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 $P_x(X_T\in B)=1$?
Note: I also asked this on Mathematics Stack Exchange.
Consider the following real-valued process: let $U$ be exponentially distributed, and let $X_t = t$ when $t < U$ and $X_t = \partial$ otherwise (where $\partial$ is the cemetery state; or, in fact, any extra state which is a trap for $X$). In other words, $X_t$ is the uniform motion to the right which is killed at a uniform rate. It is quite standard to see that $X_t$ is a nice Markov process (e.g. a Hunt process).
Consider $B = (0, \infty)$ and $X_0 = 0$. Obviously, $\mathbb{P}_0(X_t \in B \text{ for some $t$}) = 1$, because $X_t$ is never killed right away.
On the other hand, there is no stopping time $\tau$ such that $X_\tau \in B$ with probability one. Indeed, suppose $\tau$ exists. Clearly, $\mathbb{P}_0(X_\tau \ne \partial) = 1$, so that $\mathbb{P}_0(\tau < U) = 1$. By the strong Markov property, $Y_t = X_{\tau + t} - X_\tau$ has the same law as $X_t$ (but see below). Since $U - \tau$ is the life-time of $Y_t$, $U - \tau$ has the same distribution as $U$, which means that $\tau = 0$. However, $X_0 \notin B$, a contradiction.
As usual, there are some technical details hidden in the above description:
Everything is up to "events of probability zero": in fact the law of $Y_t$ is only equivalent to that of $X_t$ (rather than identical).
The strong Markov property is somewhat non-standard. One may switch to the usual formulation (that $X_{\tau + t}$ is the same as $X_t$ started with the distribution of $X_\tau$), but the description becomes less clear in this case.
A completely rigorous construction would be as follows. Let $\Omega$ consist of pairs $\omega = (x, u)$, where $x \geqslant 0$ is the starting point and $u \geqslant 0$ is the value of $U$, so that $U(\omega) = u$ and $X_t(\omega) = x + t$ when $t < u$, $X_t(\omega) = \partial$ otherwise. Let $\mathbb{P}_x$ be the exponential distributon on the fibre $\{x\} \times [0, \infty)$. The $\sigma$-algebra $\mathcal{F}_t$ (before completion) is generated by events of the form $\{s < u\} \cap \{x + s > a\}$, $a \in \mathbb{R}$, $s \in [0, t]$. It follows that for every $A \in \mathcal{F}_t$ and $x \geqslant 0$ the set $A^x := \{u \ge 0 : (x, u) \in A\}$ either contains $[t, \infty)$ or it has empty intersection with $[t, \infty)$.
For $t > 0$ write $A_t = \{\tau \geqslant t\}$. Clearly, $\mathbb{P}_0(X_\tau \ne \partial) = 1$, so that $\mathbb{P}_0(\tau < U) = 1$. Consequently, $\mathbb{P}_0(A_t, U \le t) = 0$, and thus $A_t^0 \cap [0, t]$ has zero Lebesgue measure.
Since $A_t$ is in the completion of $\mathcal{F}_t$, the set $[t, \infty)$ is either disjoint from $A_t^0$ or it is contained in $A_t^0$, up to a set of zero Lebesgue measure. In the first case, $A_t^0$ has zero Lebesgue measure, and consequently $\mathbb{P}_0(A_t) = 0$. In the latter, $A_t^0 = [t, \infty)$ up to a set of zero Lebesgue measure, and consequently $\mathbb{P}_0(A_t) = \mathbb{P}_0(U \geqslant t)$.
It follows that $\mathbb{P}_0(\tau \geqslant t)$ is equal to either $0$ or $\mathbb{P}_0(U \geqslant t)$. Thus, with probability $\mathbb{P}_0$ one, we have $\tau = \min\{\tau, t_0\}$ for some $t_0 \in [0, \infty]$. A contradiction: either $\mathbb{P}_0(\tau = 0) > 0$ (if $t_0 = 0$) or $\mathbb{P}_0(\tau = U) > 0$ (if $t_0 \ne 0$).
