\geq 1$). Then it holds $$\begin{split}
                                                        &X_t \geq x_0, \ t \geq 0, \ \mathbf{P}\text{-a.s.} \phantom{\Bigr)}\\
                              \Bigl(\text{resp., } &X_t \leq x_0,  \ t \geq 0, \ \mathbf{P}\text{-a.s.}, \\
                                                                 &X_t \leq x_0,  \ t \geq 0, \ \mathbf{P}\text{-a.s.} \text{ if } \nu(\{x_0\}) \leq -1 \text{ and }
                                                     X_t \geq x_0,  \ t \geq 0, \ \mathbf{P}\text{-a.s.} \text{ if } \nu(\{x_0\}) \geq 1 \Bigr)
                 \end{split}$$ i.e., the point $x_0$ is reflecting.

\(ii) Assume $\nu(\{x_0\}) > 1/2$ (resp., $\nu(\{x_0\}) < -1/2$, $|\nu(\{x_0\})| > 1$). Then it holds $$X_t = x_0, \qquad t \geq 0, \ \mathbf{P}\text{-a.s.},$$ i.e., the point $x_0$ is absorbing. In particular, $b(x_0) = 0$ must be fulfilled.

**1)** Let $(X,\mathbb{F})$ be a solution of Eq. (\[eqn:SDE\_mvd\]) started at $x_0 \in \mathbb{R}$. At first we reduce the problem to the case $x_0 = 0$. For the process $(X - x_0, \mathbb{F})$ it clearly holds $$\begin{split}
        X_t -x_0 &= \int_0^t b_{x_0} (X_s - x_0) \, {\mbox{\upshape d}}B_s 
                                     + \int_\mathbb{R} L^{X-x_0}(t,y-x_0) \, \nu({\mbox{\upshape d}}y), \qquad t < S_\infty^X, \ \mathbf{P}\text{-a.s.},
    \end{split}$$ where $b_{x_0}(x) := b(x+x_0)$, $x \in \mathbb{R}$, and we have used the relation $L^X(t,x) = L^{X-x_0}(t,x-x_0)$, $t<S_\infty^X$, $x \in \mathbb{R}$, $\mathbf{P}$-a.s. which can be easily deduced for example by exploiting (\[eqn:loc\_time\_as\_limit\]). Introducing 