If $M$ is a left-continuous martingale, under which additional assumption are we able to conclude $\sup_{t\ge0}\left|M_t\right|\in L^1(\text P)$?

Question

Let

• $(\Omega,\mathcal A,\operatorname P)$ be a probability space
• $(\mathcal F_t)_{t\ge0}$ be a filtration on $(\Omega,\mathcal A)$
• $M$ be a left-continuous $\mathcal F$-martingale on $(\Omega,\mathcal A,\operatorname P)$

By Doob's martingale inequality, $$\operatorname E\left[\left(\sup_{t\in[0,\:T]}\left|M_t\right|\right)^2\right]=\operatorname E\left[\sup_{t\in[0,\:T]}\left|M_t\right|^2\right]\le4\operatorname E\left[\left|M_T\right|^2\right]\tag1\;\;\;\text{for all }T\ge0\;.$$ If $M$ is $L^2$-bounded, i.e. $$\sup_{t\ge0}\left\|M_t\right\|_{L^2(\operatorname P)}<\infty\tag2\;,$$ then $(1)$ can be extended by Doob's martingale convergence theorem in the sense that there is an $M_\infty\in L^2(\operatorname P)$ with $$\operatorname E\left[\left(\sup_{t\ge0}\left|M_t\right|\right)^2\right]=\operatorname E\left[\sup_{t\ge0}\left|M_t\right|^2\right]\le4\operatorname E\left[\left|M_\infty\right|^2\right]\tag3\;.$$ In that case, $$\sup_{t\ge0}\left|M_t\right|\in L^2(\operatorname P)\subseteq L^1(\operatorname P)\;.$$

Is there an assumption weaker than $L^2$-boundedness which ensures that $\sup_{t\ge0}\left|M_t\right|\in L^1(\operatorname P)$?

Answer 1

For a local martingale $M$, $\sup_{t>0}|M_t|\in L^1(P)\iff \sqrt{[M,M]_\infty}\in L^1(P)$, please cf. S. N. Cohen and R. J. Elliott, Stochastic Calculus and Applications, 2nd Ed.(2015), Th.11.5.5, p.249.