Proving properties for the Poisson-process. Define a Poisson process as a Levy process where the increments have a Poisson distribution with parameter $\lambda$*"length of increment".
I want to prove these properties:
It has almost surely jumps of value 1.
It is almost surely increasing.
When it changes, the change it is almost surely integer-valued.
It is almost surely positive. I think this will follow from the above.
Earlier I tried to prove that the jumps is almost surely of value 1 here:
Proving that the Poisson process has a.s. jumps of value 1.
But looking back on the proof I think it is wrong. Because that proof is built on that the if for instance $N((k+1)*T/n)-N(kT/n)$ in our entire partition is either 0 or 1, then this event is contained in the event that all our jumps are of value either 0 or 1. However it could theoretically move up and down more in each interval.
So do you see how to prove the properties I wrote in the start? My only idea is to some way look at partitions that approaces 0 in length, and use that then the probability that $N((k+1)*T/n)-N(kT/n)$ is either 0 or 1 approaches 1. But the problem is that even if $N((k+1)*T/n)-N(kT/n)=1$ the process may have had two jumps where it went down 2.4 and then up 3.4.
Another problem is the increasing part. For any interval, there is a positive probability that the end value of the process at the interval minus the start value is positive, is 1.(Since the distribution here is Poisson) However, this does not say that the process is increasing throughout the interval?
How do we show that it behaves the way we know it does, a.s.?
Or can we maybe only say something about the process if we fix the interval first? If we fix a given interval, we can say with probability 1 that it increases and that the increasing value is of integer value? But we can't prove that the process has a.s. increasing sample paths? So that in essence we can only prove things about finitely many points, but not the sample path as a whole? But then, what happens with jumps of value 1 a.s.? Maybe we can not even prove this?
UPDATE: Cleaner version of the quesiton
Maybe it is not so easy to see what I am wondering about in the quesiton above. I'll try to explain it a little more precise.
Assume that you only have the interval [0,T]. Lets say that you have a sequence of partitions that converges to 0. Then you only look at the respective values of the process in the partition points. You can then prove that these values are increasing with probability 1. But if you looked at the entire sample path, could you then prove that the path is increasing with probability 1?
In each partition you can prove with probability 1 that the function only changes with integer values at the partition points. Can you prove this property for the entire sample-path a.s.?
You can prove that as the parition intervals goes to zero, the probability of the jumps at two points next to each other are bigger than 1, converges to 0.(I have only prove this myself for partitions of equal length but I assume it holds in the general case). Can this property be extended to the entire sample path?, so when the function changes value, it does so by increasing its value to 1. (as)
So in summary: If you look at the finite dimensional distribution of a Poisson process, where you look at points close to each other, you have that with probability 1 it is increasing, and the jumps are integer
-valued. And you have with a probability close to 1(depending on the partition) that the jumps are of value 1. Can these properties in some way be extended to the entire sample-path?.
 A: Let $(X_t)_{t \geq 0}$ be a Poisson process with intensity $\lambda$.
Step 1: $(X_t)_{t \geq 0}$ has almost surely increasing sample paths.
Proof: Fix $s \leq t$. Since $(X_t)_{t \geq 0}$ has stationary increments and $X_{t-s}$ is Poisson distributed, we have
$$\mathbb{P}(X_s>X_t) = \mathbb{P}(X_t-X_s < 0) = \mathbb{P}(X_{t-s}<0)=0,$$
i.e. $X_s \leq X_t$ almost surely. As $\mathbb{Q}_+$ is countable, this implies
$$\mathbb{P}(\forall q \leq r, q,r \in \mathbb{Q}_+: X_q \leq X_r)=1.$$
Since $(X_t)_{t \geq 0}$ has càdlàg sample paths, this already implies
$$\mathbb{P}(\forall s \leq t: X_s \leq X_t)= 1.$$

Step 2: $(X_t)_{t \geq 0}$ takes almost surely only integer values.
We have $\mathbb{P}(X_q \in \mathbb{N}_0)=1$ for all $q \in \mathbb{Q}_+$. Hence, $\mathbb{P}(\forall q \in \mathbb{Q}_+: X_q \in \mathbb{N}_0)=1$. As $(X_t)_{t \geq 0}$ has càdlàg sample paths, we get
$$\mathbb{P}(\forall t \geq 0: X_t \in \mathbb{N}_0)=1.$$
(Note that $\Omega \backslash \{\forall t \geq 0: X_t \in \mathbb{N}_0\} \subseteq \{\exists q \in \mathbb{Q}_+: X_q \notin \mathbb{N}_0$ and that the latter is a $\mathbb{P}$-null set.)

Step 3: $(X_t)_{t \geq 0}$ has almost surely integer-valued jump heights.
We already know from step 1 and 2 that there exists a null set $N$ such that $X_t(\omega) \in \mathbb{N}_0$ and $t \mapsto X_t(\omega)$ is non-decreasing for all $\omega \in \Omega \backslash N$. Consequently, we have
$$X_s(\omega) - X_t(\omega) \in \mathbb{N}_0$$
for any $s \geq t$ and $\omega \in \Omega \backslash N$. On the other hand, we know that the limit
$$\lim_{t \uparrow s} (X_s(\omega)-X_t(\omega)) = \Delta X_s(\omega) $$
exists. Combing both considerations yields $\Delta X_s(\omega) \in \mathbb{N}_0$. (Check that the following statement is true: If $(a_n)_{n \in \mathbb{N}} \subseteq \mathbb{N}_0$ and the limit $a:=\lim_n a_n$ exists, then $a \in \mathbb{N}_0$.) Since this holds for any $s \geq 0$, we get
$$\Delta X_s(\omega) \in \mathbb{N}_0 \qquad \text{for all $\omega \in \Omega \backslash N$, $s \geq 0$},$$
i.e. $$\mathbb{P}(\forall s \geq 0: \Delta X_s \in \mathbb{N}_0)=1.$$

Step 4: $(X_t)_{t \geq 0}$ has almost surely jumps of height $1$.
By step 3, it suffices to show that $$\mathbb{P}(\exists t \geq 0: \Delta X_t \geq 2)=0.$$
Since the countable union of null sets is a null set, it suffices to show
$$p(T) := \mathbb{P}(\exists t \in [0,T]: \Delta X_t \geq 2)=0$$
for all $T>0$. To this end, we first note that
\begin{align*} \Omega_0 \cap \{\exists t \in [0,T]: \Delta X_t \geq 2\} &\subseteq \Omega_0 \cap \bigcup_{j=1}^{kT} \{X_{\frac{j}{k}}-X_{\frac{j-1}{k}} \geq 2\} \\ &\subseteq \bigcup_{j=1}^{kT} \{X_{\frac{j}{k}}-X_{\frac{j-1}{k}} \geq 2\} \end{align*}
for all $k \in \mathbb{N}$ where $$\Omega_0 := \{\omega; s \mapsto X_s \, \, \text{is non-decreasing}\}.$$ Using that $\mathbb{P}(\Omega_0)=1$ (by Step 1) and the fact that the increments $X_{\frac{j}{k}}-X_{\frac{j-1}{k}}$ are independent Poisson distributed random variables with parameter $\lambda/k$, we get
$$\begin{align*} p(T) &=\mathbb{P}(\Omega_0 \cap \{\exists t \in [0,T]: \Delta X_t \geq 2\}) \\ &\leq \sum_{j=1}^{kT} \mathbb{P}(X_{\frac{j}{k}}-X_{\frac{j-1}{k}} \geq 2) \\ &= kT \mathbb{P}(X_{\frac{1}{k}} \geq 2) = kT \left(1-e^{-\lambda/k} \left[1+\frac{\lambda}{k} \right]\right) \\ &= \lambda T \frac{1-e^{-\lambda/k} \left(1+\frac{\lambda}{k} \right)}{\frac{\lambda}{k}}. \end{align*}$$
Letting $k \to \infty$, we find
$$p(T) \leq \lambda T \frac{d}{dx} (-e^{-x}(1+x)) \bigg|_{x=0} = 0.$$
