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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 $f(x)\in \mathcal{M}_b \Rightarrow \mathcal{P}f(x)\in\mathcal{C}_b$.

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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 $x \in \bar{A}$.

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    @Nate: I guess that this example can be also obtained for compact sets as I want? Also I am not sure that this process is even (weak) Feller continuous.2011-06-17
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    @Gortaur: Sure, if you want compact, you could use $[-\infty, \infty]$ instead. (Or $[-1,1]$ and make the endpoints absorbing.) You are right that this process is not Feller.2011-06-17
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    @Nate: since you're using this notions of "absorbing" points, maybe you are also familiar with absorbing sets which I call invariant in my question?2011-06-17
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    @Nate: could you help me to understand how does this sequence is constructed from Urysohn's lemma? I guess that we made a sequence of closed sets $B_n$ to put $f_n$ to be an Urysohn's function of $\bar{A}$ and $B_n$, but I am not sure which sets $B_n$ can be used and why $f_n \downarrow 1_{\bar{A}}$?2011-06-18
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    @Gortaur: Actually, in this case we could do something easier: set $g(x) = \min(d(x, \bar{A}), 1)$, and let $f_n(x) = (1-g(x))^n$.2011-06-19
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    @Nate: As I understand, we use here that if $x\notin \bar{A}$ then $d(x,\bar{A}) >0$. It's not true in general, but we can use that $\bar{A}$ is a compact. Am I right?2011-06-20
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    @Gortaur: Yes. In fact all you need is that $\bar{A}$ is closed.2011-06-20
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    @Nate: Sorry, but how can we only use that this set is closed? And I am also curious how to construct the sequence of sets when the space in normal but not metrizable.2011-06-20
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    @Gortaur: Proposition: If $B$ is closed and $d(x,B) = 0$ then $x \in B$. Proof: Exercise :) using the definition $d(x,B) = \inf\{d(x,y) : y \in B\}$.2011-06-20
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    @Nate: It should be since $B^c$ is open and if $x\in B^c$ then $d(x;B)>0$. Have I done your exercise correct? What about the second question of mine?2011-06-21
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    @Gortaur: Thinking about it more, you need the space to be "perfectly normal", which means it is normal and every closed set is $G_\delta$ (i.e. a countable intersection of open sets). (Every metric space certainly has this property.) Then if $\bar{A} = \bigcap U_n$ with $U_n$ open, let $g_n$ be a Urysohn function for $\bar{A}$ and $U_n^c$, and let $f_n = \min\{g_1, \dots, g_n\}$.2011-06-21
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    @Gortaur: Conversely, if $F$ is closed and there is a sequence of continuous functions $f_n$ with $f_n \downarrow 1_{F}$, then $F = \bigcap_n f_n^{-1}((1/2, 2))$ so $F$ is $G_\delta$. Thus, if $F$ is not $G_\delta$ this argument will not work. There exist spaces which are compact, normal and first countable but not perfectly normal, i.e. contain a closed set which is not $G_\delta$. There are several examples in Steen and Seebach, *Counterexamples in Topology*.2011-06-21
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    @Nate: thank you very much.2011-06-21