How can we make the "time of first return" function well-defined? Context: dynamics and recurrence of measure-preserving systems

43 Views Asked by At

First, some preliminary definitions for those not familiar with the notations of the text on ergodic theory which I am following:

Let $(X,\Sigma,\mu;\varphi)$ be a measure-preserving system - $\varphi:X\to X$, the dynamic, satisfies $\mu(\varphi^{-1}(A))=\mu(A),\,\forall A\in\Sigma$ and $(X,\Sigma,\mu)$ is a probability space - and let $\Sigma(X)$ denote the measure algebra, where one equates $A\sim B\iff\mu(A\triangle B)=0$ and $\Sigma(X)=\Sigma_{/\sim}$.

The induced map $\varphi^\ast:\Sigma(X)\to\Sigma(X)$ maps, where $[]$ denotes the equivalence class, $[A]\mapsto[\varphi^{-1}(A)]$, which is well-defined as $\varphi$ preserves measure.

In chapter $6$, when investigating the recurrence of sets under the dynamic, the book writes:

Let $A\in\Sigma(X)$. Define: $$\begin{align}B_0&=\bigcap_{n\in\Bbb N}\varphi^{\ast n}(X\setminus A)\\B_1&=\varphi^{\ast}(A)\\B_2&=\varphi^{\ast n}(A)\cap\bigcap_{k=1}^n\varphi^{\ast k}(X\setminus A)\end{align}$$

The set $B_0$ represents the equivalence class of all points which never arrive in $A$ under the dynamic, and $B_1,B_2,\cdots,B_n,\cdots$ are the equivalence classes of the sets of all points arriving in $A$ for the first time at time step $n$.

Then does $(B_n)_{n\in\Bbb N_0}$ form an essentially disjoint union of $X$ in the measure algebra and $(A_n=A\cap B_n)_{n\in\Bbb N}$ is an essentially disjoint decomposition of $A$ (we know that $A\subseteq\bigcup_{n\in\Bbb N}B_n$ by poincare's recurrence theorem).

All equalities are in the sense of the measure algebra, i.e. up to a null set within an equivalence class.

What troubles me is this definition, made on page $97$ (page $115$ of the .pdf):

For a set $A\in\Sigma$, define: $$n_A:A\to\Bbb N,\,n_A(x)=n\,\text{if $x\in A_n$}$$

Which is the "time of first return" function for $A$.

Issues:

  • The $(A_n)$ are an essentially disjoint decomposition, but in terms of pointwise operation there will be potentially much overlap - $n_A(x)$ is not uniquely defined
  • $n_A(x)$ is not even defined as a natural number everywhere - the recurrence theorem gives that we know almost all points of $A$ return to $A$, but there will potentially be a nonempty subset of $A$ for which $n_A$ does not exist as a finite natural as those points are never in $A$.

Furthermore, for a non-null set $A$, they define the conditional probability measure $\mu_A(B)=\frac{\mu(B)}{\mu(A)}$ for $B\subseteq A$ measurable, and in theorem $6.16$ they talk about:

$$\int_An_A\,\mathrm{d}\mu_A$$

And they equate it with:

$$\sum_{j=1}^\infty j\cdot\mu(A_j)$$

Which leads me to believe that they really meant this:

$$n_A:\Sigma(A)\to\Bbb N,\,[B]\mapsto\{n\in\Bbb N:B\cap A_n\neq[\emptyset]\}$$

Because now considered as a setwise operation on the measure algebra, it is well defined as the quibbles about "almost all" and "essentially disjoint" disappear in the measure algebra - the output is never an empty set, and consists of all $n$ associable to the return times of almost all the points of $B$. However, this does leave a quibble over integration: $\int_An_A\,\mathrm{d}\mu_A$ is now no longer defined in the usual way under my construction, unless we consider $\Sigma(A)$ a measurable space with measure $\mu_A([A])=\mu_A(A)$, I think.

This is all a bit murky, which is what comes of using "$=$" when one does not mean true equality.

How should I interpret $n_A$? How is it supposed to be well-defined?

Many thanks.

1

There are 1 best solutions below

1
On BEST ANSWER

Quick answer $n_A$ is not defined everywhere but it doesn't matter (and it is a map from $A$ to $\mathbb{N}$).

The $(_)$ are an essentially disjoint decomposition, but in terms of pointwise operation there will be potentially much overlap -$n_A(x)$ is not uniquely defined.

I don't understand why you say that. A point can be at most in one of the $B_n$ so there is no ambiguity.

And yes $n_A$ is defined for almost all point. For the rest of the point, let call this set $D$, you can choose your favorite value (can be $\frac{\sqrt{2}}{\pi}$) and say $n_A$ take this value on it.

Indeed, what really matter in ergodic theory are not the value of function but rather their integral. And as $D$ is a null set you have $$ \int_A n_A d\mu_A=\int_{A/D} n_A d\mu_A. $$ So the value it take on $D$ is irrelevant.