How can one prove that a Poisson process ($X_t$) has a right-continuous modification?

That is, we say that $Y_t$ is a modification of $X_t$ if for each $t $,$$\ \mathsf{P}(X_t = Y_t) = 1,$$ and $Y_t$ is a (almost everywhere) right-continuous process if for each $t$, $$\mathsf{P}\left(\lim_{s\rightarrow t+0} Y_s = Y_t\right) = 1.$$

Here is what I have so far:

Let $\Omega_1 = \{\omega\ |\ X_s < X_t\ \forall s, t \in \mathbb{Q}\}$

Perhaps $\mathsf{P}(\Omega_1) = 1$. Then let

$$ Y_t(\omega) = \begin{cases} \lim_{s \rightarrow t+0,\ s\in\mathbb{Q}}X_s(\omega) & \omega \in \Omega_1\\ 0 & \omega \notin \Omega_1 \end{cases} $$

$Y_t$ is limit of countable number of random variables, so it is random variable. But I don't know why $Y_t$ is right continuous and why it is modification.


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