Any adapted and right continuous process $X_t$ is progressively measurable.
For the above statement, I found proof in several books. They all have similar argument as follows. For a given $t > 0$ and $n \in \mathbb N$, define the following funciton sequence $$ X_n(s) := X \left( \frac{(k+1)t}{2^n} \right), \ \ \mathrm{if} \ \frac{kt}{2^n} <s\leq \frac{(k+1)t}{2^n}. $$ It is clear that $X_n$ is left continuous. My question is why use left continuous functions to approximate right continuous function, please? Or maybe it does not matter? At the beginning, I thought it is typo. However, all proof I read defines $X_n$ to be left-continuous. Thank you!.
To see the point,
Let's fix $t>s>0$ we want to prove that $\lim _{n\to \infty}X_n(s)=X(s)$.
Observe that $$X_n(s)= X(t_{k^{n}+1})$$ where $$k^n=\inf_{k\geq 0} \{k.t/2^n>s\}$$
The point here is that $t_{k^n+1}$ is decreasing to $s$. So that $\lim _{n\to \infty}X_n(s)=\lim _{n\to \infty} X(t_{k^n+1})=X(s)$ because $X$ is right continuous.
If you had taken the following right continuous versions of the dicretization :
$$X_n(s) := X \left( \frac{(k+1)t}{2^n} \right), \ \ \mathrm{if} \ \frac{kt}{2^n} \leq s< \frac{(k+1)t}{2^n}$$
Then the process is not progressively mesurable (it is "looking" into the future at times $s=kt/2^n$)
If alternatively you chose :
$$X_n(s) := X \left( \frac{kt}{2^n} \right), \ \ \mathrm{if} \ \frac{kt}{2^n} \leq s< \frac{(k+1)t}{2^n}$$
Then this one is progressively measurable but $\lim_{n\to \infty} X_n(s) =X(s^-)$ (left continuous limit of $X$) because the equivalent of $t_{k^n+1}$ is increasing to $s$ in this case. So $X_n$ doesn't converge to $X$ at discontinuity point of the trajectory.
Best regards