b^2(X_s) \, {\mbox{\upshape d}}s = b^2(0) \, t, \qquad t \geq 0, \ \mathbf{P}\text{-a.s.},$$ which is only possible if $b(0)=0$.\
**4)** Now we prove (i) and (ii) for the case of the left local time in Eq. (\[eqn:SDE\_mvd\]). We recall that we only need to consider the initial value $x_0 = 0$. So, we treat the equation $$X_t = \int_0^t b(X_s) \, {\mbox{\upshape d}}B_s + \int_\mathbb{R} L_-^X(t,y) \, \nu({\mbox{\upshape d}}y),$$ where we assume $\nu(\{0\}) \leq -1/2$. Let $(X,\mathbb{F})$ be a solution of this equation. Introducing $\widetilde{b}(x) = b(-x)$, $x \in \mathbb{R}$, and the Wiener process $\widetilde{B} = -B$, for $-X$ it holds $$-X_t = \int_0^t \widetilde{b}(-X_s) \, {\mbox{\upshape d}}\widetilde{B}_s - \int_\mathbb{R} L_-^X(t,y)\,\nu({\mbox{\upshape d}}y), 
    \qquad t < S_\infty^X, \ \mathbf{P}\text{-a.s.}$$ We define $\widetilde{\nu}(A) := -\nu(-A)$,[^4] $A \in \mathscr{B}([-N,N])$, $N \in \mathbb{N}$. Then, since $S_\infty^X = S_\infty^{-X}$ and since $L_-^X(t,y) = L_+^{-X}(t,-y)$, $t < S_\infty^X$, $y \in \mathbb{R}$, $\mathbf{P}$-a.s., which follows immediately from (\[eqn:loc\_time\_as\_limit\]), we get $$-X_t = \int_0^t \widetilde{b}(-X_s) \, {\mbox{\upshape d}}\widetilde{B}_s + \int_\mathbb{R} L_+^{-X}(t,y)\,\widetilde{\nu}({\mbox{\upshape d}}y), 
    \qquad t < S_\infty^{-X}, \ \mathbf{P}\text{-a.s.}$$ Hence, $(-X, \mathbb{F})$ is a solution of an equation of type (\[eqn:SDE\_mvd\]) with right local time started at zero and for the drift measure $\widetilde{\nu}$ it holds $\widetilde{\nu}(\{0\}) = 