Levy-Ito decomposition intuition So a Levy process $(X_t)_{t\geq0}$ can be decomposed into three parts
$$X_t = \mu t + \sigma^2B_t + L_\nu(t)$$ where $L_\nu(t)$ is "a compound Poisson process with Levy measure $\nu$".
I know the Levy measure of a set $A$ is the expected number of steps of $X_t$ having step size in $A$ in a time interval t. I.e. $\nu(A) = \mathbb{E}[N(1,A)]$ where $N(t,A)$ is the Poisson random measure of a set of counts $A$ in a time interval $t$. In the paper I'm looking at, $$L_\nu(t) = \int_{|h|>1} h N_\nu (t,dh) + \int_{|h|\leq 1} h \big(N_\nu(t,dh) - t\nu(dh)\big).$$
I understand the first two terms of the decomposition, linear drift and Brownian motion, but not the third. Roughly, the third component seems to contribute information about how fast the process is moving. But I don't understand the details -- like, why are we adding $L_\nu(t)$ rather than scaling somehow? And why are we integrating over $h$ in this way?
 A: $L_\nu(t)$ tracks the (partially compensated) sum of the jumps of $X$ that have occurred up to time $t$. Because $\nu(h:|h|>1)<\infty$, there are finitely many jumps in $[0,t]$ of magnitude $>1$. Thus the integral $\int_{|h|>1}hN_\nu(t,dh)$ is just a (random) finite sum.
As for the other integral, the Lévy measure $\nu$ satisfies $\int_{|h|\le 1}h^2\nu(dh)<\infty$, but it may happen that $\nu(h:|h|\le 1)=\infty$ or even $\int_{|h|\le 1}|h|\nu(dh)=\infty$. The first infinity signals that $N_\nu(t,\{|h|\le 1\})=\infty$ a.s.; i.e., that there are infinitely many small jumps up to time $t$. The second integral in your formula for $L_\nu(t)$ is, more accurately,
$$
\lim_{n\to\infty}\int_{1/n<|h|\le 1}h\left(N_\nu(t,dh)-t\nu(dh)\right).
$$
The compensating term $t\nu(dh)$ is just the mean of $N_\nu(t,dh)$. The  integral above converges absolutely, and the $n\to\infty$ limit can be understood to hold in the $L^2$ sense. The $L^2$ convergence is made possible by $\int_{|h|\le 1}h^2\nu(dh)<\infty$.
The point of the compensation is that although $\int_{|h|\le 1}|h|N_\nu(t,dh)$ may diverge, it does so like $t\int_{|h|\le 1}|h|\nu(dh)$, so if the singularity is subtracted off then convergence can occur.
The Poisson process $N_\nu$ is determined by the jumps of $X$:
$$
N((u,v]\times A]=\#\{s>0: u<s\le v, X_s-X_{s-}\in A\}
$$
(What is your source for information about Lévy processes?)
