implies that ${\operatorname{Re}}\psi$ is unbounded, hence it must not be a characteristic exponent of compound Poisson process, and consequently, ${\operatorname{Re}}\psi(\xi) >0$ for $\xi \neq 0$. Furthermore, since $e^{-x}{\leqslant}(1+x)^{-1}$ for $x{\geqslant}0$, we obtain that $\Big\lvert e^{-t\psi(\cdot)} \Big\rvert = e^{-t{\operatorname{Re}}\psi(\cdot)}$ is integrable. Thus, by the Fourier inversion formula, the transition density $p(t,\cdot)$ of $X_t$ exists for all $t>0$ and is given by $$p(t,x) = \frac{1}{2\pi} \int_{{\mathbb{R}}} {\operatorname{Re}}\Big[ e^{-t\psi(\xi)-i\xi x} \Big]\,d\xi, \quad x \in {\mathbb{R}}.$$ If, following Pruitt [@Pruitt81], we define *the concentration function* by setting for $r>0$, $$h(r) = \frac{\sigma^2}{r^2} + \int_{{\mathbb{R}}} \Bigg( 1 \wedge \frac{|s|^2}{r^2} \Bigg)\nu({ds}),$$ then by [@TG14 Lemma 4], for all $x \in {\mathbb{R}}$, $$h(1/|x|) \approx \psi^*(|x|).$$ Lastly, also due te Pruitt we introduce the compensated drift part by setting $$b_r = \gamma+\int_{{\mathbb{R}}} z \big( {{\bf 1}}_{|z|<r}-{{\bf 1}}_{|z|<1} \big)\,\nu(dz).$$ We now turn to introduction of basic objects from the potential theory. For any $x \in {\mathbb{R}}$, ${\mathbb{P}}^x$ and ${\mathbb{E}}^x$ will denote the distribution and the expectation for the process $\mathbb{X}+x$, with the standard notation ${\mathbb{P}}^0={\mathbb{P}}$ and ${\mathbb{E}}^0={\mathbb{E}}$. We also write ${\mathbb{E}}^x [A;Y]={\mathbb{E}}^x {{\bf 1}}_{A}Y$. By ${\tau_D}$ we denote *the first exit time* from an open set $D$, i.e. $${\tau_D}= \inf\{ t>0\colon X_t \notin D \}.$$ For a closed set 