All,
In my lecture, I read the following statement. We are working on a probability space $(\Omega, \mathcal{F}, \mathbb{P})$. Let $\mathcal{F'}$ be a sigma-field generated by, say, two random variables $\mathcal{F'} = \sigma(X, Y)$. We are in a context in which we have been able to prove that for all events $B \in \mathcal{F}$, $\mathbb{P} (B \mid \mathcal{F'})$ is $X$-measurable, that is $\mathbb{P} (B \mid \mathcal{F'}) = \phi(X) \text{ a.s.}$ for some measurable function $\phi$. The statement says that it "readily implies that $\mathbb{P} (B \mid \mathcal{F'}) = \mathbb{P} (B \mid X)$".
Intuitively, this is quit clear: we only need the information provided by $X$ to compute $\mathbb{P} (B \mid \mathcal{F'})$.
More precisely, I have written the following to understand more rigorously this implication. By definition of conditional expectation, for all events $A \in \mathcal{F'}$, \begin{equation*} \begin{split} \mathbb{E}[1_B 1_A] & = \mathbb{E}[1_A \mathbb{P} (B \mid \mathcal{F'})] \\ & = \mathbb{E}[1_A \phi(X)]. \end{split} \end{equation*} In particular, we can identify $\phi$ for events $A' \in \sigma(X) \subset \mathcal{F'}$, since \begin{equation*} \begin{split} \mathbb{E}[1_B 1_{A'}] & = \mathbb{E}[1_{A'} \mathbb{P} (B \mid X)], \end{split} \end{equation*} and by unicity of cond. exp., $\phi(X) = \mathbb{P} (B \mid X)$.
I question this way of proceeding since $\mathbb{P} (B \mid \mathcal{F'})$ would "coincide" with $\mathbb{P} (B \mid X)$ when restricted to $\sigma(X)$ event if we do not have $\mathbb{P} (B \mid \mathcal{F'}) = \phi(X)$.
Thanks for any comment
