A levy measure $\nu$ on $\mathbb R^{d}$ is a measure satisfying $$\nu\{0\} = 0, \ \int_{\mathbb R^{d}} (|y|^{2}\wedge 1) \nu(dy) <\infty.$$

A Levy process can be characterized by triples $(b, A, \nu)$ by Levy-Ito decomposition, then $$X_{t} = bt + W_{A}(t) + \int_{B_{1}} x \tilde N(t, dx) + \int_{B_{1}^{c}} x N(t, dx)$$ where $N(t, B)$ is poisson measure with $\mathbb E N(1, B) = \nu(B)$ for a set $B$ bounded below, and $\tilde N(t, dx) = N(t, dx) - t \nu(dx)$ is its compensated one.

[Q1.] Is there always a Levy process corresponding to $(0, 0, \nu)$ for an arbitrary Levy measure $\nu$? END

Remark: The answer seems to be NO. Consider $\nu(dx) = x^{-2} I(x>0) dx$, it is a Levy measure. But if there was an associated process $X_{t}$, then it's subordinator. However, $\mathbb E[X_{1}] = \int_{0}^{\infty} x \nu(dx) = \infty$. Motivated form this example, I'd like to know this question below:

[Q2.] If $(0, 0, \nu)$ is a triplet of a Levy process $X$, is it always true? $$ \lim_{r\to 0^{+}}\int_{B_{1}\setminus B_{r}^{c}} x \nu(dx) < \infty.$$ Moreover, if $\nu(B_1^c) = 0$, then $$\mathbb E X_1 = \lim_{r\to 0^{+}}\int_{B_{1}\setminus B_{r}^{c}} x \nu(dx).$$ END.

Remark: If $\nu(dx) = x^{-2} dx$, then it corresponds to 1-stable process, and $ \lim_{r\to 0^{+}}\int_{B_{1}\setminus B_{r}^{c}} x \nu(dx) = 0$, while $ \int_{B_{1}} x \nu(dx) $ is not well-defined.