Is there a continuous-time analogon to show that $\{\{t \geq 0: X_{t}=x\}\; \text{ is unbounded set}\}$ is a $0-1$ event for a Markov process $(X_{t})_{t\geq 0}$ on a discrete state space $E$
My attempt: assume that $P_{x}(\{\{t \geq 0: X_{t}=x\}\; \text{ is unbounded set}\})<1$
Then with positive probability, there exists $A$ with $P_{x}(A)>0$ such that $\forall \omega \in A$ we have $\exists T(\omega)>0$ such that for all $t > T(\omega)$, $X_{t}(\omega)\neq x$, I want to introduce some kind of stopping time, I will choose: $R_{x}:=\inf\{t \geq 0: X_{t}=x\}$.
$P_{x}(\{\{t \geq 0: X_{t}=x\}\; \text{ is unbounded set}\})=\mathbb E_{x}[1_{\{t \geq 0: X_{t+R_{x}}=x\}\; \text{ is unbounded set}}1_{\{R_{x}<\infty\}}]=\mathbb E_{x}[\mathbb E_{x}[1_{\{t \geq 0: X_{t+R_{x}}=x\}\; \text{ is unbounded set}}\lvert \mathcal{F}_{R_{x}}]1_{\{R_{x}<\infty\}}]=\mathbb E_{x}[P_{x}(1_{\{t \geq 0: X_{t}=x+R_{x}\}\; \text{ is unbounded set}}\lvert \mathcal{F}_{R_{x}})1_{\{R_{x}<\infty\}}]=\mathbb E_{x}[P_{X_{R_{x}}}(1_{\{t \geq 0: X_{t}=x\}\; \text{ is unbounded set}})1_{\{R_{x}<\infty\}}]=P_{x}(\{\{t \geq 0: X_{t}=x\}\; \text{ is unbounded set}\})P_{x}(R_{x}<\infty)$
Could this fact help me at all?
I guess I can assume cadlag sample paths so that we may use the strong Markov property.
You have to modify $R_x$ and define it as $R_x=\inf\{t \ge 1 : X_t=x\}$. If $P(R_x<\infty)=1$ then the set $\{t \ge 0 :X_t=x\}$ is unbounded a.s., while if $P(R_x<\infty)<1$ then the set $\{t \ge 0 :X_t=x\}$ is bounded a.s. (Indeed it can be covered by a finite number of unit intervals).