Time continuous discrete states Poisson processes, why is this piecewise constant?

Question

So, I'm studying time continuous discrete states stochastic processes and specifically I started looking into my notes about Poisson's processes. In my notes (from my university class), we define it this way:

Let $\big(\Omega, \mathcal E, \mathbb P\big)$ be a probability space and $\big\{\mathsf N_t\big\}_{t\geq 0}$ a stochastic process with values in $\mathbb Z$, then you call $\big\{\mathsf N_t\big\}_{t\geq 0}$ a Poisson process if

1. $\mathsf N_0(\omega) = 0$ for almost every $\omega\in\Omega$
2. For almost every $\omega\in\Omega$ the functio $f\colon [0,+\infty]\mapsto \mathbb Z$ defined as $f(t) = \mathsf N_t(\omega)$, is monotonic non-decreasing and right continuous
3. For each $n\in\mathbb N$ and $0\leq t_0 < t_1 < \dots < t_n$, for each $1 \leq k \leq n$ the random variables $\mathsf N_{t_k} - \mathsf N_{t_{k-1}}$ are independent
4. $\forall\:t > s \geq 0$ the random variables $\mathsf N_{t} - \mathsf N_{s}$ are distributed with Poisson distribution with parameter $\alpha(t-s)$, where $\alpha$ is the intensity of the process.

Then previously in the class defined a random variable $\mathsf N_t(\omega)$ that counts the number of events occurring in the time interval $(0,t]$ while $N_{s,t}$ is just about the same except that it counts the number of events occurring in $(s,t]$. We stated the family $N_{s_i,t_i}$ meets all the previously listed properties $\iff$ $\big\{\mathsf N_t\big\}_{t\geq 0}$ is a Poisson process with intensity $\alpha = \mathbb E[\mathsf N_{0,1}]$

Now, up to this point I still can understand what I'm supposed to, but then we stated that the function $f$ defined as in $(2)$ of the previous properties list has to be piecewise constant (I hope it's the correct translation since my notes aren't in english, the correct translation might be step function but I'm not sure). Why is that so?

Answer 1

As mentioned, any right-continuous function $f \colon [0,\infty) \to \mathbb{Z}$ is piecewise constant. I don't know any reference to this fact but it is easy to prove. Let $t \in [0,\infty)$. By definition of right-continuity, for every $\epsilon >0$, there exists $\delta >0$ such that $$\forall s \in [0,\infty), \quad \bigg(0 \leq s-t < \delta \quad \implies \quad \lvert f(t)-f(s) \rvert < \epsilon\bigg).$$ Now take $\epsilon = 1/2$. Then there exists $\delta >0$ such that for every $s \in [t,t+\delta)$, $\lvert f(t)-f(s) \rvert < 1/2$. Since $f$ is integer-valued, this implies that $f(t) = f(s)$ for every $s \in [t,t+\delta)$.