Can a $C^1$ function with non-singular differential map a non-zero measure set to a zero-measure set?

Question

Let $f:\mathbb R^n\to\mathbb R ^n$ be a continuously differentiable function with nowhere singular differential. Can there exist a set of non-zero measure $A$ such that the measure of $f(A)$ is zero?

The inverse function theorem rules out simple examples like an $f$ that maps a disc to a line segment, but it leaves open the possibility of an $A$ with non-zero measure but empty interior. For example, could there be a smooth function on $[0, 1]$ mapping the irrational numbers to the Cantor set?

Answer 1

Let $S$ be a set with nonzero measure. Then $S \cap K$ has nonzero measure for some compact $K$ and thus we can assume that $S$ itself lies inside an compact set. By compactness we can also assume that $f$ is injective.

Since $f$ is $C^1$ there are constant $c, C$ so that $$ 0< c\le |\nabla f(s)| \le C$$ for all $s\in S$. Thus we have

$$ C^{-1} |f(S)|\le |S| \le c^{-1}|f(S)|.$$

This implies in particular that $f(S)$ also has non-zero measure.

Answer 2

No. In fact by the change of variable formula, denoting the Jacobian of $f$ as $J_f$ (which is such that $|\det J_f|>0$ on $A$ by your hypothesis), you can see that $\lambda(f(A))=\int_{f(A)}d\lambda=\int_{A}|\det J_f|d\lambda>0$