Is it possible to construct an $L^1$ dominating random variable for a non-negative uniformly integrable martingale?

Question

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.

Answer 1

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 $\mathbb{E}(\sup_{n \in \mathbb{N}} |M_n|) \leq \mathbb{E}Y<\infty$).

Answer 2

In the dominated case, one has the a.s. convergence of $E[M_n\mid\mathcal B]$ to $E[M_\infty\mid \mathcal B]$ for each sub-$\sigma$-field of the probability space on which these random variables are defined; this result can be found in Doob's 1953 book on stochastic processes. (Here $M_\infty:=\lim_n M_n$.) The $L^1$ convergence of these conditional expectations follows easily because $\{M_n\}$ is UI. There is a sort of converse in an interesting paper of Blackwell and Dubins ("A converse to the dominated convergence theorem") https://projecteuclid.org/euclid.ijm/1255644957 , to the effect that (after enlarging the probability space if necessary) if $\sup_n M_n$ is not integrable, then there exists a sub-$\sigma$-field $\mathcal B$ such that the convergence of $E[M_n\mid\mathcal B]$ to $E[M_\infty\mid \mathcal B]$ fails a.s.