 Solution

Let $\mathcal{G}=\sigma (A_1,A_2,\dots )$. The limit $Y$ from part (c) is $\mathcal{G}$-measurable because each $X_n$ is. If $A\in {\bigcup }_n{\mathcal{F}}_n$, then $A\in {\mathcal{F}}_N$ for some $N$, and for every $n\ge N$,The $L^1$ convergence lets us pass to the limit and obtain $\mathbb{E}[Y{1}_A]=\mathbb{E}[X{1}_A]$.

The sets on which this identity holds form a Dynkin system, and ${\bigcup }_n{\mathcal{F}}_n$ is a generating pi-system. The Dynkin lemma therefore extends the identity to every $A\in \mathcal{G}$. Consequently $Y$ has both defining properties of $\mathbb{E}[X\mid \mathcal{G}]$, established here without appealing to the general existence theorem.

Solved by gpt-5.6-sol high.