Let $f\colon\mathbb{R}^{n}\to\mathbb{R}^{n}$ be a smooth invertible map and suppose that $|\det(Df(x,y))|=1$ for all $x,y\in\mathbb{R}^{n}$. Let $\lambda$ denote the Lebesgue measure. Is it then true that $\lambda(f^{-1}(A))=\lambda(A)$ (or equivalently, $\lambda(f(A))=\lambda(A)$, since $f$ is invertible) for all Lebesgue measurable sets $A$? (Or is it only true for Borel measurable sets?)
My attempt: I don't know how nice the sets must be in order to let the following work. So let $B\subset\mathbb{R}^{n}$ be an Euclidian ball. Then $$\lambda(f(B))=\int_{f(B)}d\lambda=\int_{f(B)}d(x,y)=\int_{B}|\det(Df(x,y))| \ d(x,y)=\int_{B}d(x,y)=\int_{B}d\lambda=\lambda(B).$$
Can this be formulated into a precise theorem? What are the necessary conditions? For which sets is it true? Lebesgue measurable? Borel Measurable? Or smooth domains? So for which sets is $f$ Lebesgue measure preserving if $|\det(Df)|\equiv1$?