TITLE: Markov strong property exercise
QUESTION [6 upvotes]: Let $(X_n)$ be a Markov chain with Q being its transition matrix.

Let $T=\inf\{n \ge 0:X_n \in A\}$ and Let $u(x)=P_x(T<+\infty)$.
Prove that $u$ verifies the system :
$$
\begin{cases}{}
u(x)=1 &\text{if } x\in A   \\ 
u(x)=Pu(x) &\text{if } x\notin A
\end{cases}
. 
$$
My attempt and understanding :
My understanding is that $T$ represents "the first time we get to the subset $A$" and $u(x)$ represents "the probability of hitting the subset $A$ starting from a $X=x$" ( because $P_x(T=+\infty)$ should represent the probability of never getting in the subset $A$).
That being said, the first part of the system makes sense because if $(X=x) \in A$ then $T=0$ ( smallest $n$ ) and we are already in $A$ so the probability of getting to $A$ should be equal to $1$.
The problem is the second part, how am I going to use markov strong property to prove that?

REPLY [3 votes]: Let $P=(p_{ij})$ and let $I$ be the state space. Suppose $X_0=i\not\in A$, so $T\geq1$. By the usual Markov property we have $$\mathbb P_i(T<\infty\mid X_1=j)=\mathbb P_j(T<\infty)=u(j).$$
Using the law of total probability, we can condition on the possible values of $X_1$ to get
\begin{align*}
u(i)=\mathbb P_i(T<\infty)=\sum_{j\in I}\mathbb P_i(T<\infty\mid X_1=j)\cdot\mathbb P_i(X_1=j)=\sum_{j\in I}p_{ij}\cdot u(j)=(Pu)(i).
\end{align*}