Let $(E,d)$ be a locally compact separable metric space. Let $\mathcal{D}=\mathcal{D}([0,\infty),E)$ denote the space of right continuous functions on $[0,\infty)$ having left limits and taking values in $E$ that is equipped with the Skorohod topology.
Let $A$ be an open subset of $E$ and define $\tau_A\colon \mathcal{D} \to [0,\infty]$ by \begin{align*} \tau_A(\omega)=\inf\{t \in [0,\infty) \mid \omega(t) \in E \setminus A\} \end{align*} with convention that $\inf \emptyset=\infty.$
Is $\tau_A$ a continuous fucntion with respect the Skorohod topology ?
For $\omega,\omega'\in \mathcal{D}$, we define $r(\omega,\omega')=2^{-k}\min\{1,\sup_{t \in [0,k]}|\omega(t)-\omega'(t)|$}.
Let $\omega\colon [0,\infty) \to E$ be a continuous function. Let $\omega^{(n)} \in \mathcal{D}$, $n \in \mathbb{N}$, satisfy $\lim_{n \to \infty}r(\omega^{(n)},\omega)=0$. Then, can we show that $\lim_{n\to \infty}\tau_A(\omega^{(n)})=\tau_A(\omega)$ ?
These seem intuitively correct. But I couldn't give a proof.
