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:

for any $A\subset \mathbb R^+$ holds $\mathsf P[Y_1\in A|Y_0 = x] = \mathsf P[X_1\in A|X_0 = x]$;

for $x\geq0$ we pur $\mathsf P[Y_1 = a|Y_0 = x] = \mathsf P[X_1\leq 0|X_0 = x]$;

$\mathsf P[Y_1 = a|Y_0 = a] = 1$.

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.