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