The setup of my question is the following: Suppose that we have a measurable space $(\Omega,\mathcal{F})$ and a filtration $\mathbf{F} = (\mathcal{F}\_t)\_{t \geq 0}$ on it. Let $\mathcal{P}(\mathbf{F})$ be the predictable $\sigma$-algebra, that is, the $\sigma$-algebra generated by all real-valued left-continuous processes adapted to the filtration $\mathbf{F}$. Let $\mathcal{O}(\mathbf{F})$ be the optional $\sigma$-algebra, that is, the $\sigma$-algebra generated by all real-valued right-continuous process with left-limits adapted to the filtration $\mathbf{F}$.


I was trying to find conditions on the underlying measurable space $(\Omega,\mathcal{F})$ or the filtration $\mathbf{F}$, under which the predicatble $\sigma$-algebra $\mathcal{P}(\mathbf{F})$ coincides with the optional $\sigma$-algebra $\mathcal{O}(\mathbf{F})$. For example, if the filtered space $(\Omega,\mathcal{F})$ supports a process $X=(X\_t)\_{t \geq 0}$ with values in a metric space $(E,d)$ for example, such that the paths $\mathbb{R}\_+ \ni t \mapsto X\_t \in E$ are continuous, and the filtration $\mathbf{F} = (\mathcal{F}\_t)\_{t \geq 0}$ is generated by $X$, that is,
$$\mathcal{F}\_t = \sigma(X\_r : 0 \leq r \leq t),$$
does the equality
$$\mathcal{P}(\mathbf{F}) = \mathcal{O}(\mathbf{F})$$
hold?


If this condition is too weak, does there exist some other general condition under which the predictable $\sigma$-algebra coincides with the optional $\sigma$-algebra?



When does the predictable $\sigma$-algebra $\mathcal{P}$ coincide with the optional $\sigma$-algebra $\mathcal{O}$?
(Answering your comment)


Off the top of my head, I'd look in vol.2 of *Probabilités et Potentiel* (Dellacherie & Meyer) or in *Limit Theorems for Stochastic Processes* (Jacod & Shiryaev). Another convenient resource is the blog <https://almostsure.wordpress.com> of Geo. Lowther.


The key is that for a bounded rc martingale $M$, the predictable projection ${}^p\!M$ coincides with the left limit process $(M\_{t-})$. Thus, if $\mathcal P =\mathcal O$ then each such $M$ coincides with its predictable projection and so is left continuous, hence continuous.


Conversely, if each bounded rc martingale is continuous, then each such martingale is predictable. In particular, for a bounded r.v. $Z$ the rc version of the martingale $t\mapsto E[Z|\mathcal F\_t]$ must be predictable. Using this, Theorem IV-62 of Dellacheie & Meyer cited above yields that every stopping time is predictable, and this in turn implies that $\mathcal O =\mathcal P$.


