Can real-valued Markov processes with continuous surjective sample paths admit a non-trivial "forward-invariant" set? 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.
 A: The answer to Q1 is Yes, and the answer to Q2 is No.
If $\mathbb{P}(X_0 \in A) > 0$ then we are done by taking $\tau = 0$.  So suppose $\mathbb{P}(X_0 \in A^c) = 1$.  
Call $y$ a right endpoint of $A$ if $y \in A$ and there exists $x < y$ with $(x,y) \subset A^c$.  Let $A^+$ be the set of all right endpoints of $A$; then $A^+$ is countable.  Likewise let $A^-$ be the set of all left endpoints, which is also countable.  Finally, let $D \subset A$ be countable and dense in $A$.  Set $C = A^+ \cup A^- \cup D$.
Claim.  If $a < b$ and $[a,b] \cap A \ne \emptyset$ then $[a,b] \cap C \ne \emptyset$.  That is, any nontrivial closed interval that meets $A$ must meet $C$.
Proof of claim.  If $(a,b) \cap A \ne \emptyset$ then by density $(a,b)$ contains a point of $D$.  Otherwise, we have $(a,b) \subset A^c$.  Then we must have either $a \in A$, in which case $a \in A^-$, or else $b \in A$, in which case $b \in A^+$.  In all cases $[a,b]$ meets $C$.
We now know that for almost every $\omega$, we have that $X_0(\omega) \notin A$ and there exists $t$ such that $X_t(\omega) \in A$.  Suppose $X_0(\omega) < X_t(\omega)$; then by the claim, $[X_0(\omega), X_t(\omega)]$ intersects $C$, so by continuity there exists $s$ such that $X_s(\omega) \in C$.  The same holds if $X_0(\omega) > X_t(\omega)$.
Enumerate $C$ as $C = \{c_1, c_2, \dots\}$ and let $\sigma_n = \inf\{t : X_t = c_n\}$.  We have just shown that $\mathbb{P}\left(\bigcup_{n=1}^\infty \{\sigma_n < \infty\}\right) = 1$.  So by countable additivity we may choose $N$ so large that $\mathbb{P}\left(\bigcup_{n=1}^N \{\sigma_n < \infty\}\right) > 0$.  Set $\sigma = \sigma_1 \wedge \dots \wedge \sigma_N$; then $\mathbb{P}(\sigma < \infty) > 0$. Since $\{c_1, \dots, c_N\}$ is closed, on $\{\sigma < \infty\}$ we have $X_\sigma \in \{c_1, \dots, c_N\}$.  Choose $M < \infty$ large enough that $\mathbb{P}(\sigma < M) > 0$.  Finally set $\tau = \sigma \wedge M$.  Then $X_\tau$ is well defined, and on the event $\{\sigma < M\}$ we have $X_\tau = X_\sigma \in \{c_1, \dots, c_N\} \subset A$.  So $\mathbb{P}(X_\tau \in A) \ge \mathbb{P}(\sigma < M) > 0$.  (In fact by choosing $N,M$ sufficiently large we can make $\mathbb{P}(X_\tau \in A)$ arbitrarily close to 1.)
For Q2, let $X_t$ be standard Brownian motion, which satisfies properties 1,2,3  with $\xi = 0$.  Let $B = \mathbb{R} \setminus \{1\}$, so that $0 \in B$.  Since Gaussian measure is absolutely continuous to Lebesgue measure, $P_t(x,B) = 1$ for any $x$ and any $t > 0$, and $P_0(x,B) = 1$ when $x \in B$.  So Q2 is not satisfied.  (We could have chosen $B$ to be any set of full Lebesgue measure that contains 0.)
Something seems to be wrong with your "strong Markov property" logic.
