If the Jacobian of $f$ is not equal $1$ on any set of positive measure (that can happen), then the mapping will not be measure preserving on any set of positive measure so a decomposition is not possible.

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

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) 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

If the Jacobian of $f$ is not equal $1$ on any set of positive measure (that can happen), then the Lebesgue measure will not be preserved on any set of positive measure so a decomposition is not possible.

If the Jacobian of $f$ is not equal $1$ on any set of positive measure (that can happen), then the mapping will not be measure preserving on any set of positive measure so a decomposition is not possible.

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