(I first recall the definitions, but specialists can probably go directly to the question.)


A *twist map* of the annulus $A=(\mathbb R/\mathbb Z)\times \mathbb R$ is an orientation preserving homeomorphism $f=(f\_1,f\_2):A\to A$ that satisfies the "twist condition": for every $x\_1$, the function $f\_1(x\_1,x\_2)$ is strictly monotone in $x\_2$. Here $f\_1$ and $f\_2$ are the two coordinate components of $f$, and monotonicity in $\mathbb R/\mathbb Z$ should be taken to mean the monotonicity of a lifting of the corresponding function to $\mathbb R$. 


Twist maps of the annulus have been studied a lot in the measure-preserving case, notably in the development of Mather-Aubry theory. For a map $f$ to be *measure preserving* means that for every Lebesgue-measurable set $U$, $\mu(U)=\mu(f^{-1}(U))$, where $\mu$ denotes Lebesgue measure.


A twist map $f$ is *topologically conjugate* to another map $g$ if there exists a homeomorphism $\phi : A\to A$ such that $f=\phi^{-1}\circ g\circ \phi$.


**Question.** Do there exist twist maps of the annulus that are *not* topologically conjugate to a map that preserves the measure? In particular, are there any twist maps where Mather-Aubry theory does *not* hold?


Thank you in advance!



Twist maps of the annulus
I assume that the expression "that preserves the measure" means that preserves area (a.k.a. the Lebesgue, or Haar measure) on $\mathbb{R}/\mathbb{Z}\times\mathbb{R}$.


In this case the answer is no. The most simple example of a twist map which is not topologically conjugate to an area-preserving one is given by a dissipative twist diffeomorphism (i.e. $0<|\mathrm{det} f\_x|<1$, for every $x\in A$) exhibiting a non-trivial attractor. These attractors were first studied by Birkhoff, so now they are called "Birkhoff attractors" in the literature. 


