Question on Doob's martingale convergence theorem

Question

Let $(\Omega,\mathcal F,\mathbb P)$ be a probability space and $(\mathcal F_k)_{k\in\mathbb N}$ a filtration of $\mathcal F$ such that $\mathcal F=\sigma(\mathcal F_k\mid k\in\mathbb N).$ Let $X:\Omega\rightarrow\mathbb R^d$ be $\mathcal F$-measurable. By Doob's convergence theorem, we know that $E[X\mid\mathcal F_k]\rightarrow E[X\mid\mathcal F]$ $\mathbb P$-a.s. and in $L^1,$ for $k\rightarrow\infty.$ If for some $k_0\in\mathbb N$ we have $E[X\mid\mathcal F_{k_0}]=0$ $\mathbb P$-a.s., do we have then that $E[X\mid\mathcal F]=0$ $\mathbb P$-a.s. ?

Answer 1

Define $X':=X-\mathbb E[X\mid F_{k_0}]$ for $X$ an integrable $\mathcal F$-measurable random variable: we do not have in general that $\mathbb E[X'\mid\mathcal F]=0$ a.s. , otherwise $X$ would be almost everywhere equal to a $\mathcal F_{k_0}$-measurable random variable, which is not necessarily the case.

Answer 2

We do not. Let $X$ be any nontrivial integrable random variable on $(\Omega, \mathcal{F}, P)$ with $EX = 0$. Set $\mathcal{F_k} = \{\Omega, \emptyset\}$ for $k \le 42$ and $\mathcal{F}_k = \mathcal{F}$ for $k > 42$. Then $E[X \mid \mathcal{F}_{42}] = EX = 0$ a.s, but $E[X \mid \mathcal{F}] = X$ which is not zero.