difference of convex functions, due to the generalized Itô formula for the right local time and since ${\mbox{\upshape d}}g (y) = -2\, g(y-) \, \mu({\mbox{\upshape d}}y)$, for $Y = G(Z)$ it holds $$\label{eqn:spacetrans_good_points}
 \begin{split}
    Y_t &= \int_0^t g(Z_s) \, {\mbox{\upshape d}}M_s 
                             + \int_0^t g(Z_s-) \int_\mathbb{R} L_+^{Z}({\mbox{\upshape d}}s,y) \, \mu({\mbox{\upshape d}}y) \\
                        &\phantom{=======}+ \int_0^t g(Z_s-) \int_\mathbb{R} L_+^{Z}({\mbox{\upshape d}}s,y) \, {\text{\large{$\mathds{1}$}}}_{A_2}(y)  \, \nu({\mbox{\upshape d}}y)
                             - \int_\mathbb{R} L_+^{Z}(t,y) \, g(y-) \, \mu({\mbox{\upshape d}}y)\\
                        &= \int_0^t g(Z_s) \, {\mbox{\upshape d}}M_s  
                                +  \int_\mathbb{R} L_+^{Z}(t,y) \, g(y-) \, {\text{\large{$\mathds{1}$}}}_{A_2}(y)  \, \nu({\mbox{\upshape d}}y),
 \end{split}$$ $t < S_\infty^X$, $\mathbf{P}$-a.s. Clearly, since we have $Z_t \leq 0$, $t \geq 0$, and since $G$ maps $(-\infty,0]$ into $(-\infty,0]$, it follows $Y_t \leq 0$, $t \geq 0$. Using that $g$ is strictly positive, we conclude $$\label{eqn:aux_2}
    \int_0^t g(Z_s) \, {\mbox{\upshape d}}M_s \leq Y_t \leq 0, \qquad t < S_\infty^X\, \ \mathbf{P}\text{-a.s.}$$ The process $\int_0^\cdot g(Z_s) \, {\mbox{\upshape d}}M_s$ being a non-positive continuous local martingale up to $S_\infty^X$ starting at zero must be zero $\mathbf{P}$-a.s. which implies $\int_0^t g^2(Z_s)\, {\mbox{\upshape d}}{    \ensuremath{\langle M \rangle}}_s = 0$, $t < S_\infty^X$, $\mathbf{P}$-a.s., and thus $M_t = 0$, $t < S_\infty^X$, $\mathbf{P}$-a.s. Hence, $Z$ is a process of locally bounded variation on $[0,S_\infty^X)$ and we 