# Hausdorff-dimension of non measurable sets?

The hausdorff-outer-measure is defined for all subsets of a metric space. The hausdorff measure is defined as the restriction to caratheodory measurable sets.

I actually don't know how the set of hausdorff measurable sets look like but since n-dimensional hausdorff measure and n-dimensional lebesgue measure coincidence when n is an integer there should be non-measurable sets for hausdorff measure.

However Hausdorff-dimension is often defined for all sets.

What is the Hausdorff-dimension of non measurable sets? Or how is the dimension for such sets even defined?

• According to Dodson-Christensen, a non-measurable set in $\mathbb{R}^n$ has Hausdorff dimension $n$. – Conifold Feb 12 at 11:38
• You need to be a bit careful with the term measurable set here. The notion of measurable depends on the measure used, so in this case also on the index of the Hausdorff measure. Every set is 0-Hausdorff measurable, but in contrast to that the n-Hausdorff measurable sets are exactly the Lebesgue-measurable sets which are far less. – mlk Feb 12 at 11:45

Use the Hausdorff outer measure only. For a given set $$A$$, it is still true that the $$s$$-dimensional Hausdorff outer measures satisfy: there is $$s_0$$ so that $$s$$-dimensional outer measure is zero if $$s and $$+\infty$$ if $$s.