Proof that poisson process has right continuous modification 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.
 A: I am not an expert in Poisson process, but I think literature rarely talked about the right-continuous modification of Poisson process, because all paths of poisson process are right continuous with left limits, and I think this has a name as "Cadlag".
If the process has been right continuous, then there is no need to modify it. 
Below is a standard definition of Poisson process:

A poisson process with parameter $\lambda>0$ is a stochastic process $X$ satisfying the following properties:
$(1)$ $X_{0}=0$ almost surely;
$(2)$ The paths of $X_{t}$ are right continuous with left limits;
$(3)$ If $s<t$, then $X_{t}-X_{s}$ is a Poisson random variable with parameter $\lambda(t-s)$;
$(4)$ If $s<t$, then $X_{t}-X_{s}$ is independent of $\sigma(S_{u}, u\leq s),$ that is, all increments are independent.

However, this definition brings up a key problem  --- the existence of such a process. Then many literature appears, to show the existence, some are constructive, some are measure theoretic.
There are also many other equivalent definitions, to answer your question, let us work with the minimal definition:

Poisson process is a process with independent increments that have Poisson distribution with parameter proportional to the length of time increments. 

I am gonna show you a construction, and by the construction, the Poisson process is right continuous with left limits. 
Below is the construction:
Let $(\Omega,\mathcal{F},\mathbb{P})$ be the probability space we are working with. Then we construct a sequence of i.i.d $(\mathbb{R}_{+},\mathcal{B}(\mathbb{R}_{+}))-$measurable random variables $X_{n},$ $n=1, 2, \cdots$ on this probability space, such that $X_{n}=_{d}\exp(\lambda)$ for each $n$. That is, for each $n$, $$\mathbb{P}(X_{n}>t)=e^{-\lambda t},\ \text{t}\ \geq 0.$$
Put $S_{0}=0$ and $S_{n}=\sum_{i=1}^{n}X_{i}$. Clearly, $(S_{n})_{n=0}^{\infty}$ are non-decreasing (increasing) sequence of random variables. Also, since each $X_{i}$ is $(\mathcal{F}\mapsto\mathcal{B}(\mathbb{R}_{+}))-$measurable, so is $S_{n}$. 
Define $N_{t}:=\max\{n:S_{n}\leq t\}$. We will show that this process is poisson process, using the definition provided in the exercise. 
Firstly note that each $N_{t}$ has a Poisson distribution with parameter $\lambda t$ and each $S_{n}$ has the gamma distribution $\Gamma(n,\lambda)$. 
To show the increments have Poisson distribution with parameter proportional to the length of time increments, it suffices to show that for $t\geq s$, $N_{t}-N_{s}$ has a Poisson$-\lambda(t-s)$ distribution. Firstly, we observe that $$\mathbb{P}(N_{t}-N_{s}=j)=\sum_{i\geq 0}\mathbb{P}(N_{s}=i, N_{t}-N_{s}=j)=\sum_{i\geq 0}\mathbb{P}(S_{i}\leq s, S_{i+1}>s, S_{i+j}\leq t, S_{i+j+1}>t)\ \ \ \ \ (*).$$
Firstly for $i,j>1$, recall that the density $f_{n,\lambda}$ of the gamma $\Gamma(n,\lambda)$ distribution is $$f_{n,\lambda}(x)=\dfrac{\lambda^{n}x^{n-1}e^{-\lambda x}}{\Gamma (n)},\ \ n \neq 1.$$ Then, consider the equation $(*)$ with a change of variable $u=s_{2}-(s-s_{1})$,  we can compute 
\begin{align*}
\mathbb{P}(N_{s}=i, N_{t}-N_{s}=j)&=\mathbb{P}(S_{i}\leq s, S_{i+1}>s, S_{i+j}\leq t, S_{i+j+1}>t)\\
&=\int_{0}^{s}\int_{s-s_{1}}^{t-s_{1}}\int_{0}^{t-s_{2}-s_{!}}e^{-\lambda(t-s_{3}-s_{2}-s_{1})}f_{j-1,\lambda}(s_{3})ds_{3}\lambda e^{-\lambda_{s_{2}}}ds_{2}f_{i,\lambda}(s_{1})ds_{1}\\
&=\int_{0}^{s}\int_{0}^{t-s}\int_{0}^{t-s-u}e^{-\lambda(t-s-u-s_{3})}f_{j-1,\lambda}(s_{3})ds_{3}\lambda e^{-\lambda u}due^{-\lambda (s-s_{1})}f_{i,\lambda}(s_{1})ds_{1}\\
&=\mathbb{P}(S_{j}\leq t-s, S_{j+1}>t-s)\mathbb{P}(S_{i}\leq s, S_{i+1}>s)\\
&=\mathbb{P}(N_{t-s}=j)\mathbb{P}(N_{s}=j)\ \ \ (**).
\end{align*}
We can get the same conclusion for $i=0,1$, $j=1$. Then, $(*)$ and $(**)$ together imply $$\mathbb{P}(N_{t}-N_{s}=j)=\mathbb{P}(N_{t-s}=j)\ \text{for}\ j>0.$$ Now, summing over $j$ and subtracting $1$, we get the relation for $j=0$ and so we proved all the cases.
Now, to show the increment is independent, let us consider $\sigma(N_{u}, u\leq s)$. It follows from the standard measure theoretic probability that the collection $$\Big\{A\in\mathcal{F}:\exists n\in\mathbb{Z}_{+}, t_{0}\leq t_{1}\leq\cdots\leq t_{n}, t_{k}\in [0,s], i_{k}\in\mathbb{Z}_{+}, k=0,\cdots, n\ \text{such that}\ A=\{N_{t_{k}}=i_{k}, k=0,\cdots, n\}\Big\}$$ is a $\pi-$system for this $\sigma-$algebra.
Therefore, to show the independence, it suffices to show that for each $n$, for each sequence $0\leq t_{0}\leq\cdots\leq t_{n}$ and $i_{0},\cdots,i_{n},i$ that $$\mathbb{P}(N_{t_{k}}=i_{k}, k=0,\cdots, n, N_{t}-N_{s}=i)=\mathbb{P}(N_{t_{k}}=i_{k}, k=0,\cdots, n)\mathbb{P}(N_{t}-N_{s}=i),$$ but this is similar to our computation in $(**)$ so we are done. 
Therefore, the $N_{t}$ we constructed is a Poisson Process, using the minimal definition.
And you can see that, since by the construction of $N_{t}$, it must be right continuous (almost surely) and has left limit.
