I have been learning some measure theory for my research and encountered the following in a theorem in a paper on Lévy-driven Ornstein-Uhlenbeck processes
$$\int_{|z|\geq 1} \log{(1+|z|) \nu(dz)} <\infty$$
where $\nu(z)$ is the Lévy measure of the Lévy process.
Given $\nu(z)$, how can I show this integral is finite?
I know how to solve it if the integration region was $\mathbb{R}$ but I don't know how to deal with the integration region $|z|\geq 1$.
Moreover, I'm interested in the case $\nu(z) = \lambda \delta(z-k)$ (the Lévy measure for a Poisson process with jumps of size $k>0$. Thanks for your help!
Edit: I found a paper in which they mention that to prove the integral is finite it is sufficient to prove $\mathbb{E}[\log{(1 \lor |z|)}]$ is finite. See my answer below.
I found the answer to my question in another paper. The expression
$$\int_{|z|\geq 1} \log{(1+|z|) \nu(dz)} <\infty$$
is equivalent to showing that $$\int_{\mathbb{R}} \log{(1 \lor |z|) \nu(dz)} <\infty$$ $$\mathbb{E}[\log{(1 \lor |z|)}] <\infty$$
Hence, for $\nu(z) = \lambda \delta(z-k)$, $$\int_{|z|\geq 1} \log{(1+|z|) \nu(dz)} <\infty$$
for $k$ finite since $\mathbb{E}[\log{(1 \lor |z|)}] = \log{(1 \lor |k|)}$.