obtain $$L_\pm^{Z}(t,y) = 0, \qquad t < S_\infty^X, \ y \in \mathbb{R}, \ \mathbf{P}\text{-a.s.}$$ Therefore, it holds $Z_t = 0$, $t < S_\infty^X$, $\mathbf{P}$-a.s., which means $X_t \geq 0$, $t \geq 0$, $\mathbf{P}$-a.s. and (i) is proven.\
**3)** Now, still considering the case of the right local time in Eq. (\[eqn:SDE\_mvd\]), we show (ii) which is why we assume $\nu(\{0\}) > 1/2$. From (i) we derive $X_t \geq 0$, $t \geq 0$, $\mathbf{P}$-a.s., and hence via (\[eqn:loc\_time\_zero\_outside\_compact\_interval\]) $$X_t = \int_0^t b(X_s) \, {\mbox{\upshape d}}B_s + \int_\mathbb{R} {\text{\large{$\mathds{1}$}}}_{[0,+\infty)}(y) \, L_+^X(t,y) \, \nu({\mbox{\upshape d}}y), 
    \qquad t < S_\infty^X, \ \mathbf{P}\text{-a.s.}$$ Setting $$A_1 := \{y \in [0,+\infty) : \nu(\{y\}) \geq 1/2\}, \quad A_2 := \{y \in [0,+\infty) : \nu(\{y\}) = 1/2\}$$ and defining the set function $\mu({\mbox{\upshape d}}y) := {\text{\large{$\mathds{1}$}}}_{[0,+\infty) \setminus A_1} (y) \, \nu ({\mbox{\upshape d}}y)$, we can write $$X_t = \int_0^t  b(X_s) \, {\mbox{\upshape d}}B_s + \int_\mathbb{R} L_+^X(t,y) \, \mu({\mbox{\upshape d}}y) 
                    + \int_\mathbb{R} {\text{\large{$\mathds{1}$}}}_{A_2}(y) \, L_+^X(t,y) \, \nu({\mbox{\upshape d}}y),$$ $t < S_\infty^X, \ \mathbf{P}\text{-a.s.}$, since the right local time $L_+^X$ of $X$ vanishes on $A_1 \setminus A_2$ by Lemma \[lemma:loc\_time\_zero\]. Note that $\mu$ satisfies (\[eqn:cond\_atoms\]). Let $g$ be the unique solution of (\[eqn:integral\_eqn\]) with respect to $\mu$ and $G(x) := \int_0^x 