as a surrogate to establish this result.
Note that if $p \in [0,1] \setminus \{\frac{1}{2}\}$, then this proves that the action $\{R_{j}\}_{j \in \mathbb{Z}_{2}^{\oplus \mathbb{N}}}$ on $(\mathbb{Z}_{2}^{\mathbb{N}},\mathscr{B},\mu_{p})$ is a non-measure-preserving, "ergodic" group action.  Personally that seems interesting in itself.  Taking a step back, the fact that we used $\mu_{2}$ to prove $A$ is in the tail $\sigma$-algebra (which holds entirely independent of the choice of measure) is also quite nice.
