Let $(M_t)_{t\geq 0}$ be a cadlag uniformly integrable (UI) martingale with $M_t\rightarrow M_\infty$ a.s. as $t\rightarrow\infty$. If $\Bbb E[M_\infty^2]<\infty$, is it true that $\Bbb E[M_t^2]<\infty$ for all $t$?
The natural temptation is to consider $(M_t^2)_{t\geq 0}$ with $M_t^2\rightarrow M_\infty^2$. If $\Bbb E[M_t^2]<\infty$ for all $t$ then $(M_t^2)_{t\geq 0}$ is a UI submartingale, and we in fact have $\Bbb E[M_t^2]\leq\Bbb E[M_\infty^2]<\infty$. However, I don't see how to make progress here. If necessary we can impose more constraints, e.g. for $(M_t)$ to be continuous.
Since $(M_t)_{t \geq 0}$ is a uniformly integrable martingale, it follows from the martingale convergence theorem that $M_t \to M_{\infty}$ in $L^1(\mathbb{P})$ (and almost surely); in particular, $(M_t)_{t \geq 0}$ is a closable martingale, i.e.
$$\mathbb{E}(M_{\infty} \mid \mathcal{F}_t) = M_t, \qquad t \geq 0.$$
By Jensen's inequality (for conditional expectations) this gives
$$M_t^2 = \big[ \mathbb{E}(M_{\infty} \mid \mathcal{F}_t) \big]^2 \leq \mathbb{E}(M_{\infty}^2 \mid \mathcal{F}_t),$$
and so, by the tower property,
$$\mathbb{E}(M_t^2) \leq \mathbb{E}(M_{\infty}^2)<\infty$$
for all $t \geq 0$.