Let $X$ be a homogeneous Markov process in a continuous time with value in the set $E$. Suppose that for some $T>0,x\in E, A\subset E$ we have
$$
P\_x[X\_t\in A] = 0
$$
for all $t\in [0,T]$ but
$$
P\_x[X\_{t'}\in A] >0
$$
for some $t'> T$. I wonder how to find an example of such a process $X$. The simple one 
$$
dX\_t = f(X\_t)dt
$$
in not interesting because it is purely deterministic.



Counterexample Markov process
This question was already asked [here](https://mathoverflow.net/questions/50337/). Homogenous Markov chains won't work because the probability distributions of $X\_t$ and $X\_s$ are mutually absolutely continuous for every positive $t$ and $s$. For a homogenous diffusion with nonzero diffusion term, $P[X\_{2t}\in A]\ne0$ would require that $P\_y[X\_t\in A]\ne0$ for $y$ in a set of positive measure with respect to the probability distribution of $X\_t$ while $P\_x[X\_t\in A]=0$. Unless I am missing something, this is impossible.


The idea of the proof for finite Markov chains is as follows. Assume that $P\_x[X\_t\in A]$ is positive for a given positive $t$. The transitions of the chain can be wriiten as 
$$
P\_y[X\_t=z]=(\mathrm{e}^{tQ})(y,z),
$$ 
where $Q$ is the infinitesimal generator of the process. Choose $u$ such that $u\ge -Q(y,y)$ for every state $y$, then $uI+Q$ has nonnegative coefficients. Now,
$$
P\_x[X\_t\in A]=\mathrm{e}^{-ut}(\mathrm{e}^{t(uI+Q)})(x,A)
$$ 
is a linear combination with positive coefficients of nonnegative terms $(uI+Q)^n(x,A)$, hence $(uI+Q)^n(x,y)$ is positive for at least one given $n$. 


Now, for every positive time $s$, $P\_x[X\_s\in A]$ is also a linear combination with nonnegative coefficients of $(uI+Q)^k(x,y)$, in particular $P\_x[X\_s\in A]$ is at least $\mathrm{e}^{-us}(s^n/n!)(uI+Q)^n(x,A)$, hence is positive.


