Predictable process is adapted

Question

If I start with the definition of a predictable process as a measurable mappings on the predictable $\sigma$-algebra generated by sets like
$$
(s,t]\times F, \quad s<t, \quad F \in \mathcal{F}_s
$$
and
$$
\{0\}\times F,\quad F \in \mathcal{F}_0,
$$
is it obvious why such a process, $X(t,\omega)$, would be adapted to the above filtration? Form my reading, I am told that this is the case, but I don't quite see it.

Answer 1

Yes. Set $ ​​t\ge 0 $, note that $$ X\colon [0,\infty)\times \Omega \to \mathbb{R} $$ is predictable, $$(t,\cdot)\colon \Omega \to [0,\infty)\times \Omega$$ is $\mathcal{F}_t$ measurable and $$X_t=X \circ (t,\cdot).$$ In fact, $${(t,\cdot)}^{-1}\left( (u,v]\times F \right) = F $$ when $t>u$ (with $F\in \mathcal{F}_u\subset \mathcal{F}_t$) and $$ {(t,\cdot)}^{-1}\left( (u,v]\times F \right)=\emptyset$$ when $t\le u$.