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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:

  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.

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.

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  • $\begingroup$ I've just edited the question to slightly weaken Q1. (It was stronger than necessary before.) I would assume that the answer to Q1 is a well-known fact - does anyone know? $\endgroup$ – Julian Newman Mar 28 '15 at 0:20
  • $\begingroup$ I've also edited Q2. (In its previous form, the answer was trivially no, as pointed out in Nate Eldredge's answer below.) $\endgroup$ – Julian Newman Mar 28 '15 at 11:42
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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.

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  • $\begingroup$ Thank you very much for this answer. I realise I was getting things muddled in my head as I wrote Q2. I've now corrected it (so I think it now asks the right thing!). Presumably the answer to Q2 is now yes - and I probably don't actually need Q1 to prove it: Suppose it's false, with $B$ being a counterexample. By Fubini, the set of times $t$ at which $X_t\in B$ is almost surely a Lebesgue-full set. But fix $c\in\mathbb{R}\setminus B$ and let $\tau$ be the hitting time associated to $\{c\}$; the strong Markov property at $\tau$ (+ Fubini) should contradict this. Is this right? $\endgroup$ – Julian Newman Mar 28 '15 at 10:32
  • $\begingroup$ PS: Just out of curiosity, does anyone know if the answer to Q1 generalises to higher dimensions, or even to general Polish spaces? $\endgroup$ – Julian Newman Mar 28 '15 at 11:40

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