- \nu(\{0\}) \geq 1/2$. Using step 2) and 3) of the proof yields $X_t \leq 0$, $t < S_\infty^X$, $\mathbf{P}$-a.s. if $\nu(\{0\}) \leq 1/2$ and $X_t = 0$, $t < S_\infty^X$, $\mathbf{P}$-a.s. as well as $b(0)=0$ if $\nu(\{0\}) < -1/2$.\
**5)** For the proof in case of the symmetric local time in Eq. (\[eqn:SDE\_mvd\]) all tools are already presented above. But one must be careful in adapting them. We provide the idea but the details are left to the reader. Again without loss of generality we assume $x_0 = 0$. Let $(X,\mathbb{F})$ be a solution to $$X_t = \int_0^t b(X_s) \, {\mbox{\upshape d}}B_s + \int_\mathbb{R} \hat{L}^X(t,y) \, \nu({\mbox{\upshape d}}y).$$ We first assume that $|\nu(\{0\})| > 1$ or $\nu(\{0\}) = 1$. Similar as above in step 2) for $Z_t = X_t \wedge 0$, $t \geq 0$, we deduce $$Z_t = \int_0^t {\text{\large{$\mathds{1}$}}}_{(-\infty,0)}(Z_s) \, b(Z_s) \, {\mbox{\upshape d}}B_s 
                + \int_\mathbb{R} \hat{L}^Z(t,y) \, {\text{\large{$\mathds{1}$}}}_{(-\infty,0)}(y)  \, \nu({\mbox{\upshape d}}y), \qquad t < S_\infty^X, \ \mathbf{P}\text{-a.s.},$$ Furthermore, with the help of the sets $$\begin{gathered}
    A_1 := \{y \in (-\infty,0) : |\nu(\{y\})| \geq 1\}, \quad  
    A_2 := \{y \in (-\infty,0) : \nu(\{y\}) = 1\}, \\
    A_3 := \{y \in (-\infty,0) : \nu(\{y\}) = -1\},\end{gathered}$$ the set function 