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?
I believe the answer can be found as Lemma 12.1 in
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.