the drift measure $\nu_{x_0}$ via $\nu_{x_0}(B) := \nu(B+x_0)$,[^2] $B\in\mathscr{B}([-N,N])$, $N \in \mathbb{N}$, we obtain $$\begin{split}
        X_t -x_0 &= \int_0^t b_{x_0} (X_s - x_0) \, {\mbox{\upshape d}}B_s 
                                     + \int_\mathbb{R} L^{X-x_0}(t,y) \, \nu_{x_0}({\mbox{\upshape d}}y), \qquad t < S_\infty^X, \ \mathbf{P}\text{-a.s.}
    \end{split}$$ Hence, $(X-x_0,\mathbb{F})$ is also a solution to an equation of type (\[eqn:SDE\_mvd\]) but started at zero and the drift measure satisfies $\nu_{x_0}(\{0\}) = \nu(\{x_0\})$. Therefore, without loss of generality we assume $x_0 = 0$ in the following parts of the proof.\
**2)** Now let us consider the case of the right local time in Eq. (\[eqn:SDE\_mvd\]). To prove (i) we assume $\nu(\{0\}) \geq 1/2$, we set $Z_t = X_t \wedge 0$, $t \geq 0$. We apply Tanaka’s formula (see (\[eqn:gen\_ito\_formula\]) for $f(x) = -x^- = x \wedge 0$, $x \in \mathbb{R}$) for the left local time to obtain $$Z_t = \int_0^t {\text{\large{$\mathds{1}$}}}_{(-\infty,0)}(X_s) \, b(X_s) \, {\mbox{\upshape d}}B_s 
                                     + \int_0^t {\text{\large{$\mathds{1}$}}}_{(-\infty,0)}(X_s) \int_\mathbb{R} L_+^X({\mbox{\upshape d}}s,y) \, \nu({\mbox{\upshape d}}y) - \frac{1}{2} L_-^X(t,0),$$ $t < S_\infty^X, \ \mathbf{P}\text{-a.s.}$ Lemma \[lemma:loc\_time\_zero\] shows that the left local time of $X$ in zero vanishes and we can write $$\begin{split}
    Z_t &= \int_0^t {\text{\large{$\mathds{1}$}}}_{(-\infty,0)}(Z_s) \, b(Z_s) \, {\mbox{\upshape d}}B_s 
                                        + \int_\mathbb{R} \int_0^t {\text{\large{$\mathds{1}$}}}_{(-\infty,0)}(X_s) \, L_+^X({\mbox{\upshape d}}s,y) \, \nu({\mbox{\upshape d}}y) 