If $F$ is generated by a countable partition, what is $E(X|F)$ for a random variable $X$?
so we say that a $\sigma$-algebra $F$ is said to be generated by a partition if there is some partition $\{B_i\}$ of $\Omega$ so that every set $A$ in $F$ is a union of some parts in the partition, and every such union is in $F$.
I can't see why the fact that the partition is countable matter when we look at the expectancy.
Does anyone know?
Every member of your $\sigma$-algebra $F$ is countable union of elements from the partition. If function $f$ takes two different values in two different points $a,b$ belonging to the same element $A$ of the partition, $$f(a)=a', \quad f(b)=b' \neq a'$$ then inverse images $f^{-1}(\{a'\})$ and $f^{-1}(\{b'\})$ can not belong to $F$ (they would have to contain some elements of $A$ but not the others).
This means that every $F$-measurable function has to be constant on every member of the partition. In particular, conditional expectation can be written as $$E(X \mid F)=\sum_{i=0}^\infty c_i \chi_{A_i}$$ where $\mathcal C = \{A_1,A_2,\ldots\}$ is partition which generated $F$. Since $$E(E(X \mid F) \chi_{A_i}) = E(X \chi_{A_i}) = c_i P(A_i)$$ we get that $c_i = \displaystyle \frac{E(X \chi_{A_i})}{P(A_i)}$.