What is the Lebesgue measure of the Koch and the Minkowski curves?

Question

Are the Koch curve and the Minkowski curve Lebesgue measurable? (I believe they are.) If so, what are their measures? (Intuitively, it would seem to be zero.) I unfortunately can't seem to find much on this question, so I'd also be grateful for any helpful resource.

Answer 1

Suppose that a compact set $K$ in $\mathbb R^2$ is composed of $m$ copies of itself: $$ K = \bigcup_{i=1}^m K_i. $$ Suppose further that each $K_i$ is geometrically similar to $K$ and scaled by the factor $r_i$. Finally, let us suppose that the intersection between any two distinct $K_i$s has Lebesgue measure zero. Then, by the additive and scaling properties of two dimensional Lebesgue measure $m$, we have $$ m(K) = \sum_{i=1}^m m(K_i) = m(K)\sum_{i=1}^m r_i^2. $$ Now, the only way this can happen is if either $$ m(K) = 0 \: \: \text{ or } \: \: \sum_{i=1}^m r_i^2 = 1. $$ In the examples given, the sets are self-similar with distinct copies intersecting in finite sets and it is easy to check directly that the summation condition is not satisfied. So we must have $m(K) = 0$.

There's an interesting connection between this and fractal dimension in that the condition $$ \sum_{i=1}^m r_i^2 = 1 $$ implies that the fractal dimension of set should be two, which is not the case here.

Answer 2

Just repeating the comments to get this answered.

One reference is Amann, Escher's Analysis III. In the first chapter, they deal with general measures, in particular Hausdorff measure and Lebesgue measure. In general, for every $s>0$, we can consider the $s$-dimensional Hausdorff measure $\mathcal{H}^s$ on $\Bbb{R}^n$ (or other metric spaces). One interesting property is that if $A$ is a (Hausdorff)-measurable subset of $\Bbb{R}^n$ (eg. every Borel set is Hausdorff and Lebesgue measurable) with $\mathcal{H}^s(A)$ finite, then for every $t>s, \mathcal{H}^t(A)=0$.

Another fact is that on $\Bbb{R}^n$ the Hausdorff $n$-measure, $\mathcal{H}^n$ and the Lebesgue measure $\lambda^n$ are multiples of each other (because they are both locally finite-translation invariant measures on the Borel/Lebesug $\sigma$-algebra of $\Bbb{R}^n$). So, for the sets you mention, their Hausdorff dimension is $<2$. Hence, $\mathcal{H}^2(A)=0$ (actually in the notes below, they show the Hausdorff dimension is $<2$ by using the fact that $\mathcal{H}^2(A) = 0$), and hence $\lambda^2(A)=0$. (all of these facts are there in the book either as part of the text or part of the exercises). This set of online notes also discusses these topic pretty quickly.