Definition: A bounded measurable function $u$ on $\bar{D}$ is called a weak solution of the Neumann problem $N(D ; q, \varphi)$ if, for all $x \in \bar{D}$, $$ M_{\varphi}^u(t)=u\left(X_t\right)-u\left(X_0\right)+\frac{1}{2} \int_0^t \varphi\left(X_s\right) L(d s)+\int_0^t q u\left(X_s\right) d s $$ is a continuous $P^x$ martingale. Here $x \in \partial D$, and $L$ is the boundary local time of X.
Assume that $u \in C^2(D) \cap C^1(\bar{D})$. Let $x \in D$ and define as usual $$ \begin{aligned} & B_{\varepsilon}(x)=\left\{y \in R^d:\|x-y\| \leqq \varepsilon\right\}, \\ & \tau_{\varepsilon}=\inf \left\{t>0:\left\|X_t-X_0\right\|=\varepsilon\right\} . \end{aligned} $$ For a fixed $t>0$ $$ \sup _{0 \leqq u \leqq t} E^x\left[\left(\int_0^u|q|\left(X_s\right) d s\right)^2\right] \leqq 2 E^x\left[e_{|q|}(t)\right]<\infty . $$ Thus, from the definition and the fact that $E^x\left[L(t)^2\right]<\infty$ we get that $\left\{M_{\varphi}^u(s)\right.$, $0 \leqq s \leqq t\}$ is a continuous $L^2$-bounded martingale. By Doob's optional stopping theorem, we have $E^x\left[M_{\varphi}^u\left(t \wedge \tau_{\varepsilon}\right)\right]=0$. Now if $\varepsilon$ is sufficiently small, then $B_{\varepsilon}(x) \subset D$ and $\tau_{\varepsilon}<\tau_D$. Hence, for $0 \leqq t \leqq \tau_{\varepsilon}$, we have $L(t)=0$.
Question: This is taken from Probablistic approach to the Neumann problem by Pei Hsu, I am not sure boundary local time $L(t)=0$, is it \begin{align*} L_{t}=\int_{0}^{t}1_{\left\{X_{s} \in \partial D \right\}}\,dL_{s}, \end{align*}, and $X_s \notin \partial D$ in this case?
