Recently, I have been studying stochastic processes and came across an interesting question about cadlag modifications of processes and I couldn't find the answer on the internet. My question is written below.
Suppose that we have the probability space $(D[\mathbb{R}^+, S]), \mathcal{B}_S, \mathbb{P}) $ and the canonical process $ X_t(\omega) = \omega(t) $. Here, $D[\mathbb{R}^+, S] $ denotes the set of all left continuous with right limits functions (ladcag paths) defined on $\mathbb{R}^+ = [0,+\infty)$, with values in a finite set $S$. We equip $D[\mathbb{R}^+, S])$ with the usual cylinder $\sigma$-algebra. My question is, do we have a cadlag modification $Y_t$ of $X_t$? That is, is there a process $Y_t$, defined on the same probability space such that
$$ \forall t \geq 0, \mathbb{P}(Y_t = X_t) = 1 \quad \text{and} \quad \mathbb{P}(Y = (Y_t)_{t\geq 0} \ \text{has a cadlag path}) = 1? $$
My first thought was to define for each $\omega$
$$Y_t(\omega):= \begin{cases} \omega(t), \quad \text{if} \quad t \in C_\omega, \\
\lim_{s \rightarrow t^+}\omega(s), \quad \text{otherwise}, \end{cases} $$
where $C_\omega := \{t \geq 0, \omega \ \text{is continuous at t} \}$. In otherwords, I modified the "left jump points" of any ladcag process to "right jump points" so that its path becomes cadlag. But this scheme had me stuck at proving
$$ 1 = \mathbb{P}(Y_t = X_t) = \mathbb{P}(\omega: Y_t(\omega) = \omega(t)) = \mathbb{P}(\omega: \omega \ \text{is continuous at t}), \quad \forall t \geq 0. $$
Thank you for all your help.