\ \mathbf{P}\text{-a.s.} \phantom{\Bigr)}\\
                        \Bigl(\text{resp., } L_+^X(t,x) = 0, \qquad & t < S_\infty^X, \ x \in \{\nu \leq -1/2\},  \ \mathbf{P}\text{-a.s.}, \\
                                                                 L_+^X(t,x) = 0, \qquad & t < S_\infty^X, \ x \in \{|\nu| > 1 \text{ or } \nu = -1\}, \ \mathbf{P}\text{-a.s.} \Bigr)
        \end{split}$$

Using (\[eqn:int\_wrt\_loc\_time\]) and (\[eqn:loc\_time\_and\_variation\_process\]), we see $$\begin{split}
    L_+^X(t,x)-L_-^X(t,x) &= 2 \int_0^t {\text{\large{$\mathds{1}$}}}_{\{x\}} (X_s) \int_\mathbb{R} L^X({\mbox{\upshape d}}s, y) \, \nu({\mbox{\upshape d}}y) \\
                                                &= 2 \, L^X(t, x) \, \nu(\{x\}), \qquad t < S_\infty^X, \ x \in \mathbb{R}, \ \mathbf{P}\text{-a.s.}
\end{split}$$ and it follows $$\begin{split}
        \bigl(1-2\nu(\{x\})\bigr) \, L_+^X(t,x) &= L_-^X(t,x), \qquad t < S_\infty^X, \ x \in \mathbb{R}, \ \mathbf{P}\text{-a.s.} \phantom{\Bigr)}\\
        \Bigl(\text{resp., }
        \bigl(1+2\nu(\{x\})\bigr) \, L_-^X(t,x) &= L_+^X(t,x), \qquad t < S_\infty^X, \ x \in \mathbb{R}, \ \mathbf{P}\text{-a.s.}, \\
         \bigl(1-\nu(\{x\})\bigr) \, L_+^X(t,x)  &= \bigl(1+\nu(\{x\})\bigr)\,L_-^X(t,x), \qquad t < S_\infty^X, \ x \in \mathbb{R}, \ \mathbf{P}\text{-a.s.} \Bigr)\\
    \end{split}$$ which, together with the non-negativity of the local times, implies the claims.

The following theorem is the main result of the present note.

\[theorem:reflection\_and\_absorbing\] Let $(X,\mathbb{F})$ be a solution of Eq. (\[eqn:SDE\_mvd\]) with right (resp., left, symmetric) local time started at $x_0 \in \mathbb{R}$. Then the following statements are satisfied:

\(i) Assume $\nu(\{x_0\}) \geq 1/2$ (resp., $\nu(\{x_0\}) \leq -1/2$, $|\nu(\{x_0\})| 