# One-dimensional Hausdorff measure of preimages

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 one-dimensional Hausdorff measure of $$f^{-1}(E)$$ is zero whenever $$E$$ is of zero one-dimensional Hausdorff measure zero. Note that f is not assumed to be a homeomorphism.

• What is $\Omega$ here? Dec 5, 2012 at 15:02
• 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 1-D lebesgue measure, but rather the 1-D Hausdorff measure $H^1$. Dec 5, 2012 at 17:32
• 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. Dec 6, 2012 at 8:11
• This is the worst question title in the history of Math Overflow. It provides literally no information about what the question is about. Dec 6, 2012 at 12:14
• @arsmath after almost 7 years of hard thinking I changed to a more informative title
– YCor
Nov 15, 2019 at 9:10

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.).
• In the one-dimensional case suggested in this answer, the image measure (also called push-forward) $f_\ast\lambda (E) = \lambda(f^{-1}(E))$ is dominated by $\lambda$ hence, by Radon-Nikodym, 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 one-dimensional Hausdorff measure is not $\sigma$-finite and Radon-Nikodym is not applicable. Dec 7, 2012 at 13:05