Mean-Square Convergence of Sub-Sigma-Algebras

Let $(\Omega,\mathcal{F},P)$ be a probability space, let $(\mathcal{G}_n)_{n\in\mathbb{N}}$ be a sequence of sub-$\sigma$-algebras of $\mathcal{F}$, and let $\mathcal{G}$ be a sub-$\sigma$-algebra of $\mathcal{F}$.

Definition. The sequence $(\mathcal{G}_n)_{n\in\mathbb{N}}$ converges in mean square to $\mathcal{G}$, written $\mathcal{G}_n\to\mathcal{G}$ in mean square, if for every square-integrable random variable $X$ on $(\Omega,\mathcal{F},P)$ and every choice of conditional expectations $Y_n$ of $X$ given $\mathcal{G}_n$ ($n\in\mathbb{N}$) and $Y$ of $X$ given $\mathcal{G}$, the real sequence of mean-square distances $(\lVert Y_n-Y\rVert_{2})_{n\in\mathbb{N}}$ has limit $0$. In the notation of Conditional Expectation of a Square-Integrable Random Variable:

Independence of the choices. For fixed $X$ and $n$, any two conditional expectations of $X$ given $\mathcal{G}_n$ (respectively given $\mathcal{G}$) are almost surely equal by the uniqueness assertion of Existence and Uniqueness of Conditional Expectation for Square-Integrable Random Variables, hence lie at mean-square distance $0$ from one another by the null-equivalence statement of Square-Integrable Random Variables and the Mean-Square Inner Product. By the triangle inequality, the displayed limit therefore holds for one choice of the conditional expectations if and only if it holds for every choice; the definition is stated over all choices in accordance with the notational convention of Conditional Expectation of a Square-Integrable Random Variable.

Basic example. If $(\mathcal{G}_n)_{n\in\mathbb{N}}$ is nondecreasing, meaning $\mathcal{G}_n\subseteq\mathcal{G}_{n+1}$ for every $n$, then $\mathcal{G}_n\to\sigma\bigl(\bigcup_{n}\mathcal{G}_n\bigr)$ in mean square, where the limit is the generated $\sigma$-algebra of the union: this is precisely Levy's upward theorem in mean square.

Caution. If $\mathcal{G}_n=\sigma(U_n)$ and $\mathcal{G}=\sigma(U)$ are the $\sigma$-algebras generated by random variables $U_n$ and $U$, then convergence of $U_n$ to $U$ — even pointwise on all of $\Omega$ and in mean square simultaneously — does not imply $\sigma(U_n)\to\sigma(U)$ in mean square; a limit of $\sigma(U_n)$-measurable random variables can fail to be $\sigma(U)$-measurable.