both $\{X_i\}_{i\in I}$ (the index set $I$ is uncountable) and $\xi$ are integrable r.v. in $(\Omega,\mathcal F, \mathbb P)$
how to prove: there is a countable subet $J\subset I$ s.t. $$\mathbb E[\xi|\sigma(X_i,i\in I)]=\mathbb E[\xi|\sigma(X_i,i\in J)]\tag{1}$$
$\sigma(X_i,i\in I)$ means the $\sigma-$algebra generate by $(X_i,i\in I)$
I think the only tool I can use is the monotone class theorem for functions, so I collect all the rv $\xi$ that satisfy (1) to a set $\mathcal H$, if we can prove $\mathcal H$ contains all the measurable functions, then we get the conclusion.
but I finially found it doesn't solve the problem, since I can't fine why $(1)$ holds for indicator function $1_A$, where $A\in\mathcal F$.
Just an idea:
If $X_i \in L^2$ then using the fact that $L^2$ is separable Hilbert space there exist countable subset $J$ such that for any $Y \in L^2$ $$Y=\sum_{j \in J} \langle Y,X_j \rangle X_j$$
from this it should be possible to deduce that $$\sigma\left( X_i,i \in I\right) = \sigma\left( X_j,j \in I\right).$$