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Let $X$ be a strong Markov process in the continuous time with a state space $\mathbb R^n$. Consider a reachability problem for this process, i.e. $$ v(x):=\mathsf P_x(X_t\in A\text{ for some }0\leq t<\infty) $$ for an poen set $A$. In the discrete time there is an well-known (Bellman) integral equation on $v(x)$. It is necessary condition on $v$, the actual solution is given through the supremum over all solutions of this equation over function bounded with $0$ and $1$.

I am interested if there are similar results in the continuous time. In one paper I've read that if $X$ is a diffusion process then $$ \mathcal Av(x) = 0,\text{ for }x\in A^c $$ and $v(x) = 1$ for $x\in\partial A$. Unfortunately, there were no strict conditions on the process $X$ as well as a strict proof of such a characterization.

I also asked in on MSE.

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@Gortaur : Hi I think you might have a look at a paper from Mijatovic and Pistorius "Continuously Monitored Barrier Options under Markov Processes" where if I remember well some results you might be interested in are either proved or referenced. Regards – The Bridge Aug 10 '11 at 12:07
Maybe this isn't what you're asking, but it seems to me that for a diffusion the fact that $\mathcal{A}v(x)=0$ doesn't really need a proof – ShawnD Mar 20 '12 at 16:26
@ShawnD: well, to claim that something in mathematics doesn't need it proof is rather strong, isn't it? For example, it might happen that $v(x) = 1_{\mathbb R^n\setminus \{0\}}(x)$ even for a diffusion. – Ilya Mar 22 '12 at 10:58

This is probably a consequence of the Kolmogorov backward/forward equations, by noting that the function $v$ does not depend on time. See my answer to your other question for references on Kolmogorov equations (I realize this is a bit self-promoting on MO, but I have just seen both questions and they are intimately connected, so I plan to answer both anyway).

But as you said, this is a bit folklore, and I haven't seen a rigorous treatment either - if you have found a source of rigorous proof of this, please also let me know. Intuitive it's quite clear.

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