A Lévy 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 Lévy process can be characterized by triples $(b, A, \nu)$ by
Lévy-Itô 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 a 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.


[Q.] If $(0, 0, \nu)$ is a triplet of a Lévy process $X$ whose first moment is finite, is the following 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.
[Q1.] Is there always a Lévy process corresponding to $(0, 0, \nu)$ for an
arbitrary Lévy measure $\nu$? 

Remark:  Consider $\nu(dx) = x^{-2} I(x>0) dx$, it is 
a Lévy measure. But if there was an associated process $X_{t}$, then  $\mathbb E[X_{1}] = \int_{0}^{\infty} x \nu(dx) = \infty$.