Lévy measure $\nu$. Recall that any Lévy measure satisfies the following condition: $$\int_{{\mathbb{R}}} \big(1 \wedge x^2\big)\nu({dx})<\infty.$$ If we let $\psi$ be its characteristic exponent, that is $${\mathbb{E}}e^{-\xi X_t} = e^{-t\psi(\xi)}, \quad \xi \in {\mathbb{R}},$$ then it is well known that $\psi$ is of the form $$\psi(\xi) = \sigma^2 \xi^2-i\gamma \xi - \int_{{\mathbb{R}}} \big( e^{i\xi z}-1-i\xi z{{\bf 1}}_{|z|<1} \big)\,\nu(dz), \quad \xi \in {\mathbb{R}},$$ where $\sigma >0$ and $\gamma \in {\mathbb{R}}$. Note that since we do not assume symmetry of $X_t$, both $\nu$ and $\psi$ need not be symmetric. The triplet $(\sigma, \gamma, \nu)$ uniquely determines the Lévy process and is therefore called *the generating triplet* of $\mathbf{X}$. Let us notice that $${\operatorname{Re}}\psi(\xi) = \sigma^2 \xi^2+ \int_{{\mathbb{R}}} \big( 1-\cos \xi z \big)\,\nu(dz), \quad \xi \in {\mathbb{R}},$$ and $${\operatorname{Im}}\psi (\xi) = -\gamma\xi + \int_{{\mathbb{R}}} \big( \xi z {{\bf 1}}_{|z|<1} -\sin \xi z \big)\,\nu(dz), \quad \xi \in {\mathbb{R}}.$$ Observe that ${\operatorname{Re}}\psi$ is symmetric even if $\mathbf{X}$ is not symmetric. If we assume that $\int_{(-1,1)^c} |z|\,\nu(dz)<\infty$, then we can also write $$\label{eq:Im_reprezentation}
{\operatorname{Im}}\psi(\xi) = -\gamma_1 \xi +\int_{{\mathbb{R}}} \big( \xi z-\sin \xi z \big)\,\nu(dz),$$ where $\gamma_1=\gamma+\int_{(-1,1)^c}z\,\nu(dz)$. In particular, if ${\mathbb{E}}X_1=0$ then $\gamma_1=0$.

Our standing assumption is finiteness of the following integral: $$\label{eq:main_assumption}
\int_0^{\infty} \frac{d\xi}{1+{\operatorname{Re}}\psi(\xi)}<\infty.$$ Such condition 