Conditional expectation on countably generated $\sigma$-algebra

Question

Let $X$ be a random variable on $\mathscr L^1(\Omega, \mathscr F, \mathbb P)$.

Let $\{J_n\}_{n=1}^{\infty}$ be disjoint events with positive probability. Let $\mathscr J = \sigma(\{J_n\}_{n=1}^{\infty})$.

Q1 Is $\sum_{J \in \mathscr J} E[X|J]1_J$ a version of $E[X|\mathscr J]$?

Q2 Is $\sum_{n=1}^{\infty} E[X|J_n]1_{J_n}$ a version of $E[X|\mathscr J]$?

Both questions: It can be shown that $E[X|\mathscr J]$ is integrable. It is obvious that $E[X|\mathscr J]$ is $\mathscr J$-measurable.

Q1 It remains to show that $E[\sum_{J \in \mathscr J} E[X|J]1_J1_{J*}] = E[X1_{J*}]$ for all $J* \in \mathscr J$.

$$LHS = E[\sum_{J \in \mathscr J} E[X|J]1_J1_{J*}]$$

$$ = E[E[X|J*]1_{J*}1_{J*}]$$

$$ = E[E[X|J*]1_{J*}] = RHS$$

Q2 It remains to show that $E[\sum_{n=1}^{\infty} E[X|J_n]1_{J_n}1_{J*}] = E[X1_{J*}]$ for all $J* \in \mathscr J$.

Does it suffice to show the equality for $J* \in \sigma(J_1), \sigma(J_2), ...$

and then conclude by Dynkin's lemma or uniqueness lemma that we have the equality for all $J* \in \sigma(\sigma(J_1), \sigma(J_2), ...) \stackrel{?}{=} \mathscr J$

?

Answer 1

Q1: As I already pointed out in my comment, there is a measurability issue (because of the summation over possibly uncountably many events $J$) and therefore the mapping is, in general, not a version of $\mathbb{E}(X \mid \mathcal{J})$.

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Q2: If the sets $(J_n)_{n \in \mathbb{N}}$ cover the whole space, i.e. $\bigcup_{n \in \mathbb{N}} J_n = \Omega$, then the assertion follows from the following statement which is a direct consequence of a well-known uniqueness of measure theorem.

Proposition Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space, and let $\mathcal{J} \subseteq \mathcal{F}$ be a sub-$\sigma$-algebra. Let $\mathcal{G}$ be a generator of $\mathcal{J}$, i.e. $\sigma(\mathcal{G}) = \mathcal{J}$, such that $\mathcal{G}$ is $\cap$-stable and there exists $(G_n)_{n \in \mathbb{N}} \subseteq \mathcal{G}$ such that $G_n \uparrow \Omega$. If $X \in L^1(\mathcal{F})$ and $Y \in L^1(\mathcal{J})$ are such that $$\forall G \in \mathcal{G}: \quad \int_G X \, d\mathbb{P} = \int_G Y \, d\mathbb{P},$$ then $Y$ is a version of $\mathbb{E}(X \mid \mathcal{J})$.

You will have to apply this theorem for the generator

$$\mathcal{G} := \left\{ \bigcup_{i=1}^k J_{n_i}; k \in \mathbb{N}, n_i \in \mathbb{N} \right\}$$

which consists of all sets which can be obtained as finite unions of the sets $(J_n)_{n \in \mathbb{N}}$.

If $\bigcup_{n \in \mathbb{N}} J_n = \Omega$ does not hold true, then the assertion is, in general, wrong; it is not difficult to construct counterexamples. (Consider for instance $\mathcal{J} = \{A,A^c\}$ and $J_n := A$ for $n \in \mathbb{N}$.)