\\
                                 &= \int_0^t {\text{\large{$\mathds{1}$}}}_{(-\infty,0)}(Z_s) \, b(Z_s) \, {\mbox{\upshape d}}B_s 
                                        + \int_\mathbb{R} 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.},
\end{split}$$ where we used (\[eqn:int\_wrt\_loc\_time\]) and the easy to check relation $L_+^X(t,y) = L_+^Z(t,y)$, $t < S_\infty^X$, $y < 0$, $\mathbf{P}$-a.s. (use e.g. (\[eqn:loc\_time\_as\_limit\])). Introducing the sets $$A_1 := \{y \in (-\infty,0) : \nu(\{y\}) \geq 1/2\}, \quad A_2 := \{y \in (-\infty,0) : \nu(\{y\}) = 1/2\}$$ and the set function $\mu({\mbox{\upshape d}}y) := {\text{\large{$\mathds{1}$}}}_{(-\infty,0) \setminus A_1} (y) \, \nu ({\mbox{\upshape d}}y)$, in the decomposition of $Z$ we can split the last integral and, using Lemma \[lemma:loc\_time\_zero\], we can write $$Z_t = \int_0^t {\text{\large{$\mathds{1}$}}}_{(-\infty,0)}(Z_s) \, b(Z_s) \, {\mbox{\upshape d}}B_s 
                                        + \int_\mathbb{R} L_+^{Z}(t,y) \, \mu({\mbox{\upshape d}}y)
                                        + \int_\mathbb{R} L_+^{Z}(t,y) \, {\text{\large{$\mathds{1}$}}}_{A_2}(y)  \, \nu({\mbox{\upshape d}}y),$$ $t < S_\infty^X$, $\mathbf{P}$-a.s. Noting that $\mu(\{x\}) < 1/2$, $x \in \mathbb{R}$, we define the strictly positive function $g$ as the unique solution of the integral equation (\[eqn:integral\_eqn\]) with respect to $\mu$ and denote by $G(x) = \int_0^x g(y)\, {\mbox{\upshape d}}y$, $x \in \overline{\mathbb{R}}$, its strictly increasing and continuous primitive. Then, with the notation $$M_t := \int_0^t {\text{\large{$\mathds{1}$}}}_{(-\infty,0)}(Z_s) \, b(Z_s) \, {\mbox{\upshape d}}B_s, \qquad t < S_\infty^X\,,$$ and noting that $G$ restricted to $\mathbb{R}$ is the 