most recent 30 from http://mathoverflow.net 2013-06-19T19:01:15Z http://mathoverflow.net/feeds/question/115504 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/115504/is-there-any-known-condition-for-the-following-property Changyu Guo 2012-12-05T14:48:25Z 2012-12-07T09:12:50Z

<p>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. </p>

http://mathoverflow.net/questions/115504/is-there-any-known-condition-for-the-following-property/115694#115694 Daniel Spector 2012-12-07T09:12:50Z 2012-12-07T09:12:50Z

<p>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 1-D Hausdorff measure in the range) would have positive Lebesgue measure, and therefore infinite 1-D Hausdorff measure.</p> <p>A good start might be to investigate the situation on maps $f:[0,1] \to \mathbb{R}$ with the Lebesgue measure in both places. </p> <p>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.).</p>