Consider a real-valued Markov process $X$ with a transition density $f(x,y)$, i.e.
$$
\mathsf P[X\in A|X\_0 = x] = \int\limits\_A f(x,y)\,dy.
$$
For this process I want to find 
$$
u(x) = \mathsf P[X\_n < 0\text{ for some }n\geq 0|X\_0 = x].
$$


Since for this problem does not matter the distribution of this process for $X\_0 = x<0$ I would like to modify it in the following way. I define a modification $Y$ taking values in $\mathbb R^+\cup\{a\}$ with $a<0$ such that: 


1. for any $A\subset \mathbb R^+$ holds $\mathsf P[Y\_1\in A|Y\_0 = x] = \mathsf P[X\_1\in A|X\_0 = x]$;
2. for $x\geq0$ we pur $\mathsf P[Y\_1 = a|Y\_0 = x] = \mathsf P[X\_1\leq 0|X\_0 = x]$;
3. $\mathsf P[Y\_1 = a|Y\_0 = a] = 1$.


I hope that for this process 
$$
v(x) = \mathsf P[Y\_n < 0\text{ for some }n\geq 0|Y\_0 = x]
$$
coincides with $u(x)$ but I don't know how to prove this fact.
This question is also presented [here](https://math.stackexchange.com/questions/46699/modification-of-a-markov-process-on-the-real-line).



Modification of a Markov process on the real line
Here is a realization of the process $(Y\_n)$, which shows that $u(x)=v(x)$ for every $x\ge0$. Fix $x\ge0$, consider $(X\_n)$ starting from $X\_0=x$, introduce the stopping time $\tau=\inf\{n\ge0;X\_n<0\}$ and the function $\varphi$ such that $\varphi(y)=y$ for every $y\ge0$ and $\varphi(y)=a$ for every $y<0$, and finally define 
$$
Y\_n=\varphi(X\_{n\wedge\tau}),
$$ 
for every $n\ge0$. You are done.


