Hausdorff-dimension of non measurable sets?

Question

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?

Answer 1

They key word here is the 'outer'. Start with a fixed dimension. If you define Hausdorff-outer-measure you are taking a limit of sets that all contain your original set. So if the original set is not measurable, the Hausdorff-outer-measure will give you the size of the smallest measurable set that contains your original set.

When you compute the Hausdorff dimension you still get the dimension of some limit of measurable sets that contain your original set.

Answer 2

Or how is the dimension for such sets even defined?

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 $s<s_0$ and $+\infty$ if $s<s_0$.

An alternative is to use the Hausdorff content for this.

The idea is: to compute the Hausdorff dimension, we do not need to know everything about Hausdorff measure, but only "When the the Hausdorff measure equal to zero?"