Skip to main content
6 events
when toggle format what by license comment
Jan 13, 2010 at 12:04 vote accept Sune Jakobsen
Jan 13, 2010 at 10:26 comment added Thorny If a set has positive, finite d-dimensional measure (and it is easy to create for example a self-similar Cantor set with a prescribed Hausdorff dimension), then it has measure 0 according to any d'-dimensional Hausdorff measure with d'>d and has infinite measure according to any d"-dimensional Hausdorff measure with d"<d.
Jan 13, 2010 at 9:41 comment added Matus Telgarsky the conditions require that there exists a set with lebesgue measure 0 but infinite measure according to this measure. Are you sure this holds for Hausdorff measure (with dimension in $(0,1)$)?
Jan 13, 2010 at 9:27 comment added Pete L. Clark Acknowledgment: My response is independent of Thorny's but came slightly later.
Jan 13, 2010 at 9:24 comment added Thorny And I guess adding Hausdorff measures with gauge functions would give even more examples, which aren't all that easy to works with though.
Jan 13, 2010 at 9:22 history answered Thorny CC BY-SA 2.5