In Revuz and Yor, they denote $\mathbb{H}^2$ the space of $L^2$-bounded martingales, and $H^2$ the space of continuous $L^2$-bounded martingales. They state
"... by Doob's inequality ... $M_\infty^* = \sup_t |M_t|$ is in $L^2$ if $M \in \mathbb{H}^2$; hence $M$ is u.i. and $M_t = E[M_\infty \mid \mathscr{F}_t]$ with $M_\infty \in L^2$. This sets up a one to one correspondence between $\mathbb{H}^2$ and $L^2(\Omega,\mathscr{F}_\infty,P)$, ..."
Note: the only assumption made on the filtration is that $\mathscr{F}_t$ contains all the sets of $P$-measure zero in $\mathscr{F}$. As far as I know, Doob's inequality cannot be applied in this case. More specifically my questions is:
- How do we apply Doob's inequality without knowing $M$ is right continuous?
EDIT: It appears that they state that all sub/supermartingales considered in the sequel are assumed to be cadlag. I'm not sure where the scope of that statement ends, so in the case that it extends to this question ($\mathbb{H}^2$ is cadlag martingales), my new question would be:
- If $M_\infty\in L^2$ does $t\mapsto E[M_\infty \mid \mathscr{F}_t]$ define a cadlag martingale (it is clearly a martingale)?
My thoughts on the matter...
Doob's inequalities for discrete time state that for $X$ a martingale or positive submartingale indexed by $D=\{0,1\ldots,N\}$, then for $p\geq 1$ and $\lambda > 0$ $$ \lambda^pP\left[\sup_{n\in D} |X_n| \geq \lambda\right] \leq E[|X_N|^p] $$ and for any $p>1$, $$ E\left[\sup_{n \in D} |X_n|^p\right] \leq \left(\frac{p}{p-1}\right)^p E[|X_N|^p]. $$
By taking monotone limits we can take $D$ to be countable as well: $$ \lambda^pP\left[\sup_{t\in D} |X_t| \geq \lambda\right] \leq \sup_{t \in D} E[|X_t|^p] $$ and for any $p>1$, $$ E\left[\sup_{t \in D} |X_t|^p\right] \leq \left(\frac{p}{p-1}\right)^p \sup_{t \in D}E[|X_t|^p]. $$
If $X$ is right continuous then the above holds as written for $D$ an interval of $\mathbb{R}^+$.