This is a (probably basic) question about the generator of a Markov process.

Let $(E,d)$ be a locally compact metric space. We consider a Feller process $X=(\{X_t\}_{t \ge 0},\{P_x\}_{x \in E})$ on $E$. That is, for any $t>0$, the semigroup $P_t$ of $X$ maps $C_{\infty}(E)$ into itself. Here, $C_\infty(E)$ denotes the space of continuous functions on $E$ vanishing at infinity.

We assume that the generator $L$ of $X$ is described as follows: for $f \in C_{\infty}(E)$ and $x \in E$, \begin{align*} Lf(x)=\int_{X \setminus \{x\}}(f(y)-f(x))c(x,y)\,\mu(dy). \end{align*} Here, $\mu$ denotes a Radon measure on $E$, and $c(x,y)$ a nonnegative bounded function on $E \times E \setminus \text{diag}$ with compact support. Therefore, the Feller processs $X$ is a pure jump process.

For a bounded open set $U$, we define $\tau_U=\inf\{t>0 \mid X_t \notin U\}$. The part process $X^U$ of $X$ on $U$ is defined as
\begin{align*}
X^U_t=\begin{cases}
X_t,&\quad t<\tau_U,\\
\partial ,&\quad t \ge \tau_U.
\end{cases}
\end{align*}
We assume also that $X^U$ is also Feller process. **Then, can we describe the generator $L^U$ of $X^U$?**

We write $E_x$ for the expectation under $P_x$, the law of $X$ starting from $x$. For $f \in C_{\infty}(U)$ and $x \in U$, we have \begin{align*} L^Uf(x)&=\lim_{t \to 0}\frac{E_{x}[f(X^U_t)]-f(x)}{t}\\ &=\lim_{t \to 0}\left(\frac{E_{x}[f(X_t)]-f(x)}{t}-\frac{E_{x}[f(X_t):t \ge \tau_U]}{t} \right). \end{align*} Can wa characterized the quantity $\lim_{t \to 0}E_{x}[f(X_t):t \ge \tau_U]/t$ in terms of $c(x,y)$ and $\mu$?