Given $p>1$, if a martingale $M_n$ is bounded in $L^p$ (meaning $\sup_n E(|M_n|^p)<\infty$), then it converges almost surely and in $L^p$ to some $M_\infty$. However, if the martingale is only bounded in $L^1$, then one can only state for sure that it converges almost surely.
Consider a bounded martingale (meaning there exists $A>0$ such that for all $n$, $|M_n|\leq A$ a.s). Must $M_n$ converge in $L^1$ ?
If $M_n$ is bounded, it's bounded in $L^p$ for every $p>1$, so it must converge in $L^p$ for any $p>1$. I haven't been able to prove or disprove convergence in $L^1$, hence my post.
The following is well known (e.g. see Theorem 1.3 of this) : $M_n$ converges almost surely and in $L^1$ to $M_\infty$ iff $M_n$ is uniformly integrable, i.e.:
$$\lim_{K\rightarrow\infty}[\sup_{n}E[|M_n|1_{M_n\geq K}]=0.$$
Since $|M_n|\leq A$, then yes, $M_n$ will converge to $M_\infty$ both almost surely and in $L^1$.