= Solution
Let $\mathcal G=\sigma(A_1,A_2,\ldots)$. 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 $n\geq N$,
$$
\mathbb E[X_n\mathbf1_A]=\mathbb E[X\mathbf1_A].
$$
The $L^1$ convergence lets us pass to the limit and obtain $\mathbb E[Y\mathbf1_A]=\mathbb E[X\mathbf1_A]$.
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.
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