$L^1$ Convergence of expectation of conditional expectation

Question

The underlying space is a probability space.

Suppose we have a filtration $\{\mathscr{F}_n\}_{n\in \mathbb N}$ and denote $\cup_{n\in \mathbb N} \mathscr{F}_n$ by $\mathscr{F}_\infty$.

Suppose we have a sequence of random variables $\{X_n\}_{n\in \mathbb N}$ s.t. $X_n\to X$ in $L^1$.

In class there was a statement that $\mathbb E[X_n|\mathscr F_n]\to \mathbb E[X|\mathscr F_\infty]$ in $L^1$ without proof.

I think this should be a very standard result in the theory. Can anyone give me a reference for the proof? Thanks!

Answer 1

I assume that you mean $\mathcal F_\infty=\sigma(\bigcup_{n\in\mathbb N}\mathcal F_n)$.

The result can be deduced from a combination of classic properties:

• $(\mathbb E[X\vert\mathcal F_n])_{n\in\mathbb N}$ is a martingale (very easy, see for instance Doob martingale)
• $(\mathbb E[X\vert\mathcal F_n])_{n\in\mathbb N}$ is uniformly integrable (easy consequence of the de la Vallée-Poussin theorem)
• From the above we deduce that $(\mathbb E[X\vert\mathcal F_n])_{n\in\mathbb N}$ converges in $L^1$ to some $X_\infty$ ($(\mathbb E[X\vert\mathcal F_n])_{n\in\mathbb N}$ is uniformly integrable so bounded in $L^1$, so by Doob's martingale convergence theorems it converges almost surely and therefore in probability to an integrable random variable, and uniform integrability + convergence in probability is equivalent to convergence in $L^1$).
• Up to replacing $X_\infty$ with $\liminf_{n\to+\infty}X_n$ we can suppose that $X_\infty$ is $\mathcal F_\infty$-measurable.
• We can show that $X_\infty=\mathbb E[X\vert\mathcal F_\infty]$ ($\{A\in\mathcal F_\infty\mid\mathbb E[X1_A]=\mathbb E[X_\infty1_A]\}$ is $\lambda$-system which contains $\bigcup_{n\in\mathbb N}\mathcal F_n$ and therefore $\mathcal F_\infty$ by the $\pi$-$\lambda$ theorem).
• We deduce that $(\mathbb E[X\vert\mathcal F_n])_{n\in\mathbb N}$ converges in $L^1$ to $\mathbb E[X\vert\mathcal F_\infty]$.
• By $L^1$ continuity of the conditional expectation, $$ \Vert\mathbb E[X\vert\mathcal F_n]-\mathbb E[X_n\vert\mathcal F_n]\Vert_{L^1}\le\Vert X_n-X\Vert_{L^1}, $$ hence $(\mathbb E[X_n\vert\mathcal F_n])_{n\in\mathbb N}$ converges in $L^1$ to $\mathbb E[X\vert\mathcal F_\infty]$.