g(y)\, {\mbox{\upshape d}}y$, $x \in \overline{\mathbb{R}}$, its strictly increasing and continuous primitive. Using similar arguments as in (\[eqn:spacetrans\_good\_points\]), for $Y := G(X)$ we obtain $$Y_t = \int_0^t g(X_s) \, b(X_s) \, {\mbox{\upshape d}}B_s 
                            + \int_\mathbb{R} g(y-) \, L_+^X(t,y) \, {\text{\large{$\mathds{1}$}}}_{A_2}(y) \, \nu({\mbox{\upshape d}}y), \qquad t < S_\infty^X, \ \mathbf{P}\text{-a.s.}$$ Having in mind that $0 \notin A_2$ and that $\nu$ is a finite signed measure on $\mathscr{B}([-1,1])$, we conclude[^3] $c := \inf A_2 > 0$. Therefore, $\tau_c := \inf\{t \geq 0 : X_t = c\}$ is a strictly positive $\mathbb{F}$-stopping time. For the stopped process $Y_t^{\tau_c} := Y_{\tau_c \wedge t}$, $t \geq 0$, due to (\[eqn:loc\_time\_zero\_outside\_compact\_interval\]) it holds $$Y_t^{\tau_c} = \int_0^{\tau_c \wedge t} g(X_s) \, b(X_s) \, {\mbox{\upshape d}}B_s, 
                                             \qquad t < S_\infty^X, \ \mathbf{P}\text{-a.s.}$$ Since $G$ maps $[0,+\infty)$ into $[0,+\infty)$, the last relation means that $Y^{\tau_c}$ is a non-negative continuous local martingale up to $S_\infty^X$ started at zero, which implies immediately $Y_t^{\tau_c} = 0$, $t \geq 0$, $\mathbf{P}$-a.s. Hence, because of $\tau_c = \inf\{t \geq 0: Y_t = G(c)\}$ and $G(c) > 0$, we conclude $\tau_c = +\infty$ $\mathbf{P}$-a.s. Finally, we obtain $X_t = 0$, $t \geq 0$, $\mathbf{P}$-a.s. Moreover, we have $$0 = {    \ensuremath{\langle X \rangle}}_t = \int_0^t 