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I have found a statement in the introduction of the paper 'Sets of Recurrence and Generalized Polynomials' by Bergelson & Haland, which is

Result: Given a polynomial $p \in \mathbb{R}[x]$ such that $p(\mathbb{Z})\subseteq \mathbb{Z}$, the set $\{p(n) | n ∈ \mathbb{Z}\}$ is a set of recurrence iff $p(n)\equiv 0\;(\text{mod } a)$ has a solution for $n$ for each $a\in \mathbb{N}$,

where a set $S\subseteq \mathbb{Z}$ is called a set of recurrence if for any invertible measure preserving system $(X, \mathscr{B}, \mu, T )$ and any $A\in \mathscr{B}$ with $\mu(A)>0$ there exists $n\in S,n\neq 0$ such that $\mu(A\cap T^{-n}A) > 0$.

$\quad$ But there is another definition(for example the definition of 'Poincare sequence' in the book 'Recurrence in Ergodic Theory and Combinatorial Number Theory' by Furstenberg), which doesn't assume invertibility of the measure preserving system, i,e;

A set $S\subseteq \mathbb{Z}$ is called a set of recurrence if for any measure preserving system $(X, \mathscr{B}, \mu, T )$ and any $A\in \mathscr{B}$ with $\mu(A)>0$ there exists $n\in S,n\neq 0$ such that $\mu(A\cap T^{-n}A) > 0$.

$\quad$ My question is, if we consider the second definition for the 'set of recurrence', does the result still holds true? I mean is this a known result? If so, please share a reference. Thanks in advance.

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    $\begingroup$ Invertibility should not be an issue: as long as $(X,\mathcal B,\mu)$ is a Lebesgue space (this is not a serious restriction), every measure-preserving transformation on $X$ has a “natural extension”, that is an invertible transformation $\bar T:\bar X\to X$ preserving a measure $\bar\mu$. For each measurable set $A\subset X$, there is a corresponding $\bar A\subset \bar X$. You have recurrence in the natural extension if and only if there is recurrence in the original map. $\endgroup$ – Anthony Quas Sep 8 '18 at 20:02

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