I read the proof of lemma 2.11 here: http://www.personal.leeds.ac.uk/~mattbw/entropy/Ch2June2017.pdf#page=10.
To quote, it says:
Claim. Let $(X, B, \mu)$ be a Borel probability space, then $H_{\mu}(\mathcal C|\mathcal A) = \int H_{\mu_x^\mathcal A}(\mathcal C) d\mu$, where the integrand is infinite unless C agrees modulo $\mu_x^\mathcal A$ with a σ-algebra generated by a countable partition of finite entropy with respect to $\mu_x^\mathcal A$.
In the proof it says that we may reduce to the case where $\mathcal C = \sigma(\xi)$ where $\xi$ is a finite partition, by using the continuity and monotonicity in the first argument of $H(\cdot|\cdot)$. How is this reduction done?
It seems every countably-generated sigma algebra $\mathcal C$ is an increasing limit of $\sigma (\xi_n)$ where $\xi_n$ are finite partitions, because we can take just the first $n$ generators of $\mathcal C$ if they are and create a finite partition having all the mutual intersections of them and the complement of their union.
Then it seems we can indeed just say that the RHS of the equation is the limit of the RHS's of each element in the sequence.
But I don't see where this reasoning uses the additional information given, that the integrand is equivalent modulo each conditional measure to an algebra of a countable partition. What is missing from this reasoning to take account of possible infinities?