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If a sequence of random variables $X_n$ is defined on the space $(\Omega,\mathscr{F},\mathbb{P})$, such that $X_n$ converges to $ X$ a.s. and $X_n$ converges to $ X$ in $L^1$

Is it true that for any sub $\sigma$-field $\ \mathscr{G},\mathbb{E}[X_n|\mathscr{G}]\to\mathbb{E}[X|\mathscr{G}] \quad\mathbb{P}$-a.s. ?

I want to use Dominated Convergence Theorem to solve this problem. Since $X_n$ converges to $ X$ $\mathbb{P}$-a.s. , all we need to do is to derive $X_n$ can be bouned by a integrable rancom variable.How can we deduce this from $L^1$ convergence? I have no idea.

SiFei Yang
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1 Answers1

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As mentioned here A dominated convergence theorem for the conditional expectation $E( \cdot \mid \mathcal{F}_n)$ where $\mathcal{F}_n$ loses information over time

Given a sequence $(\mathcal{F}_n)_n$ of $\sigma$-Algebras with $\mathcal{F}_{n+1} \subset \mathcal{F}_n$ and defining $\mathcal{F}_\infty := \bigcap_n \mathcal{F}_n$. If $Y_n \to Y_\infty$ a.s. and $|Y_n| \leq Z \in L^1$ then $$ E(Y_n \mid \mathcal{F}_n) \to E(Y_\infty \mid \mathcal{F}_\infty) \text{ a.s.} \tag{*}$$

However, here we don't have the bound $|Y_n| \leq Z$. And indeed there is a counterexample for it Does almost sure convergence and $L^1$-convergence imply almost sure convergence of the conditional expectation?.

The main idea is

  1. Take $Y_{n}$ the typographic-sequence converging to zero in $L^1$ but not a.s..
  2. Consider iid $\xi_{n}$ with $P(\xi_{n}=2^{n})=2^{-n}$ and $P(\xi_{n}=0)=1-2^{-n}$. Then by Borel-Cantelli the $Z_{n}=\prod_k \xi_k$ is zero for $k\geq N(\omega)$. It also satisfies $E[Z_{n}]=1$
  3. Set $\mathcal{G}=\sigma(Y_{n})$ and $X_{n}=Z_{n}Y_{n}$.
  4. Then $E[X_{n}]=1E[Y_{n}]\to 0$ and $X_{n}=Z_{n}Y_{n}\to 0$ almost surely by Borel-Cantelli.
  5. However, $E[X_n|\mathcal{G}]=Y_{n}\not\rightarrow 0$.

Also, one should note that boundedness in $L^1$ does not imply uniform integrability: Set for $\omega\in \Omega=[0,1]$

$$X_{n}(\omega):=n 1_{\omega\leq 1/n}.$$

Then $E[X_{n}]=1$ but also $E[X_{n}1_{X_{n}>K}]=1$ for all $n> K$.

Thomas Kojar
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