Is there a version of Poincare recurrence theorem on attracting set? In this paper, on the left-side on page 2, it has a statement of poincare recurrence theorem, which I quoted here:
Statement 1: Our algorithm relies
on the Poincaré recurrence theorem, which states that a trajectory on an attracting set will sooner or later visit the same regions of the state space.
I found it is difficult to find which version of poincare recurrence theorem match this statement. The poincare recurrence theorem stated in many textbooks in ergodic theory and dynamical system is about volume preserving system, and the recurrence property applies to any bounded region in the phase space. I quote the definition from the book Mathematical methods of classical mechanics from V.I. Arnold here:
Statement 2: Let $g$ be a volume-preserving continuous one to one mapping which maps a bounded region $D$ of euclidean space onto itself: $gD=D$. Then in any neighborhood $U$ of any point of $D$ there is a point $x\in U$ which returns to $U$, i.e. $g^nx\in U$ for some $n>0$.
Could someone help to clarify what the first statement of poincare recurrence is about? maybe there exists a version of the theorem that specialized for attracting set?
 A: That version of the Poincaré recurrence theorem that you cite is just one special case of a much more general theorem.
The Wikipedia page for Poincaré recurrence seems to have a good statement of the general theorem. Here is a brief description.
In your setting $g$ restricts to an area preserving self-map $g : D \to D$. So you can generalize $D$ to any set $\Sigma$ equipped with a $\sigma$-algebra and a finite measure $\mu$. And then you can generalize $g : D \to D$ to any measure preserving transformation $f : \Sigma \to \Sigma$ meaning that for any measurable subset $A \subset \Sigma$ its inverse image $f^{-1}(A)$ is measurable and $\mu(f^{-1}(A)) = \mu(A)$. The conclusion of the theorem is that the subset
$$\{x \in A \mid \exists n \ge 1, f^n(x) \in A\}
$$
has full measure $\mu(A)$ (for any measurable $A \subset \Sigma$).
Notice that one-to-one is not even mentioned in this more generalized version. In the special case that $f$ is one-to-one, the measure preserving requirement $\mu(f^{-1}(A))=\mu(A)$ is equivalent to $\mu(f(B))=\mu(B)$ (identifying $B$ with $f^{-1}(A)$).
Notice also that continuity is not mentioned. All that is important is that continuity$+$area preserving implies that the transformation is measure preserving.
