Conditional expectation by $\sigma (G_n,Y)$ when $Y$ is $G_\infty$-measurable Let $G_n$ be a filtration (an increasing sequence of sigma-algebras), $Y$ a random variable that is $G_\infty$-measurable, and $X$ a random variable. Is it true that in $L^2$-norm,
$$ \mathbb{E}[X \mid \sigma (G_n,Y)]- \mathbb{E}[X \mid G_n] \stackrel{n \rightarrow + \infty}{\rightarrow} 0 $$
 A: 
Lemma Let $X \in L^2$ and $(\mathcal{G}_n)_{n \in \mathbb{N}}$ a filtration. Then $$\lim_{n \to \infty} \mathbb{E}(X \mid \mathcal{G}_n) = \mathbb{E}(X \mid \mathcal{G}_{\infty}) \quad \text{in} \, L^2$$ where $\mathcal{G}_{\infty} := \sigma(\mathcal{G}_n; n \in \mathbb{N})$.

Proof: Set $X_n := \mathbb{E}(X \mid \mathcal{G}_n)$. Obviously, $(X_n,\mathcal{G}_n)_{n \in \mathbb{N}}$ is a martingale. Since the martingale is bounded in $L^2$, the martingale convergence theorem implies $X_n \to Y$ in $L^2$ for some random variable $Y \in L^2$. It remains to show that $Y= \mathbb{E}(X \mid \mathcal{G}_{\infty})=:X_{\infty}$. By the tower property, we have $$\mathbb{E}(Y \mid \mathcal{G}_n) = X_n = \mathbb{E}(X \mid \mathcal{G}_n) = \mathbb{E}(X_{\infty} \mid \mathcal{G}_n)$$
Hence, for any $G \in \bigcup_{n \in \mathbb{N}} \mathcal{G}_n$: $$\int_G Y \, d\mathbb{P} = \int_G X_{\infty} \, d\mathbb{P}$$ Since $\bigcup_{n \in \mathbb{N}} \mathcal{G}_n$ is a generator of $\mathcal{G}_{\infty}$ which is stable under intersections, this implies
$$\int_G Y \, d\mathbb{P} = \int_G X_{\infty} \, d\mathbb{P}$$
for any $G \in \mathcal{G}_{\infty}$. By choosing $G := [\pm (X_{\infty}-Y)>0]$, we find $X_{\infty} = Y$ a.s. This finishes the proof.

The lemma shows that $\mathbb{E}(X \mid \mathcal{G}_n) \to \mathbb{E}(X \mid \mathcal{G}_{\infty})$ in $L^2$. Applying the lemma for the filtration $\mathcal{H}_n := \sigma(\mathcal{G}_n,Y)$ yields $$\mathbb{E}(X \mid \sigma(\mathcal{G}_n,Y)) = \mathbb{E}(X \mid \mathcal{H}_n) \to \mathbb{E}(X \mid \mathcal{H}_{\infty}) = \mathbb{E}(X \mid \mathcal{G}_{\infty})\quad \text{in} \, L^2$$ using that $\mathcal{H}_{\infty} = \mathcal{G}_{\infty}$ since $Y$ is $\mathcal{G}_{\infty}$-measurable. Consequently,
$$\mathbb{E}(X \mid \sigma(\mathcal{G}_n,Y))-\mathbb{E}(X \mid \mathcal{G}_n)\to 0 \quad \text{in} \, L^2.$$
