Let $X$ be a continuous time stochastic process, and denote by $\mathcal F\_t$ its natural filtration. We define $\mathcal F\_z = \mathcal F\_0$ for all $z \leq 0$.


$X$ is said to be *strongly predictable* if there exists some $r > 0$, and an $\mathcal F\_{t-r}$ adapted process $Y$ such that


$$\lim\_{T \to \infty} \mathbb E \left [\int\_{T}^\infty |X\_t - Y\_t| \, dt \right ] = 0.$$


**Question:** Suppose $X$ is a strongly predictable martingale. Is it true that $X\_t$ converges almost surely as $t \to \infty$?



A martingale convergence theorem
$\newcommand{\F}{\mathcal{F}}$The answer is yes.


Indeed, by time rescaling, without loss of generality (wlog) $r=1$.


Take any random variable (r.v.) $Z$ with $EZ=0$, and let $Z'$ be an independent copy of $Z$. Then, by Jensen's inequality, for any real $z$ we have $E|Z|\le E|Z-Z'|=E|(Z-z)-(Z'-z)|\le2E|Z-z|$, so that
\begin{equation\*}
 E|Z|\le2E|Z-z|. \tag{1}
\end{equation\*}


Letting $E\_t:=E(\cdot|\F\_t)$ and using (1) with $Z=X\_t-X\_{t-1}$ and real $t\ge1$, we get
\begin{equation\*} 
\begin{aligned}
 E\_{t-1}|X\_t-X\_{t-1}|&\le2E\_{t-1}|(X\_t-X\_{t-1})-(Y\_t-X\_{t-1})| \\ 
 &=2E\_{t-1}|X\_t-Y\_t|,
\end{aligned}
\end{equation\*}
since $Y\_t-X\_{t-1}$ is $\F\_{t-1}$-measurable.
So, $E|X\_t-X\_{t-1}|\le2|X\_t-Y\_t|$. So, for any real $u\ge0$ and any natural $T$,
\begin{equation\*}
 S\_u:=\sum\_{n=T}^\infty E|X\_{n+u}-X\_{n-1+u}|\le2\sum\_{n=T}^\infty E|X\_{n+u}-Y\_{n+u}|
\end{equation\*}
and hence
\begin{equation\*}
\begin{aligned}
 \int\_0^1 du\, S\_u &\le2\sum\_{n=T}^\infty \int\_0^1 du\, E|X\_{n+u}-Y\_{n+u}| \\ 
& =2\int\_T^\infty dt\, E|X\_t-Y\_t|<\infty
\end{aligned}
\end{equation\*}
if $T$ is large enough, by your displayed condition. So, $S\_u<\infty$ for some $u\in[0,1]$. By time shift, wlog $u=1$. So,
\begin{equation\*}
 \sum\_{n=T}^\infty E|X\_{n+1}-X\_n|=S\_1<\infty. \tag{2}
\end{equation\*}
So, $(X\_n)$ converges in $L^1$ and hence $(E|X\_n|)$ is bounded. So, by [Doob's martingale convergence theorem](https://en.wikipedia.org/wiki/Doob%27s_martingale_convergence_theorems#Statement_for_discrete-time_martingales),
\begin{equation\*}
 X\_n\to Y \tag{3}
\end{equation\*}
as $n\to\infty$ almost surely (a.s.) for some (real-valued) r.v. $Y$.


Next, by [Doob's martingale inequality](https://en.wikipedia.org/wiki/Doob%27s_martingale_inequality),
\begin{equation\*}
 P(\max\_{t\in[n,n+1]}|X\_t-X\_n|>h)\le\frac{E|X\_{n+1}-X\_n|}h
\end{equation\*}
for any real $h>0$.
So, by the [Borel–Cantelli lemma](https://en.wikipedia.org/wiki/Borel%E2%80%93Cantelli_lemma) and (2), for each real $h>0$ a.s. there will be only finitely many natural $n$ such that $\max\_{t\in[n,n+1]}|X\_t-X\_n|>h$. That is, $\max\_{t\in[n,n+1]}|X\_t-X\_n|\to0$ a.s. as $n\to\infty$.


Thus, by (3), $X\_t\to Y$ a.s. as $t\to\infty$, as claimed.


