If a strong Markov process reaches a Borel set a.s., can it be restarted from that set? Let $X$ be a strong Markov process on $E$, and $B\in \mathcal B(E)$. Suppose that, for some $x\in E$, 
$$
 P_x(\exists t\ge0 \text{ such that } X_t\in B)=1.
$$
My question: Does there exist a stopping time $T$ such that $P_x(X_T\in B)=1$?
 A: The answer is no. Here is a counter-example. Let $E=\mathbb R$, $x=0$, $B=(1,2)$, and $X$ move as the Brownian motion until time $T_B = \inf\{t>0 : X_t\in B\}$, after which time $X$ drifts upwards with rate 1 until being killed at an exponentially distributed time (of some rate $q>0$) and sent to a cemetary state $\Delta$. By definition $X_t\notin B$ for all times $t<T_B$. Since $X$ has a continuous path up to the killing time, $X_{T_B} = 1\notin B$ almost surely. For any $t>0$,
$$
P_x(X_{T_B+t}\in B) < P_x(X_{T_B+t}\in (0,\infty)) = e^{-qt} < 1.
$$
But two weaker results hold. First, the answer is yes if $B$ is closed. By the assumption in the question, $P_x(T_B<\infty)=1$. A strong Markov process is right-continuous, and therefore $X_{T_B}\in\overline B$ almost surely [1, Theorem 11.4 and discussion below Definition 11.1]. If $B$ is closed, then we can take $T = T_B$.
If $B$ is not closed, the following holds:
$$
P_x(\text{there exists stopping time $T$ such that }X_T\in B)=1.
$$
We saw above that $P_x(T_B<\infty)=1$. In addition, [1, Theorem 10.19] gives that there exists an increasing sequence of compact sets $K_n\subseteq B$ such that $P_x(T_{K_n}\downarrow T_B) = 1$. In particular, there is almost surely a choice of $K_n$ such that $T_{K_n}<\infty$, and this proves the weaker result.
[1]: Blumenthal, Getoor. "Markov Processes and Potential Theory" 1968.
