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
$$ \mathbb{P}(\exists \, t \geq 0 \textrm{ s.t. } X\_t \in A) \ = \ 1. $$



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


1. $X\_0$ is (almost surely) equal to some constant value $\xi\,$;
2. for almost all $\omega$ the set $\{X\_t(\omega):t \geq 0\}$ is equal to the whole of $\mathbb{R}\,$;
3. $(X\_t)\_{t \geq 0}$ 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.



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> > 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?
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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.



Can real-valued Markov processes with continuous surjective sample paths admit a non-trivial "forward-invariant" set?
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.


