This is continuation of the question stated here.
Let $\left( {{X_\alpha }:\alpha \in A} \right)$ be a finite space Markov chain (discrete or continuous), consisting of only transient and absorbing states. Let $S$ be a set of all absorbing states, ${T_S}$ the appropriate hitting time and $\mathbb{P}\left( {{X_0} \in S} \right) = 0$. It can be shown that $\mathbb{P}\left( {{T_S} < \infty } \right) = 1$, that ${{T_S}}$ is a stopping time and that ${X_{{T_S}}}$ is a random variable on $\left\{ {{T_S} < \infty } \right\}$ (${X_{{T_S}}}$ is defined as ${X_{{T_S}}}\left( \omega \right) = {X_\alpha }\left( \omega \right)$ if ${T_S}\left( \omega \right) = \alpha $).
Furthermore, we know that the set of $\omega \in \Omega $ for which $\mathop {\lim }\limits_{\alpha \to \infty } {X_\alpha }$ exists is an event, let's denote it by $C$.
I conclude that $\left\{ {{T_S} < \infty } \right\} \subseteq C \Rightarrow 1 = \mathbb{P}\left( {{T_S} < \infty } \right) = \mathbb{P}\left( C \right)$ so ${X_\alpha }$ converges almost surely.
My question: Does ${X_\alpha }\mathop \to \limits^{{\text{a}}{\text{.s}}{\text{.}}} {X_{{T_S}}}$? If not, does ${X_\alpha }\mathop \to \limits^\mathcal{D} {X_{{T_S}}}$?
Let $Y=X_{T_S}$ and $$C=\left[\lim\limits_{\alpha\to\infty}X_\alpha=Y\right].$$ For every fixed $\alpha$, on the event $[T_S\leqslant\alpha]$, $X_\beta=Y$ for every $\beta\geqslant\alpha$ hence $X_\beta\to Y$ when $\beta\to\infty$. This proves that $[T_S\leqslant\alpha]\subseteq C$.
This inclusion holds for every $\alpha$ and $\bigcup\limits_\alpha[T_S\leqslant\alpha]=[T_S\ \text{finite}]$ hence $[T_S\ \text{finite}]\subseteq C$. Since $P(T_S\ \text{finite})=1$, this implies that $P(C)=1$, that is, $X_\alpha\to Y$ almost surely.
The almost sure convergence implies the convergence in distribution, hence $X_\alpha\to Y$ in distribution as well.