I want to show that given a Lévy measure $l(dx)$ such that
$$\int_\mathbb R x^2\land 1l(x)<\infty$$
Then for the quadratic variation of the Lévy process $X$ almost surely $$\sum_{s\leq t} \Delta X(s)^2<\infty.$$
And the quadratic variation is a Lévy process. It should work in any dimension as well.
The definition of a Lévy process is:
$$X(t)=\int_0^t \int_0^\infty x N(dx,ds)$$
Where $N$ is a Poisson process with intensity $\lambda(t)=l(dx)ds$.
Any suggestions how to show this? I find the definition difficult to use or even understand. Here is an attempt:
$$\sum_{s\leq t} \Delta X(s)^2 = \int_0^t \int_0^\infty x^2 N(dx,ds)<\infty$$
Almost surely since $$\int_\mathbb R x^2\land 1 l(x)< \int_\mathbb{R} (1-e^{-1 \land x^2})d \lambda$$
Although I have not managed to yet show it. The quadratic variation is clearly a Lévy process using the usual definition since all the properties still hold, but using the above definition, how to actually find the Lévy measure?
