I am reading "Conformal fractals" by Przytycki, Urbański (available legally on the first author's site https://www.impan.pl/~feliksp/ksiazka1.pdf ) and have trouble understanding the sketch of the proof of Theorem 1.6.7, which starts on page 47 with the words "The method of construction [...]". This is the Rokhlin Disintegration Theorem about the existence of conditional measures with respect to a Rokhlin-measurable partition. I have seen the proof in some other sources, but I'd like to understand what the authors of this book had in mind.
They write:
Having chosen a basis $(B_n)$ of the Lebesgue space, for every finite intersection $$B = \bigcap B_i^{\varepsilon_{n_i}}$$ one considers $\phi _B = E(1_{B}|\tilde{A})$....
and
It is not hard to prove that for a.e. A, for each B, $\phi_B (A)$ as a function of $B$ generates the Lebesgue space $A$ , with $\mu_A (B) = \phi_B (A)$. Uniqueness of $\phi_B$ yields additivity.
Here $\tilde{A}$ is a sigma field generated by the partition, see notation 1.6.1.
My questions are what is the reason behind taking only a finite intersection of $B_i$, and how do the authors apply measurability of the partition in the proof? Why does uniqueness specifically yield additivity, isn't additivity obvious from basic conditional expectation properties? I guess i'm just lost at what exactly the authors had in mind.