I have both a more general question (concerning stopping times), and then a more specific application (as described in the title).

Let $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t \geq 0},\mathbb{P})$ be a filtered probability space, which we can assume to satisfy the "usual conditions" if necessary.

Let $(X_t)_{t \geq 0}$ be a real-valued progressively measurable homogeneous Markov process over the above filtered probability space, and assume that $(X_t)$ has continuous sample paths.

Let $A \in \mathcal{B}(\mathbb{R})$ be a Borel set such that $$ \mathbb{P}(\exists \, t \geq 0 \textrm{ s.t. } X_t \in A) \ = \ 1. $$

Q1. Does there necessarily exist a stopping time $\tau:\Omega \to [0,\infty]$ such that $$ \mathbb{P}(X_\tau \in A) \ > \ 0 \, ? $$

Now I fear that the answer to Q1 is *no* (although I'd like it to be *yes*), since even in general, the existence of measurable selections is a highly non-trivial issue. Nonetheless, I am hoping that the particular claim which I wanted to prove using a positive answer to Q1 might still be true:

Suppose our Markov process above satisfies the following additional properties:

- $X_0$ is (almost surely) equal to some constant value $\xi\,$;
- for almost all $\omega$ the set $\{X_t(\omega):t \geq 0\}$ is equal to the whole of $\mathbb{R}\,$;
- $(X_t)_{t \geq 0}$ is described by transition probabilities $P_t(x,\cdot)$ satisfying the following type of continuity: for every bounded continuous $g:X \to \mathbb{R}$ the map $(t,x) \mapsto \int_\mathbb{R} g(y) \, P_t(x,dy)$ is continuous.

Q2. Is it necessarily the case that the only Borel set $B \in \mathcal{B}(\mathbb{R})$ satisfying:

- $\xi \in B$
- $P_t(x,B)=1$ for all $x \in B, \, t \geq 0$
- $P_t(x,B)<1$ for all $x \in \mathbb{R} \setminus B, \, t \geq 0$
is $\mathbb{R}$ itself?

If the answer to Q1 is *yes*, then I'm sure the answer to Q2 must also be *yes*: supposing otherwise for a contradiction, just take $A$ in Q1 to be $X \setminus B$ and use the strong Markov property.

no, as pointed out in Nate Eldredge's answer below.) $\endgroup$ – Julian Newman Mar 28 '15 at 11:42