is a single point measurable
additivity is too strong; for example, it would imply that if the measure of a set. consisting of a single point is zero, then the measure of every subset of Rn would. be zero. It is not possible to deï¬ne the Lebesgue measure of all subsets of Rn in a. geometrically reasonable way. Hausdorï¬ (1914) showed that for any dimension. n â¥ 1, there is no countably additive measure deï¬ned on all subsets of Rn that is. invariant under isometries (translations and rotations) and assigns measure one to.