Let
- $(\Omega,\mathcal A,\operatorname P)$ be a probability space;
- $I\subseteq\overline{\mathbb R}$ be countable;
- $\mathcal F_t\subseteq\mathcal A$ be a $\sigma$-algebra on $\Omega$ for $t\in I$ such that $(\mathcal F_t)_{t\in I}$ is monotone and $$\mathcal F_{\sup I}:=\begin{cases}\displaystyle\sigma\left(\bigcup_{t\in I}\mathcal F_t\right)&\text{, if }(\mathcal F_t)_{t\in I}\text{ is increasing};\\\displaystyle\bigcap_{t\in I}\mathcal F_t&\text{, if }(\mathcal F_t)_{t\in I}\text{ is decreasing};\end{cases}$$
- $p\ge1$.
Let $X\in\mathcal L^1(\operatorname P)$ and $$M_t:=\operatorname E\left[X\mid\mathcal F_t\right]\;\;\;\text{for }t\in I\cup\{\sup I\}.$$ I want to show that $$\left\|M_t-M_\infty\right\|_{L^p(\operatorname P)}\xrightarrow{t\to\sup I}0\tag1.$$ I'm able to prove this when $(\mathcal F_t)_{t\in I}$ is increasing, but struggle to show this in the decreasing case.
The following observation should be crucial:
- If $(\mathcal F_t)_{t\in I}$ is increasing, then $$\overline{\bigcup_{t\in I}\mathcal L^p(\mathcal F_t,\operatorname P)}=\mathcal L^p(\mathcal F_{\sup I},\operatorname P),\tag2$$ where the closure is taken in $L^p(\operatorname P)$ and $\mathcal L^p(\mathcal F,\operatorname P):=\{f\in\mathcal L^p(\operatorname P):f\text{ is }\mathcal Ftext{-measurable}\}$ whenever $\mathcal F\subseteq\mathcal A$ is a $\sigma$-algebra on $\Omega$.
- If $(\mathcal F_t)_{t\in I}$ is decreasing, then $$\bigcap_{t\in I}\mathcal L^p(\mathcal F_t,\operatorname P)=\mathcal L^p(\mathcal F_{\sup I},\operatorname P).\tag3$$
The increasing case: $(M_t)_{t\in I\cup\{\sup I\}}$ is an $(\mathcal F_t)_{t\in I\cup\{\sup I\}}$-martingale. In particular, $$M_t=\operatorname E\left[M_{\sup I}\mid\mathcal F_t\right]\;\;\;\text{for all }t\in I\cup\{\sup I\}\tag4.$$ Let $\varepsilon>0$. By $(2)$, $$\left\|\xi-M_{\sup I}\right\|_{L^p(\operatorname P)}<\frac\varepsilon2\tag5$$ for some $\xi\in\mathcal L^p(\mathcal F_t,\operatorname P)$ with $t_0\in I$. By the pull-out property of conditional expectation, $$\xi=\operatorname E\left[\xi\mid\mathcal F_t\right]\tag6$$ and hence $$M_t=\operatorname E\left[M_{\sup I}-\xi\mid\mathcal F_t\right]+\xi\tag7$$ for all $t\in I$ with $t\ge t_0$. Since the conditional expectation is a contractive operator, $$\left\|\operatorname E\left[M_{\sup I}-\xi\mid\mathcal F_t\right]\right\|_{L^p(\operatorname P)}\le\left\|M_{\sup I}-\xi\right\|_{L^p(\operatorname P)}\tag8$$ and hence $$\left\|M_t-M_{\sup I}\right\|_{L^p(\operatorname P)}\le2\left\|\xi-M_{\sup I}\right\|_{L^p(\operatorname P)}\le\varepsilon\tag9$$ for all $t\in I$ with $t\ge t_0$. Thus, we are done.
How can we show the decreasing case? There should be an analogous argumentation available ...