For a Lévy-process, I need to prove that the probability that the trajectories are bounded on $[0,\infty)$ is either 0 or 1. Can you please help me? (The author says that this is a consequence of the zero-one law, but he does not state this zero-one law, so I do not know where it comes from. ) We are working with Lévy-processes with bounded jumps.
My attempt:
The probility that a trajectory is bounded, is the union of the probability that the trajectory is bounded by an N, that is if $A = \{\omega: |X_t(\omega)| \text{ bounded}\}=\cup_{N \in \mathbb{N}}\{\omega: |X_t(\omega)|<N, \forall t \in [0,\infty)\}=\cup_{N \in \mathbb{N}}A_N$.
So one way to finish the proof is to show that for all $N$, $P(A_N)=0$ or $P(A_N)=1$.
So we fix N. For every $k>0$ we look at the partition $\{i2^{-k} \}_{i=0}^{\infty}$. This is a finite partition(the idea is to use smaller and smaller partitions, and cádlág-properties).
If $\omega \in A_N$, then for every partition we have that $|X_{t_1}|<N, |X_{t_2}-X_{t_1}|<2N,\ldots |X_{T}-X_{t_{n(k)-1}}|<2N$, and conversely I think if this holds for every partition we are in $A_N$, I am a little surte about the converse part, is it true?.
Using the independent and stationary increments,the probability for this event is $P(|X_{2^{-k}}|<N)\cdot P(|X_{2^{-k}}|<2N)^{\infty}$. I know that the last part don't really make sense, independence only works for a finite number?
Now I am guessing that I need to take the limit in $k$. If I can show that the limit of $P(|X_{2^{-k}}|<N)\cdot P(|X_{2^{-k}}|<2N)^{\infty}$ converges to 0, then I must have that $P(A_N)=0$?, because $A_N$ is contained in the event that for a finite number of partitions points, those points are bounded? But lets say that the expression converges to 1?, then I can not say that $P(A_N)=1$?, or I must argue in some way to do it? And I also do not see how I should deal with the limits to show I get the kind of convergence I want? By stochastic continuity of the Lévy-process, it looks like I am going to get a $1^\infty$-expression, and I don't see how to use l'Hôpital here.
end attempt
Can you please help me? Either on how to finish the above argument, or another argument?(for instance by a zero-one law if you see what the author may mean?)
Update:
I have a second attempt using the Kolmgorov zero one law, but I am not able to finish it, do you see how to finish it?
start attempt: For each n, define the sigma-algebra $\mathcal{F}_n=\sigma(X_t-X_s: n\le s,t\le n+1)$. I think it follows by the independent increments that $\{\mathcal{F}_n\}$ are mutually independent?
Assume that for an $\omega$, the trajectory $X(t,\omega)$ is bounded. For any finite interval $[0,B]$ a cádlág function is bounded. So that the function is bounded is a "tail-event" in the sense that it only depends on big t.
So I must in some way connect this with the tail event $\cap_{n \ge 0}\sigma(\cup_{k\ge n}\mathcal{F}_k)$. Do you see how to do this? That is:
$A = \{\omega: \text{There is a B$(\omega)$ and $N(\omega)$ such that }|X(t,\omega)|<N(\omega), \forall t \ge B(\omega)\}\in \cap_{n \ge 0}\sigma(\cup_{k\ge n}\mathcal{F}_k)$.
Is this a correct attempt? Will this show that the Lévy-process having bounded trajectories is an event with either probability 0 or 1? And how do I finish it?
end attempt
Following your second attempt using Kolmogorov's zero one law:
Define for each $n \in \mathbb{N}$
$$\mathcal{F}_n := \sigma(X_t-X_s; n \leq s \leq t <n+1)$$
then the so-defined $\sigma$-algebras are independent since the Lévy process has independent increments. Now set
$$\mathcal{G}_n := \sigma \left( \bigcup_{k=n}^{\infty} \mathcal{F}_k \right) = \sigma(X_t-X_s; n \leq s < t).$$
If we can show that $$A:=\left\{\omega \in \Omega; \sup_{t \geq 0} |X_t(\omega)|<\infty\right\} \in \mathcal{G}_n$$ for each $n \in \mathbb{N}$, then it follows from Kolmogorov's 0-1 law that $\mathbb{P}(A) \in \{0,1\}$. Fix $n \in \mathbb{N}$. Since the sample paths of the Lévy process are càdlàg, we know that $t \mapsto X_t(\omega)$ is bounded on any compact interval. Consequently, we have
$$A = \left\{\omega \in \Omega; \sup_{t \geq n} |X_t(\omega)|<\infty \right\}$$
implying
$$A = \left\{\omega \in \Omega; \sup_{t \geq n} |X_t(\omega)-X_n(\omega)| < \infty \right\}.$$
As $\omega \mapsto \sup_{t \geq n} |X_t(\omega)-X_n(\omega)|$ is $\mathcal{G}_n$-measurable (as a (countable) supremum of measurable functions), this shows $A \in \mathcal{G}_n$.