I am reading the book Conformal Fractals: Ergodic Theory Methods by Przytycki and Urbański (the book is available legally on a site of the first author https://www.impan.pl/~feliksp/ksiazka1.pdf ) and I have trouble understanding Theorem 1.9.1.
The statement is as follows:
Suppose that $\mathcal{A}$ and $\mathcal{B}$ are two measurable partitions of a Lebesgue space $(X, \mathcal{F}, \mu)$ such that $\mathcal{A} \cap B$ is countable ($\operatorname{mod}0$ with respect to $\mu_B$) for almost every $B \in \mathcal{B}$. Then there exists a countable partition $\Gamma = \{\gamma_1, \gamma_2, ...\}$ of $X$ $(\operatorname{mod}0)$ such that each $\gamma_j \in \Gamma$ intersects almost every $B$ at not more than one point, which is then an atom of $\mu_B$, in particular
$$\mathcal{A} \vee \mathcal{B} = \Gamma \vee \mathcal{B} \,\,(\operatorname{mod}0).$$
My problem is I don't see how this theorem can be true, which leads me to believe I'm misunderstanding some other concepts. Can someone point out to me what is wrong with my "counterexample": Let $X$ be the unit interval. Take $\mathcal{A}$ to be some partition of $X$ into two sets and $\mathcal{B}$ to be the partition into $1$ element, the whole set $X$. These are measurable partitions, of course $\mathcal{A} \cap B$ is countable (even finite), but how could there possibly be a partition $\Gamma$ from the statement? It would yield a partition of the unit interval into a countable set of points. Obviously I'm doing something wrong, help would be greatly appreciated, thanks.
You seem to be right; the phrase "each $\gamma_j \in \Gamma$ intersects almost every $B$ at not more than one point, which is then an atom of $\mu_B$, in particular" is true only when $\mathcal{A}$ is the point partition of $X$ (or all fiber measures of $\mathcal{B}$ are discrete). It seems the authors are following Parry's Entropy and Generators in Ergodic Theory (see p.40, Thm.4.6); where the method of proof is to replace $X$ with $X/\mathcal{A}\vee\mathcal{B}$; for spaces over $X/\mathcal{B}$ the partition $\mathcal{A}$ of $X$ is equivalent to the point partition of $X/\mathcal{A}\vee\mathcal{B}$. The construction of $\Gamma$ follows from another lemma of Rohlin, which says that $X\to X/\mathcal{B}$ has a measurable a.e. defined section whose image has maximal measure (among all measurable a.e. defined sections); see his paper "On The Fundamental Ideas Of Measure Theory", p.43. The entropy estimate part of the statement (also in Parry) is the more subtle part of the statement.
For the record, in your example $\Gamma=\mathcal{A}$ a.e.; the reason why it's a counterexample (to the part that is inaccurate) is because $X\to X/\mathcal{B}$ has the singleton as the base, so there is one fiber whose measure has no atoms.