Predictable Implies Progressive

Question

If I define the predictable $\sigma$ algebra $\mathcal{P}_T$ in terms of the sets $(s,t]\times F$ for $0\leq s <t\leq T$ and $F \in \mathcal{F}_s$ (along with $\{0\}\times F, F\in \mathcal{F}_0$). How do I infer that a $\mathcal{P}_T$ measurable process (say $X(t,\omega)$) is progressively measurable? Is it sufficient to observe that the sets that generate $\mathcal{P}_T$, $$ (s,t]\times F,\quad F\in \mathcal{F}_s\subset \mathcal{F}_t $$ are in $\mathcal{B}([0,t])\times \mathcal{F}_t$ (and analogously for $\{0\}\times F, F\in \mathcal{F}_0$)? This would certainly imply that since the preimage of $X(\cdot,\cdot)$ is a predictable set, it can be constructed by sets of the above form, hence it is is also in $\mathcal{B}([0,t])\times \mathcal{F}_t$. Is this sufficient, or is there more to the argument?

Answer 1

Let $\mathcal{A}$ be the generating system of $\mathcal{P}_T$ specified in your question. Let $\mathcal{H}$ be the class of progressive processes. Then, by the functional monotone class theorem (see e.g. here), $\mathcal{H}$ contains all bounded predictable processes. For a general predictable process, use that $X \wedge n$ is a predictable process increasing to $X$.