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Show that unique ergodicity is a topological invariant.

Is arguing as follows an overkill (hopefully if the logic is correct — I have a feeling that there has to be a way a $T$-invariant measure has to depend on the $S$-invariant measure and vice-versa and so that they are unique together)?

Let $T$ be a uniquely ergodic transformation acting on the probability space $(X, \mathfrak{A}, \mu)$ and $S$ a transformation acting on $(Y, \mathfrak{B}, \nu)$. Let $h: (X, \mathfrak{A}) \to (Y, \mathfrak{B})$ be a homeomorphism such that $h \circ T = S \circ h$. Note that $T^{k} = h^{-1} \circ S^{k} \circ h$. Suppose that $S$ is uniquely ergodic. We wish to show that $T$ is uniquely ergodic. Let $\phi \in C(X)$.By one equivalence characterizations of unique ergodicity,we must show that $\frac{1}{n}\sum_{k=0}^{n-1}\phi(T^{k}(x))$ converges to a constant pointwise on $X$.

Since $S$ is uniquely ergodic. We have in particular for a continuous function $\psi = \phi \circ h^{-1} \in C(Y)$ and each point $h(x) = y$, the time average $\frac{1}{n}\sum_{k=0}^{n-1} (\phi \circ h^{-1})(S^{k}(h(x)) )$ converges pointwise to a constant on $Y$. But then so is $\frac{1}{n}\sum_{k=0}^{n-1}\phi(T^{k}(x))$ because $\frac{1}{n}\sum_{k=0}^{n-1}\phi(T^{k}(x))= \frac{1}{n}\sum_{k=0}^{n-1} (\phi \circ h^{-1})(S^{k}(h(x)) )$.

Alp Uzman
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Daniel Akech Thiong
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1 Answers1

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Probably by now this is clear, but here is an answer for posterity.

First, this is exercise 4.1.6. of Katok & Hasselblatt's Introduction to the Modern Theory of Dynamical Systems (p.145).

Next, although unique ergodicity is a purely measure theoretical concept (A measurable self-map $T:X\to X$ of a measurable space $X$ is uniquely ergodic if it has a unique invariant probability measure, or alternatively it has a unique invariant ergodic probability measure), topological invariance is not a measure theoretical concept, and indeed, given the context of the exercise the spaces that carry dynamics ought to be taken as compact metrizable spaces. (This is not to say that this is the most general context in which the statement "unique ergodicity is an invariant" is true.)

Thus the statement, when written out explicitly, reads:

Exercise: If $X,Y$ are compact metrizable, $T:X\to X$ and $S:Y\to Y$ are continuous, and $H:X\to Y$ is a homeomorphism such that $H\circ T=S\circ H$, then $T$ is uniquely ergodic iff $S$ is uniquely ergodic.

The proof the OP presents above is indeed perfectly valid in my opinion, though there is a much more structural proof that uses only the definition of unique ergodicity (and some abstract nonsense); this is in line with the hunch the OP has.

First note that if $A$ is compact metrizable, then the set $\operatorname{Prob}(A)$ of (Borel) probability measures on $A$, when endowed with the vague topology, is a nonempty compact convex space. If $f:A\to B$ is a continuous function betwen compact metrizable spaces, then there is an induced affine map $f_\ast=\operatorname{Prob}(f):\operatorname{Prob}(A)\to\operatorname{Prob}(B)$, called typically the pushforward under $f$; it is defined like so: For $\mu$ a probability on $A$, and $E$ a measurable subset of $B$, $f_\ast(\mu)(E)=\mu(f^{-1}(E))$ (Indeed, $\operatorname{Prob}$ is a functor, see https://ncatlab.org/nlab/show/Giry+monad .) .

In our situation $T_\ast:\operatorname{Prob}(X)\to \operatorname{Prob}(X)$ is affine, and the set of fixed points of it is exactly the set $\operatorname{Prob}(X,T)$ of probability measures on $X$ invariant under $T$. Classical theory guarantees that $\operatorname{Prob}(X,T)\neq\emptyset$.

Finally, observe that since $H:X\to Y$ is a homeomorphism, $H_\ast:\operatorname{Prob}(X)\to \operatorname{Prob}(Y)$ is an affine automorphism, and $H\circ T=S\circ H$ gives $\mu=(H^{-1})_\ast(\nu)\in \operatorname{Prob}(X,T) \iff \nu=H_\ast(\mu)\in\operatorname{Prob}(Y,S)$, so that $\operatorname{Prob}(X,T)$ is a singleton iff $\operatorname{Prob}(Y,S)$ is; this is exactly the topological invariance of unique ergodicity.

Alp Uzman
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