Closure of an invariant set Consider a complete metric compact space $X$. For each $x\in X$ we define a probability measure $T(\cdot|x)$ over a Borel sigma-algebra $\mathcal{B}(X)$. We call a set $A\subset X$ invariant if $T(A|x) = 1$ for all $x\in A$. Does it mean that if $A$ is invariant, the same holds for its closure?
I am especially interested in the case when $T$ is Feller continuous or strong Feller continuous.
More precisely, denote
$$
\mathcal{P}f(x) = \int\limits_X f(y)T(dy|x)
$$
and spaces $\mathcal{M}_b$ and $\mathcal{C}_b$ of measurable bounded and continuous bounded functions on $X$. Then Feller continuity means $f(x)\in \mathcal{C}_b \Rightarrow \mathcal{P}f(x)\in\mathcal{C}_b$ and strong Feller continuity means $f(x)\in \mathcal{M}_b \Rightarrow \mathcal{P}f(x)\in\mathcal{C}_b$.
 A: In general, no; consider a deterministic process on $\mathbb{R}$ which, when started from $x \in (-\infty, 0)$, moves left with constant speed, but when started from $x \in [0, +\infty)$ moves right.  Then $(-\infty, 0)$ is invariant but $(-\infty, 0]$ is not.
If you want a compact space, you could use $[-1,1]$ instead and make the endpoints absorbing.
If $T$ is strong Feller, the answer is yes: if $A$ is invariant, then $T(A|x) = 1$ for $x \in A$.  But $T(A | x) = \mathcal{P}1_A$ is continuous, so $T(A|x) = 1$ for $x \in \bar{A}$.  Thus for $x \in \bar{A}$, $T(\bar{A} | x) \ge T(A|x) = 1$.
If $T$ is Feller, the answer is also yes.  By Urysohn's lemma we may find a sequence $f_n$ of bounded continuous functions with $0 \le f_n \le 1$, $f_n = 1$ on $\bar{A}$, and $f_n \downarrow 1_{\bar{A}}$.  For $x \in A$, we have $\mathcal{P}f_n(x) \ge \mathcal{P}1_A(x) = 1$, and since $\mathcal{P}f_n$ is continuous, $\mathcal{P} f_n(x) \ge 1$ for all $x \in \bar{A}$.  But by monotone convergence, $\mathcal{P} f_n(x) \downarrow \mathcal{P} 1_{\bar{A}}(x) = T(\bar{A} | x)$, so $T(\bar{A} | x) = 1$ for all $x \in \bar{A}$.
