The question is very vague therefore any kind of suggestions, reference, ideas are welcome.


Suppose $S$ is an oriented surface with or without boundary. Let $m$ be an area form. Let $f$ be a diffeomorphism of $S$ isotopic to the identity. Is there a way to decompose $f$ *naturally* such that one component preserves $m$. By naturally I mean the decomposition is continuous with respect to some topology on the group of diffeomorphisms isotopic to the identity. 


(over) Expectation: There exist $A,B\subset S$ such that $A\cup B=S$, $m(A)\neq 0,$ (also $m(B)\neq 0$ if possible) and $f\_1=f\_{|A}, f\_2=f\_{|B}$ such that $f\_1$ preserves $m$. 



Nice decomposition of surface diffeomorphisms
If the Jacobian of $f$ is not equal $1$ on any set of positive measure, then the mapping will not be measure preserving on any set of positive measure so a decomposition is not possible. 


A classical result of Moser (see e.g. [Lower regularity version of Moser's theorem on volume elements](https://mathoverflow.net/questions/185794/lower-regularity-version-of-mosers-theorem-on-volume-elements)) states that for any smooth positive function $g$ on $S$ whose integral equals to the area of $S$ there is a diffeomorphism with the Jacobain equal to $g$ and clearly we can take $g$ to be equal $1$ only on a set of measure zero (on a curve).


I am not sure what the author means by preserving $m$ so I took a freedom to understand it as a measure preserving map:
<https://www.encyclopediaofmath.org/index.php/Measure-preserving_transformation>


