$\mu({\mbox{\upshape d}}y) := {\text{\large{$\mathds{1}$}}}_{(-\infty,0) \setminus A_1} (y) \, \nu ({\mbox{\upshape d}}y)$ and Lemma \[lemma:loc\_time\_zero\], we can write $$Z_t = \int_0^t {\text{\large{$\mathds{1}$}}}_{(-\infty,0)}(Z_s) \, b(Z_s) \, {\mbox{\upshape d}}B_s 
                                        + \int_\mathbb{R} \hat{L}^{Z}(t,y) \, \mu({\mbox{\upshape d}}y)
                                        + \int_{A_2} L_+^{Z}(t,y) \, \nu({\mbox{\upshape d}}y)
                                        + \int_{A_3} L_-^{Z}(t,y) \, \nu({\mbox{\upshape d}}y),$$ $t < S_\infty^X$, $\mathbf{P}$-a.s. Using the unique solution $g$ of the integral equation (\[eqn:integral\_eqn\]) in case of the symmetric local time with respect to $\mu$, its primitive $G(x) := \int_0^x g(y)\,{\mbox{\upshape d}}y$, $x \in \overline{\mathbb{R}}$, and setting $Y := G(X)$, analogously as in (\[eqn:spacetrans\_good\_points\]) we get $$Y_t    = \int_0^t g(Z_s) \, {\mbox{\upshape d}}M_s + \int_{A_2} L_+^{Z}(t,y) \, g(y)  \, \nu({\mbox{\upshape d}}y) + \int_{A_3} L_-^{Z}(t,y) \, g(y)  \, \nu({\mbox{\upshape d}}y), 
 \qquad t < S_\infty^X, \ \mathbf{P}\text{-a.s.}$$ where $M := \int_0^\cdot {\text{\large{$\mathds{1}$}}}_{(-\infty,0)}(Z_s) \, b(Z_s) \, {\mbox{\upshape d}}B_s$. Note that, because of $\mu(\{y\}) = 0$, $y \in A_2 \cup A_3$, the function $g$ is continuous in the points of $A_2 \cup A_3$. Since $\nu$ is a finite signed measure on $\mathscr{B}([-N,N])$, $N\in \mathbb{N}$, we deduce[^5] $c:= \sup A_3 <0$ and hence the $\mathbb{F}$-stopping time $$\tau_c := \inf\{t\geq 0 : Z_t = c\} = \inf\{t\geq 0 : Y_t = G(c)\}$$ is strictly positive. Clearly, due to (\[eqn:loc\_time\_zero\_outside\_compact\_interval\]) we have $$Y_t    