Let be a set that is not measurable with respect to the n-dimensional Lebesgue measure. Show that has Hausdorff dimension n.
Assuming that the Hausdorff dimension is the number , then we have for all (To see this note that the s'-Hausdorff measure of an n-cube is zero for and use subadditivity of the outer measure), and by monotonicity , so . If then , but Lebesgue measure is complete so is measurable, a contradiction.