# Extend a formula of conditional expectation from finitely to countably generated $\sigma$-algebra

I'm reading an exercise in this lecture note, i.e.,

Exercise 2. Suppose that the $$\sigma$$-algebra $$\mathscr{G}$$ is finite, that is, suppose that there is a finite measurable partition $$B_1, B_2, \ldots, B_n$$ of $$\Omega$$ such that $$\mathscr{G}$$ is the $$\sigma$$-algebra generated by the sets $$B_i$$. Show that for any $$X \in L^2(\Omega, \mathscr{F}, P)$$, $$E(X | \mathscr{G})=\sum_{i=1}^n \frac{E\left(X \mathbf{1}_{B_i}\right)}{P\left(B_i\right)} \mathbf{1}_{B_i} \quad \text {a.s.}$$

I have found a related lemma from Erhan Çınlar's Probability and Stochastics, i.e.,

Lemma: Let $$\mathcal A = (A_i)_{i\in I}$$ be a countable partition of $$X$$ such that $$\emptyset \in \mathcal A$$. Then $$\sigma (\mathcal A) = \left \{ \cup_{i \in J} A_i \,\middle\vert\, J \subset I \right\}.$$

It seems from the Lemma that Exercise 2 extends naturally to countably infinite measurable partition of $$\Omega$$. Could you confirm if my understanding is correct?

If we define:$$Y:=\sum_{i\in I}\mathbb{E}\left[X\mid A_{i}\right]\boldsymbol{1}_{A_{i}}$$then $$Y$$ is measurable wrt $$\sigma(\mathcal A)$$ and for every $$J\subseteq I$$: $$\mathbb{E}[Y\boldsymbol{1}_{\bigcup_{i\in J}A_{i}}]=\mathbb{E}\left[\sum_{i\in J}\mathbb{E}\left[X\mid A_{i}\right]\boldsymbol{1}_{A_{i}}\right]=\sum_{i\in J}\mathbb{E}\left[X\mid A_{i}\right]\mathbb{E}\boldsymbol{1}_{A_{i}}=$$$$\sum_{i\in J}\mathbb{E}\left[X\mid A_{i}\right]P\left(A_{i}\right)=\sum_{i\in J}\mathbb{E}\left(X\boldsymbol{1}_{A_{i}}\right)=\mathbb{E}\left[\sum_{i\in J}X\boldsymbol{1}_{A_{i}}\right]=\mathbb{E}[X\boldsymbol{1}_{\bigcup_{i\in J}A_{i}}]$$
Proved is now that $$\mathbb{E}\left[Y\boldsymbol{1}_{A}\right]=\mathbb{E}\left[X\boldsymbol{1}_{A}\right]$$ for every $$A\in\sigma(\mathcal A)$$ and that is enough to conclude that: $$Y=\mathbb E[X\mid \mathcal A]$$
Eventual situations in which $$P(A_i)=0$$ can be neglected without essential harm.
The lemma looks a bit weird because if $$\mathcal A$$ is a partition then by definition $$\varnothing\notin\mathcal A$$. Leaving the condition $$\varnothing\in\mathcal A$$ out gives no obstacle because we can take $$J=\varnothing$$ to get $$\varnothing$$ as an element of $$\sigma(\mathcal A)$$.