Measure of the set of periodic points of a measure preserving map

Question

Given a continuous (Lebesgue) measure preserving map $T$ from a compact convex region to itself that has an aperiodic point (i.e. a point $p$ such that $p \ne T^n(p)$ for any $n$), does the set of aperiodic points necessarily have positive measure?

It can be arbitrarily small as can be seen by taking a rubber disk of radius 1 and gluing the disk of radius $1-\epsilon$ having the same center and turning the outer ring by a small irrational amount (so we are deforming only a small annulus), but can it be zero?

Answer 1

Yes, it can be zero. Take your region for instance to be $D=[0,1]\cup\{1+\frac1n\,:\,n\in\mathbb{N}\}$. Define $T(x)=x$ for $x\in[0,1]$ and $T(1+\frac1n)=1+\frac1{n+1}$. Then clearly $T$ is continuous and measure preserving, and the set of aperiodic points is $\{1+\frac1n\,:\,n\in\mathbb{N}\}$ which has measure zero. I imagine the answer would be different if you also assumed $D$ is connected, but I haven't given that much thought.

Answer 2

One simple way is to define the compact convex region itself to have zero measure, such as a 2-d square in 3-space. For example, consider the following set $\mathcal{X}\subseteq \mathbb{R}^3$, which is a compact and convex set with measure zero:

$$ \mathcal{X} = \{(x,y,0) : x \in [0,1], y \in [0,1]\} $$

Define $T(x,y,0) = (x^2,y,0)$. Then the point $(1/2,0,0)$ is aperiodic, and the set of all aperiodic points trivially has measure zero.

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Of course, this example breaks the spirit of your question. I think you are looking for a compact, convex set $\mathcal{X} \subseteq \mathbb{R}^N$ that itself has positive measure. Just a few observations about that:

1) Define $\mathcal{A}_k = \{x \in \mathcal{X} : T^{(k)}(x)=x\}$. Then $\mathcal{A}_k$ is closed (and hence measurable).

Proof: Let $x$ be a limit point of $\mathcal{A}_k$ and let $\{x_n\}_{n=1}^{\infty}$ be an infinite sequence of points in $\mathcal{A}_k$ that converges to $x$. Then $T^{(k)}(x_n)=x_n$ for all $n$. Taking a limit as $n\rightarrow\infty$ and using continuity of $T^{(k)}$ gives $T^{(k)}(x) = x$, so $x \in \mathcal{A}_k$. So $\mathcal{A}_k$ is closed (and hence measurable).

2) The set $\cup_{k=1}^{K}\mathcal{A}_k$ is closed and measurable, its limiting set $\cup_{k=1}^{\infty} \mathcal{A}_k$ is measurable, and so the set of all aperiodic points is measurable.

Proof: The finite union of closed sets is closed (and hence measurable). The limit of an increasing sequence of measurable sets is measurable. The set of all aperiodic points is the complement of $\cup_{k=1}^{\infty} \mathcal{A}_k$, and the complement of a measurable set is measurable.

3) If $x$ is an aperiodic point in $\mathcal{X}$ and if $\{x_n\}_{n=1}^{\infty}$ is an infinite sequence of points in $\mathcal{X}$ that converges to $x$, then the smallest period of $x_n$ diverges to infinity as $n\rightarrow\infty$.

Proof: Suppose there is a positive integer $N$ such that $x_n$ has smallest period less than or equal to $N$ for an infinite number of indices $n$. Then there is an integer $k\leq N$ such that $T^{(k)}(x_{n[i]})=x_{n[i]}$ for an infinite subsequence $x_{n[i]}$. Taking a limit as $i\rightarrow\infty$ gives $T^{(k)}(x)=x$, contradicting the fact that $x$ is an aperiodic point.