Submartingale bounded in $L^2$ converges in $L^2$.

I have been wondering whether it is possible to extend the $$L^2$$-martingale convergence theorem to submartingales that are not necessarily non-negative (the case of non-negative submartingales is treated here). Thus, if $$\{X_n, F_n\}_{n=0}^\infty$$ denotes a submartingale with filtration $$F_n$$ such that $$\sup_{n} E(X_n^2)<\infty$$, can we say that $$X_n$$ converges in $$L^2$$?

A good reference suffices. Thank you in advance.

Let $$R_i,\,i\ge1,$$ be independent, nonnegative r.v. with mean $$1$$. Then $$M_n:=\prod_{i=1}^nR_i,\,n\ge0,$$ defines a nonnegative martingale w.r.t. its natural filtration $$F_n=\sigma(R_1,\ldots,R_n),\,n\ge0$$. By the martingale convergence theorem, it converges almost surely as $$n\to\infty$$ towards a nonnegative r.v. $$M_\infty$$, and further $$\mathbb E[M_\infty]\le1$$ (by Fatou's lemma).

Next, let $$X_n:=-\sqrt{M_n}$$. It is clear that $$X_n,\,n\ge0,$$ is a $$\{F_n\}_{n=0}^\infty$$-adapted process bounded in $$\mathrm L^2(\mathbb P)$$ (we have $$\mathbb E[X_n^2]=\mathbb E[M_n]=1$$ for every $$n\in\mathbb N$$). Further, by convexity of the function $$-\sqrt{\cdot}$$ and (conditional) Jensen's inequality, $$\mathbb E[X_{n+1}\mid F_n]=\mathbb E\!\left[-\sqrt{M_{n+1}}\mid F_n\right]\ge-\sqrt{\mathbb E[M_{n+1}\mid F_n]}=-\sqrt{M_n}=X_n$$ for every $$n\in\mathbb N$$. Thus $$(X_n)_{n\ge0}$$ is a submartingale bounded in $$\mathrm L^2(\mathbb P)$$.

Now $$X_n$$ converges almost surely as $$n\to\infty$$ towards $$X_\infty:=-\sqrt{M_\infty}$$. Using the Riesz-Scheffé lemma, the following assertions are therefore equivalent:

1. $$X_n\to X_\infty$$ in $$\mathrm L^2(\mathbb P)$$,
2. $$\mathbb E[X_n^2]\to\mathbb E[X_\infty^2]$$,
3. $$\mathbb E[M_n]\to\mathbb E[M_\infty]$$,
4. $$M_n\to M_\infty$$ in $$\mathrm L^1(\mathbb P)$$,
5. $$\mathbb E[M_\infty]=1$$.

By Kakutani's martingale theorem, these assertions are also equivalent to $$\prod_{i=1}^\infty\mathbb E\!\left[\sqrt{R_i}\right]>0,$$ or equivalently $$\sum_{i=1}^\infty\left(1-\mathbb E\!\left[\sqrt{R_i}\right]\right)<\infty.$$

For instance, choose the independent $$R_i$$'s such that $$R_i=\begin{cases}\frac{(i+1)^2}{i^2},&\text{with probability \frac{i^2}{(i+1)^2},}\\0,&\text{with remaining probability.}\end{cases}$$ In this case $$(X_n)_{n\ge0}$$ is a (negative) submartingale, bounded but not converging in $$\mathrm L^2(\mathbb P)$$.