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

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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. – Ilya Jun 17 '11 at 13:34
@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. – Nate Eldredge Jun 17 '11 at 15:00
@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? – Ilya Jun 17 '11 at 15:15
@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 $f_n \downarrow 1_{\bar{A}}$? – Ilya Jun 18 '11 at 14:11
@Gortaur: Actually, in this case we could do something easier: set $g(x) = \min(d(x, \bar{A}), 1)$, and let $f_n(x) = (1-g(x))^n$. – Nate Eldredge Jun 19 '11 at 14:51