Convergence in $\mathcal{L}^2$ Let $Y \in \mathcal{L}^1(\Omega, \mathcal{F}, \mathbb{P})$ and a filtration $(\mathcal{F}_n)_{n \in \mathbb{N}}$ be given. Define for all $n \in \mathbb{N}$ the random variable $X_n = \mathbb{E}[Y|\mathcal{F}_n]$. We know that there is a $X_\infty$ such that $X_n \to X_\infty$ a.s.
Now, we want to show that for $Y \in \mathcal{L}^2(\Omega, \mathcal{F}, \mathbb{P})$ we have that $X_n \to X_\infty$ in $\mathcal{L}^2$ and a condition such that $X_\infty = Y$.
Therefore, we want to show that $\mathbb{E}[|X_n - X_\infty |^2] \to 0$ as $n \to \infty$. However, I am not convinced if this approach will give the above results. Any ideas?
 A: Note that for any $n\in\mathbb{N}$
$$
E|X_n|^2 = E(E(Y|\mathcal{F}_n)^2) \leq EE(Y^2|\mathcal{F}_n)=EY^2,
$$
By conditional Jensens inequality, proving that $\sup_{n\in\mathbb{N}}E|X_n|^2<\infty$. Furthermore for any $n\geq 1$
$$
E(X_n|\mathcal{F}_{n-1})=E(E(Y|\mathcal{F_n})|\mathcal{F}_{n-1})=E(Y|\mathcal{F}_{n-1})=X_{n-1}.
$$
All in all, we have that $(X_n)$ is a $\mathcal{L}^2$-bounded martingale. Now it follows from Doob's $\mathcal{L}^p$-bounded martingale convergence theorem (see Durrett, Rick (1996). Probability: theory and examples - Theorem 5.4.5.) that $X_n$ converges almost surely and in $\mathcal{L}^2$ towards some element $X\in \mathcal{L}^2$. 
I want to show that a version of $X$ is given by $E(Y|\mathcal{F}_\infty)$ where $\mathcal{F}_\infty= \sigma \left( \bigcup_n \mathcal{F}_n\right)$. 
Now note that the almost sure limit of $X_n$ can be chosen to be $\mathcal{F}_\infty$ measurable since it can be written as $X=\lim_{n\to\infty} X_n1_F$ where $F\in \mathcal{F}_\infty$ is the almost sure set where the limit exists. As per uniqueness of limits we have that $X_n$ converges almost surely and in $\mathcal{L}^2$ to $X$. 
 Next realize that $X$ is a version of $E(Y|\mathcal{F}_\infty)$ if
$$
\int_G X dP = \int_G E(Y|\mathcal{F}_\infty) dP,
$$
for any $G\in\mathcal{F}_\infty$ (since $X$ and $E(Y|\mathcal{F}_\infty)$ both are $\mathcal{F}_\infty$-measurable. By Dynkins lemma one can also show that it suffices to prove that
$$
\int_B X dP = \int_B E(Y|\mathcal{F}_\infty) dP,
$$
for all $B\in\mathcal{F}_n$ and any $n\in\mathbb{N}$. Using the definition of conditional expectations we have that for any $N\geq n$ and $B\in\mathcal{F}_n\subset \mathcal{F}_N\subset \mathcal{F}_\infty$ then 
$$
\int_B E(Y|\mathcal{F}_\infty)dP = \int_B YdP =\int_B E(Y|\mathcal{F}_N)dP,
$$
and note that
$$
\left| \int_BE(Y|\mathcal{F}_N)dP -\int_BXdP  \right|\leq \int |E(Y|\mathcal{F}_N)-X|dP \to_N 0
$$
since $X_n$ converges in $\mathcal{L}^1$ to $X$. Lastly we note that
$$
\int_B X dP = \lim_{N\to\infty}\int_B E(Y|\mathcal{F}_N) dP= \lim_{N\to\infty} \int_B E(Y|\mathcal{F}_\infty)dP = \int_B E(Y|\mathcal{F}_\infty)dP 
$$
for any $B\in\mathcal{F}_n$ and $n\in\mathbb{N}$, proving that the equality also holds for any $F\in\mathcal{F}_\infty$. We conclude that $X=E(Y|\mathcal{F}_\infty)$ almost surely. Lastly a condition on $Y$ what would yield your wanted result would be to say that $Y$ is $\mathcal{F}_\infty$-measureable.
