Condition for a partition to be a generator in the context of Kolmogorov-Sinai entropy While studying about Kolmogorov-Sinai entropy, I found in Walkden's notes (https://personalpages.manchester.ac.uk/staff/charles.walkden/magic/lecture07.pdf) the following (page 9, just before Theorem 7.9):

Remark: To check whether a partition $\alpha$ (of a measurable space $(X,\mathcal{B},\mu)$ on which it is defined a measure preserving transformation $T$) is a strong generator it is sufficient to check that it separates almost every pair of points. That is, for almost every $x$, $y$ in the space $X$, there exists $n$ such that
$x$, $y$ are in different elements of the partition $$\bigvee_{j=0}^{n-1}T^{-j}\alpha.$$

Here I am assuming (correct me if that's not the right definition) that for the partition $\alpha$ to be a strong generator means that
$$\mathcal{B}=\sigma \left(\bigcup_{i=0}^{\infty}T^{-i}\alpha\right)\mod\mu.$$
However it is not at all clear to be why this should be true. Could someone please help me?
 A: I believe the answer can be found as Lemma 12.1 in

Coudène, Yves. Ergodic theory and dynamical systems. Springer, 2016.

Let me paraphrase the proof found there. Note that the condition you mention is equivalent to say that $\operatorname{diam}(\alpha_n(x)) \xrightarrow{n \to \infty} 0$, where we write $\alpha_n := \bigvee_{i = 0}^{n-1}T^{-1}\alpha$ and $\alpha(x)$ for the subset $A \in \alpha$ such that $x \in A$.
Now, let $U \subseteq X$ be open. By the given condition, for every $x \in U$ there exists $n \in \mathbb{N}$ such that $\alpha_n(x) \subseteq U$. Therefore we can write
$$
U = \bigcup_{n \in \mathbb{N}} \bigcup_{A \in \alpha_n, A \subseteq U} A.
$$
In words, we have written $U$ as a countable union of sets in the partitions $\alpha_n$, therefore $U$ is in the $\sigma$-algebra generated by $\alpha_\infty$ (which is  $\bigvee_{i = 0}^\infty T^{-i}\alpha = \sigma(\bigcup_{i = 0}^\infty T^{-i}\alpha)$. Thus, $\mathcal{B} \subseteq \sigma(\bigcup_{i = 0}^\infty T^{-i}\alpha)$ since $\mathcal{B}$ is the smallest $\sigma$-algebra containing the open sets, and the converse inclusion holds since $T$ is measurable and $\alpha$ is a measurable partition.
