I am a beginner in stochastic process. Here is a standard proof of a proposition from brownian motion and stochastic calculus.
1.13 Proposition. If the stochastic process $X$ is adapted to the filtration $\left\{\mathscr{F}_{t}\right\}$ and every sample path is right-continuous or else every sample path is leftcontinuous, then $X$ is also progressively measurable with respect to $\left\{\mathscr{F}_{t}\right\}$.
Proof. We treat the case of right-continuity. With $t>0, n \geq 1, k=0,1$, $\ldots, 2^{n}-1$, and $0 \leq s \leq t$, we define: $X_{s}^{(n)}(\omega)=X_{(k+1) t / 2^{n}}(\omega) \quad$ for $\quad \frac{k t}{2^{n}}<s \leq \frac{k+1}{2^{n}} t$, as well as $X_{0}^{(n)}(\omega)=X_{0}(\omega)$. The so-constructed map $(s, \omega) \mapsto X_{s}^{(n)}(\omega)$ from $[0, t] \times \Omega$ into $\mathbb{R}^{d}$ is demonstrably $\mathscr{B}([0, t]) \otimes \mathscr{F}_{t}$-measurable. Besides, by right-continuity we have: $\lim _{n \rightarrow \infty} X_{s}^{(n)}(\omega)=X_{s}(\omega), \forall(s, \omega) \in[0, t] \times \Omega$. Therefore, the (limit) map $(s, \omega) \mapsto X_{s}(\omega)$ is also $\mathscr{B}([0, t]) \otimes \mathscr{F}_{t}$-measurable.
I don't see why it is obvious that the map $(s, \omega) \mapsto X_{s}^{(n)}(\omega)$ from $[0, t] \times \Omega$ into $\mathbb{R}^{d}$ is measurable?
