= \int_0^t g(Z_s) \, {\mbox{\upshape d}}M_s + \int_{A_2} L_+^{Z}(t,y) \, g(y)  \, \nu({\mbox{\upshape d}}y) \qquad t < \tau_c \wedge S_\infty^X, \ \mathbf{P}\text{-a.s.}$$ and similarly to (\[eqn:aux\_2\]) and the lines thereafter we conclude $Z_t = 0$, $t < \tau_c \wedge S_\infty^X$, $\mathbf{P}$-a.s. By the definition of $\tau_c$ this gives $Z_t = 0$, $t < S_\infty^X$, $\mathbf{P}$-a.s. and therefore $X_t \geq 0$, $t \geq 0$, $\mathbf{P}$-a.s.\
Now we assume $|\nu(\{0\})| > 1$ or $\nu(\{0\}) = -1$. As shown in step 4), $-X$ is a solution of Eq. (\[eqn:SDE\_mvd\]) with drift measure $\widetilde{\nu}$ defined by $\widetilde{\nu}(A) = - \nu(-A)$, $A \in \mathscr{B}([-N,N])$, $N \in \mathbb{N}$. Hence, from the result just proven we obtain $-X_t \geq 0$, $\geq 0$, $\mathbf{P}$-a.s. Summarizing, this yields $$X_t = 0, \ t \geq 0, \ \mathbf{P}\text{-a.s. and necessarily $b(0)=0$ if $|\nu(\{0\})| > 1$,}$$ $$X_t \geq 0, \ t \geq 0, \ \mathbf{P}\text{-a.s. if $\nu(\{0\}) = 1$,}$$ and $$X_t \leq 0, \ t \geq 0, \ \mathbf{P}\text{-a.s. if $\nu(\{0\}) = -1$.}$$ This finishes the proof of (i) and (ii) for symmetric local time.

Introducing the $\mathbb{F}$-stopping time $\tau_x := \inf\{t \geq 0 : X_t = x\}$ for a solution $(X,\mathbb{F})$ of Eq. (\[eqn:SDE\_mvd\]), we can state more generally the 