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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?

Jack M
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    By the inverse function theorem, your function is locally a diffeomorphism, and thus maps a set of positive measure to a set of positive measure. This implicitly uses that being a null set is a local property, i.e., if $A\cap B_r (x)$ is a null set for all $x$ and arbitrary $r =r(x)$, then $A$ is a null set. To see this use that $\Bbb{R}^d$ is second countable. – PhoemueX Mar 25 '18 at 21:00
  • @PhoemueX Do you have a reference for the latter result? – Jack M Mar 25 '18 at 21:01
  • I edited my comment. Does this answer your question? – PhoemueX Mar 25 '18 at 21:04

2 Answers2

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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.

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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 $\lambda(f(A))=\int_{f(A)}d\lambda=\int_{A}|\det J_f|d\lambda>0$

xyz
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