Let $\Omega$ be an open subset of $\mathbf{R}^n$. For a mapping $f: \Omega\to \bf{R}^n$, what kind of condition ensures that the onedimensional Hausdorff measure of $f^{1}(E)$ is zero whenever $E$ is of zero onedimensional Hausdorff measure zero. Note that f is not assumed to be a homeomorphism.

1$\begingroup$ What is $\Omega$ here? $\endgroup$– Tom LeinsterDec 5, 2012 at 15:02

$\begingroup$ Since $f: \Omega \to \mathbb{R}^N$, $f^{1}:f(\Omega) \to \Omega$, so if $E \subset f(\Omega)$, we should not be measuring the 1D lebesgue measure, but rather the 1D Hausdorff measure $H^1$. $\endgroup$– Daniel SpectorDec 5, 2012 at 17:32

$\begingroup$ Yes, you are right. It is better to use the Hausdorff measure H^1. \Omega is a domain in \bR^n and f is not necessarily to be a homeomorphism here. $\endgroup$– Changyu GuoDec 6, 2012 at 8:11

11$\begingroup$ This is the worst question title in the history of Math Overflow. It provides literally no information about what the question is about. $\endgroup$– arsmathDec 6, 2012 at 12:14

1$\begingroup$ @arsmath after almost 7 years of hard thinking I changed to a more informative title $\endgroup$– YCorNov 15, 2019 at 9:10
1 Answer
There may be a name for this, but it seems like a strange condition. Such a function cannot take a constant value on any set of positive Lebesgue measure, otherwise the inverse image of that constant (having zero 1D Hausdorff measure in the range) would have positive Lebesgue measure, and therefore infinite 1D Hausdorff measure.
A good start might be to investigate the situation on maps $f:[0,1] \to \mathbb{R}$ with the Lebesgue measure in both places.
There is also a related notion, called Lusin's N property, which means $f$ takes sets of measure zero into sets of measure zero (as opposed to $f^{1}$, as you desire). This is a quality of Lipschitz functions that Sobolev functions also inherit, and is necessary to satisfy the fundamental theorem of Calculus (along with being differentiable a.e., etc.).

$\begingroup$ In the onedimensional case suggested in this answer, the image measure (also called pushforward) $f_\ast\lambda (E) = \lambda(f^{1}(E))$ is dominated by $\lambda$ hence, by RadonNikodym, it has a density. This may help to get something more concrete in particular cases. A problem with the higher dimensional case is that the onedimensional Hausdorff measure is not $\sigma$finite and RadonNikodym is not applicable. $\endgroup$ Dec 7, 2012 at 13:05