Ergodicity of a sequence of independent blocks I am stuck with the problem given below, more precisely, with the part regarding ergodicity. I have a proof, also given in what follow, but it does not seem to be correct; well, at least, it does not sound like a solid mathematical proof.
Problem (Independent blocks, Exercise 7.1.6 in Durrett):
Let $X_1, X_2, \dotsc$ be a stationary sequence. Let $n < \infty$ and let $Y_1, Y_2, \dotsc$ be a sequence so that $(Y_{nk+1}, \dotsc, Y_{n(k+1)})$, $k \geq 0$ are i.i.d. and $(Y_1, Y_2, \dotsc, Y_n) = (X_1, X_2, \dotsc, X_n)$. Finally
let $\nu$ be uniformly distributed on $\{1, 2, \dotsc, n\}$, independent of Y, and let $Z_m = Y_{\nu + m}$ for $m \geq 1$. Show that Z is stationary and ergodic.
Proof (ergodicity only):
We know that, for an i.i.d. sequence, Kolmogorov's 0-1 law implies that the tail $\sigma$-field is trivial (Example 7.1.6 in Durrett). Therefore, the tail $\sigma$-field of the process formed by the (independent) blocks of the $Y$ sequence is trivial.
This tail $\sigma$-field is
\begin{align*}
  \mathcal{T}_Y & = \cap_{k = 1}^\infty \sigma(Y_{nk + 1}, \dots, Y_{n(k + 1)}, Y_{n(k + 1) + 1}, \dots, Y_{n(k + 2)}, \dots) \\
  & = \sigma(Y_{n + 1}, \dots) \cap \sigma(Y_{2n + 1}, \dots) \cap \sigma(Y_{3n + 1}, \dots) \cap \dots \\
  & = \mathcal{F}_{n + 1} \cap \mathcal{F}_{2n + 1} \cap \mathcal{F}_{3n + 1} \cap \dots
\end{align*}
Now, let us have a look at the tail $\sigma$-field of the $Z$ sequence:
\begin{align*}
  \mathcal{T}_Z & = \cap_{k = 1}^\infty \sigma(Z_k, Z_{k + 1}, \dots) = \cap_{k = 1}^\infty \sigma(Y_{\nu + k}, Y_{\nu + k + 1}, \dots) \\
  & = \sigma(Y_{\nu + 1}, \dots) \cap \sigma(Y_{\nu + 2}, \dots) \cap \sigma(Y_{\nu + 3}, \dots) \cap \dots \\
  & = \mathcal{F}_{\nu + 1} \cap \mathcal{F}_{\nu + 2} \cap \mathcal{F}_{\nu + 3} \cap \dots
\end{align*}
It can be seen that the $Z$ sequence of $\sigma$-algebras is seamless: depending on the outcome of $\nu$, it starts from $\mathcal{F}_{\nu + 1}$ (from $\mathcal{F}_{n + 1}$ in the worst case) and has not further gaps. On the other hand, the $Y$ sequence of $\sigma$-algebras starts from exactly $\mathcal{F}_{n + 1}$ and does have gaps. The more terms involved in an intersection, the smaller the resulting set. Hence, $\mathcal{T}_Z \subset \mathcal{T}_Y$, and, therefore, the $Z$ sequence is ergodic.
The idea here is to prove the ergodicity property of the $Z$ sequence by showing that the tail $\sigma$-field of this sequence is trivial, and this is done by showing that this tail $\sigma$-field is contained in the tail $\sigma$-field of the sequence formed by the independent blocks of the $Y$ sequence.
I would appreciate any suggestions.
Regards,
Ivan
 A: The sequence $(Y_k)$ is $n$-dependent because the blocks of $n$ are independent. Using the fact that $\nu$ is independent of $Y$, we get that $Z$ is $2n$-independent. Then we use a similar argument as in the zero-one law. 
A: I would be grateful if somebody could confirm that the following proof is correct.
Proof (ergodicity only):
By definition, the $Z$ sequence is ergodic if
$$
  P((Z_1, Z_2, \dotsc) \in B) \in \{0, 1\}
$$
whenever $B \in \mathcal{R}^{\{0, 1, \dotsc\}}$ has
$$
  \{ (Z_1, Z_2, \dotsc) \in B \} = \{ (Z_2, Z_3, \dotsc) \in B \}. \tag{$\ast$}
$$
Let $A = \{ (Z_1, Z_2, \dotsc) \in B \}$, which is shift invariant according to ($\ast$).
We are to show now that $P(A) \in \{0, 1\}$.
By iteratively applying the shift operator to ($\ast$), $A \in \sigma(Z_k, Z_{k + 1}, \dotsc)$ for $k \geq 1$; hence,
$$
  A \in \cap_{k = 1}^\infty \sigma(Z_k, Z_{k + 1}, \dotsc) = \mathcal{T}_Z, \text{ the tail $\sigma$-field of the sequence.}
$$
Next, for any $B \in \mathcal{R}$, due to the independence of $Y$ and $\nu$,
$$
  \{ Z_m \in B \} = \{ Y_{m + \nu} \in B \} = \cup_{k = 1}^n \{ \nu = k \} \cap \{ Y_{m + k} \in B \} \subseteq \cup_{k = 1}^n \{ Y_{m + k} \in B \}.
$$
Thus,
$$
  \sigma(Z_m) \subseteq \sigma(Y_{m + 1}, \dotsc, Y_{m + n}).
$$
Consequently,
$$
  \mathcal{T}_Z \subseteq \mathcal{T}_Y = \cap_{k = 1}^\infty \sigma(Y_k, Y_{k + 1}, \dotsc), \text{ the tail $\sigma$-field of the block sequence.}
$$
Finally, by Kolmogorov's zero-one law, the tail $\sigma$-field of the block sequence, $\mathcal{T}_Y$, is trivial.
Hence, $P(A) \in \{0, 1\}$ as $A \in \mathcal{T}_Z \subseteq \mathcal{T}_Y$.
