Consider a non-negative martingale $(M_n)_{n \in \mathbb{N}}$ that is also uniformly integrable. Does there exist $Y \in L^1$ such that $$|M_n| \leq Y \ \ \text{for all } n \in \mathbb{N}?$$
I haven't been able to make much progress on this problem for the past few days, besides the fact that we can note the existence of an $L^1$ limit $M_{\infty}$ for the martingale (from the uniform integrability). I tried to construct a dominating sequence $Y_n$ based on a running maximum of $M_n$ and acquire a $Y$ via a convergence (to the martingale limit, so our candidate $Y = M_{\infty}$) in the spirit of the generalized DCT but it didn't seem to work.
Prima facie, it looks like a partial converse to what David Williams, in Probability with Martingales, calls Hunt's Lemma:
Suppose that $(X_n)$ is a sequence of random variables such that $X := \lim_n X_n$ exists, almost surely and that $(X_n)$ is dominated by non-negative $Y \in L^1$. For any filtration $\{\mathcal{F}_n\}$, we can show that $E(X_n| \mathcal{F}_n) \to E(X|\mathcal{F}_{\infty})$.
Note: NOT a homework problem.
Best Answer
No, in general there is no such dominating random variable $Y$.
There exist non-negative uniformly integrable martingales $(M_n)_{n \in \mathbb{N}}$ such that $\mathbb{E}(\sup_{n \in \mathbb{N}} |M_n|)=\infty$, see this question for an example. For such a martingale there cannot exist $Y \in L^1$ such that $|M_n| \leq Y$ for all $n \in \mathbb{N}$ (because this would imply $\mathbb{E}(\sup_{n \in \mathbb{N}} |M_n|) \leq \mathbb{E}Y<\infty$).