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This is kind of a spin-off of the question asked here. Take the interval $X:=[0,1]$ with $\mu$ being standard Lebesgue measure. Let $f$ be a measure preserving map $f:[0,1]\rightarrow [0,1]$. The Poincare Recurrence theorem tells us that if I pick a measurable set $E\subset [0,1]$, then under iterations of $f$ almost every point in $E$ returns to $E$ infinitely often, i.e.

$\mu\left({x\in E: \ \exists N, \mbox{ such that } \forall n\geq N, \ f^n(x)\notin E}\right)=0$

Call the set of exceptions above $M$. My question is:

If I specify an $f$, for what class of sets $E$ in $X$ is $M$ not dense? I am interested in two cases:

1) "dense" with respect to $X$, if $E$ is dense in $X$

2) "dense" with respect to $E$

For example, let $E=\mathbb{Q}\cap[0,1]$ and $f(E):=E+\phi$ where $\phi$ is irrational. Then $M=E$, which of course is not a contradiction since $\mu(M)=\mu(E)=0$. On the other hand, throw in the set $H:=\{n\cdot \phi: n\in \mathbb{N}\}$, so that $E':=H\cup E$. In this case we still have $M=E$. As well, it looks like the class of sets I'm interested in is $H\cup {\mbox{not a dense set in X}}$.

Note: My original motivation for asking this was to try and conceptualize the Poincare recurrence theorem for a human physicist. If I were looking at the phase plot of balls on a billiard table at a specific time, I would only be able to give imprecise measurements of both position and velocity. In this case, it seems that in order to invoke Poincare recurrence, I would need small intervals around every point to recur perfectly, in the sense that $M=\emptyset$. Perhaps this IS the case if $f$ arises from some nice ODE, but I'm interested in a more general setting. I also don't really want to require that $\mu(M)>0$, which is why I feel asking about denseness is more appropriate.

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You certainly assume $E$ to be of positive measure, no? –  Stefan Geschke Nov 18 '10 at 7:41
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If you are interested in topological notions of size, such as density, rather than measure-theoretic notions, such as being of full or almost full measure, then ergodic theory will probably not give you quite what you want. You might prefer the related field called topological dynamics. –  Noah Stein Nov 18 '10 at 12:32
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3 Answers

up vote 1 down vote accepted

If $f$ is minimal, i.e. every orbit is dense, then $E$ is either empty or dense. So it remains to decide if your set is empty. Clearly it is non-empty for positive measure sets. If $\mu(E) = 0$. Then also $$ \mu(\bigcup_{n} f^{-n} E) =0 $$ and for this set the exceptional set is non-empty (and trivially dense).

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Theorem

If $T:X\to X$ transformation which preserves the measure $\mu$ of a a probability measure on $(X,\mathcal{B}(X))$. the following statements ara equivalent:

(i) $T$ is ergodic

(ii) The only members $B$ of $\mathcal{B}$ with $\mu(T^{-1}B\Delta B)=0$ are those with $\mu(B)=0~~or~~1$

(iii) For every $A\in\mathcal{B}$ with $\mu(A)>0$ we have $\mu(\bigcup_{n\geq 1}T^{n}A)=1$

These characterizations of ergodicity has a clear geometric meaning. But what I want to show you is the following theorem

Theorem

Let be $X$ a compact metric space with enumerable basis , $\mathcal{B}(X)$ the $\sigma$-algebra of Borel subsets of $X$ ande let $\mu$ be a probability measure on $(X,\mathcal{B}(X))$ such that $\mu(U)>0$ for all open set $U$ in $X$. Suppose $T:X\to X$ is a continuous transformation which preserves the measure $¨\mu$ and is ergodic. Then almost all points of $X$ have a dense orbit under T, i.e $$x\in X: T^{n}(x) ~~ is~~ dense ~~subset ~~ of ~~X$$ has $\mu$-measure one.

I'm not sure but I think you can do some connection between this theorem and the question you proposed about Poincare Recurrence. Hope this is of some value to you, for more details and proofs of the theorems above search for the book Peter Walters: an introduction to Ergodic Theory.

Ohh man I just remembered this theorem has a topological version! Let us remember before the definition of $\omega$-limit. Let be $X$ a topological space and $T:X\to X$ a trasformation. The $\omega$-limit set of $x\in X$ is defined by (denoted by $\omega(x)$) $y\in X $ such that for all neighborhood U of $y$ the relation $T^{n}(x)\in U$ is satisfied for infinite values of n.

Theorem

Let be $X$ a metric space and $T:X\to X$ a mensurable transformation which preserves the measure $\mu$ of a a probability measure on $(X,\mathcal{B}(X)).$ Then the set

$$x: x\notin \omega(x) $$ has 0 $\mu$-measure

This theorem you will find in the book the great mathematician Ricardo Mane: Ergodic theory

I'm not sure but I think you can do some connection between this theorems and the question you proposed about Poincare Recurrence. Hope this is of some value to you

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Since you are considering $\mu$ to be the Lebesgue measure, you have that given a set $F$ with $\mu(F) = 1$, then $F$ must be dense. This is because any set with non-empty interior has positive measure, and $F^\complement$ has measure $0$. So, $F^\complement$ has empty interior and consequently, $F$ is dense.

The above easily gives a partial answer to your first question:

If $\mu(E) = 1$, then $\mu(M) = \mu(E) = 1$. As explained, $M$ is dense in $X$.

In a similar fashion, for the second question:

If $E$ is such that for any open set $A$, $$A \cap E \neq \emptyset \Rightarrow \mu(A \cap E) \neq 0,$$ then $M$ is dense in $E$.
In fact, $\mu( M^\complement \cap E ) = 0$, so, the assumption on $E$ implies that $M^\complement \cap E$ has empty interior in $E$. That is, $M$ is dense in $E$.

Now, using this fact, we can improve the answer to the first question:

If $E$ is dense in $X$, $\mu(E) > 0$ and $$A \cap E \neq \emptyset \Rightarrow \mu(A \cap E) \neq 0,$$ then $M$ is dense in $E$, and consequently, dense in $X$.
It is obvious that if $E$ is not dense in $X$, then $M$ is not dense either.

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