Continuity of the invariant measure of a perturbed Markov chain Consider an aperiodic and irreducible continuous-time Markov chain $X_\epsilon(t)$ on a countable state space $S$, where $\epsilon\ge 0$ is a "perturbation" parameter. Assume that the transition matrix $Q_\epsilon=[q_{i,j}(\epsilon)]$ is such that the $q_{i,j}(\epsilon)$'s are continuous in $\epsilon$. Let $\pi_\epsilon$ denote the invariant probability measure when the chain is perturbed by $\epsilon$.
My question is whether or not $\lim_{\epsilon\downarrow 0} \pi_\epsilon(i)=\pi_0(i)$ for all $i\in S$. If not, under which conditions the previous limit holds true? This should be a classic question but I am not very successful in finding appropriate references. Any reference is highly appreciated.
 A: EDITED: This proof (as written) only works if the state space is finite.
This post is a partial answer only, and shows that, if the state space is finite, and if the limit $\lim_{\epsilon\downarrow 0} \pi_\epsilon$ exists, then it equals $\pi_0$.
Define $T_\epsilon = Q_\epsilon - I$, then by definition $\pi_\epsilon$ is the unique left probability vector (i.e. vector whose entries sum to $1$) in the null space of $T_\epsilon$, i.e. $\pi_\epsilon T_\epsilon = 0$.  It is unique because $Q_\epsilon$ is aperiodic and irreducible.
Consider $\pi_\epsilon T_0 = \pi_\epsilon (T_0 - T_\epsilon)$.  (Equality due to $\pi_\epsilon T_\epsilon = 0$.)
As $\epsilon \downarrow 0, (T_0 - T_\epsilon) \rightarrow 0$ by continuity of $Q_\epsilon$.  This, coupled with the fact that all the entries of $\pi_\epsilon$ are bounded (specifically, $\forall \epsilon, \forall i, \pi_\epsilon(i) \in [0,1]$), and there are a finite number of entries, allows us to conclude that $\pi_\epsilon (T_0 - T_\epsilon) \rightarrow 0$, and consequently:
$$\lim_{\epsilon \downarrow 0} (\pi_\epsilon T_0) = 0.$$
This also means that, if $\lim_{\epsilon \downarrow 0} \pi_\epsilon$ exists, we can exchange $\lim$ and $\sum$ (because the state space is finite) and write:
$$(\lim_{\epsilon \downarrow 0} \pi_\epsilon) T_0 = 0.$$
Given that $Q_0$ is also aperiodic and irreducible, the only left probability vector in the left null space of $T_0$ is $\pi_0$, so $\lim_{\epsilon \downarrow 0} \pi_\epsilon = \pi_0$.
However, so far I cannot prove that the limit actually exists.  $\lim_{\epsilon \downarrow 0} (\pi_\epsilon T_0) = 0$ leaves open the possibility that $\pi_\epsilon$ actually oscillates, while getting closer and closer to the left null space of $T_0$.  The facts that (1) all $\pi_\epsilon$ are probability vectors and (2) $T_0$ only has a 1-dimensional left null space, should together somehow prevent the oscillation.  E.g. if we decompose $\pi_\epsilon = \alpha_\epsilon \pi_0 + \sigma_\epsilon$ where $\alpha_\epsilon$ is a scalar and $\sigma_\epsilon \perp \pi_0$, it should be possible to show that $\lim_{\epsilon \downarrow 0} \sigma_\epsilon = 0$, i.e. if there is any oscillation it happens purely in $\alpha_\epsilon$, but then fact (1) should prevent $\alpha_\epsilon$ from oscillating.  But I don't know how to make this argument rigorous.
