# a $L^1$ convergence for backward martingale

I have a question which may be naive, but I can not find the related result in the classical reference such as "Foundations of Modern Probability" and "Probability"(Billingsley). So if someone knows the result please let me know, many thanks!

Let $\{M_n\}_{n\geq 0}$ be a backward martingale, i.e.

$$E[M_n|\mathcal{F}_{n+1}]=M_{n+1},~ \forall n\geq 0$$

where $\mathcal{F}_n:=\sigma(M_k,~ k\geq n)$.

Now suppose

$$\lim_{n\to\infty}M_n=M,~ a.s.$$

My question is whether we have

$$\lim_{n\to\infty}E[|M_n-M|]=0$$

Clearly by Fatou Lemma and Jensen's inequality $M\in L^1$. If we suppose $M_0\in L^p$ with $p>1$ then it is just an application of Doob's inequality, but when $p=1$ I do not know how to prove it. Thanks a lot for your help!

• It's known to be true (see e.g. www.stat.duke.edu/courses/Spring07/sta205/lec/topics/mg.pdf). I don't have a proof in mind right now. Jan 23, 2014 at 16:12
• OK. So it's a corollary of the fact that the family of all conditional expectations of a single $L^1$ random variable is uniformly integrable. (In this case they're expectations of $M_1$). Jan 23, 2014 at 16:39

First remark: the limit always exists. Consider indeed $U_n$ the number of upcrossings of $[a,b]$ by $M_0,\dots,M_n$. Then $(b-a)\mathbb E(U_n)\leqslant \mathbb E(M_0-a)^+$.
Using towering property of conditional expectation, we obtain $\mathbb E(M_0\mid\mathcal F_n)=M_n$. If $M_0\in\mathbb L^1$, then the family $\{\mathbb E(M_0\mid \mathcal G),\mathcal G\subset\mathcal F \}$ is uniformly integrable and we conclude convergence in $\mathbb L^1$.