I am trying to understand progressive measurability in general, so here's the question (the idea is taken from my notes):
Show that an adapted process $(X_t)$ whose paths are all right-continuous (or are all left-continuous) is progressively measurable.
We present the argument for a right-continuous $X$. For $t > 0, n ≥ 1, k = 0,1,2...,2^n-1$ let $X_0^{(n)}(\omega)=X_0(\omega)$ and $X_s^{(n)}=X_{\frac{k+1}{2^n}t}(\omega)$ for $\frac{k}{2^n}t<s\leq\frac{k+1}{2^n}t$
We claim that $(X_s^{(n)}: s\leq t)$ is $\mathcal{B}([0,t])\otimes \mathcal{F}_t)$ measurable.
Let $A \in \mathcal{B}(\mathbb{R})$. We want to show that $\{(s,\omega):X_s^{(n)}(\omega)\in A\} \in \mathcal{B}([0,t])\otimes \mathcal{F}_t)$
But since $X_s^{(n)}$ takes finitely values, then we have that
$$\{(s,\omega):X_s^{(n)}(\omega)\in A\} = \bigcup_{k \in \text{ the set of possible values for s}}\{(k,\omega):X_k^{(n)}(\omega)\in A\}$$
Since for a fixed $k$ we have that $\{(k,\omega):X_k^{(n)}(\omega)\in A\} \in \mathcal{F}_k$ then this union is a finite union of sets of the form $k \otimes A_k$ where $A_k\in \mathcal{F}_k$ hence the whole thing is in the product sigma algebra as desired
Is this proof ok?
In general, any references where a dummy can understand progressive measurability are welcome