# Exact value of Hausdorff measure of two dimensional Cantor set

Let $$\mathcal{C}$$ denote the classical Cantor set, then it is well-known that $$\mathcal{C}$$ has Hausdorff dimension $$\alpha = \ln 2 /\ln 3$$, and its $$\alpha$$-dimensional Hausdorff measure is $$\mathcal{H}_\alpha(\mathcal{C}) = 1$$.

For $$\mathcal{C}\times \mathcal{C}\subset \mathbb{R}^2$$, it is fairly easy to show its Hausdorff dimension is $$2\alpha$$, and $$0< \mathcal{H}_{2\alpha}(\mathcal{C}\times \mathcal{C}) <\infty$$. I am interested in knowing the exact value of $$\mathcal{H}_{2\alpha}(\mathcal{C}\times \mathcal{C})$$.

The proof for 1D case $$\mathcal{H}_\alpha(\mathcal{C}) = 1$$ cannot carry directly into higher dimension.

This answer claims the measure is $$1$$ and refers to Falcon's book The Geometry of Fractal Sets. However, after really looking into the book, it does not say anything about Hausdorff measure (only Hausdorff dimension) for any non-1D set (at least at the position quoted).

Nonetheless, it might be possible that the Hausdorff measure does not admit a simple closed-form, as expected for many fractals in $$\mathbb{R}^n$$ with $$n>1$$. Any idea/reference is much appreciated.

Sketch for a possible solution: we can build $$C\times C$$ in the same way that we build $$C$$, that is, starting from $$C_0:=[0,1]$$, $$C_1:=[0,1/3]\cup[2/3,1]$$ and so on then $$C\times C=\bigcap_{k=0}^\infty C_k\times C_k\tag1$$

Set $$A_k:=C_k\times C_k$$, then $$A_k$$ is composed by $$4^k$$ squares of side length $$3^{-k}$$, this means that each square have diameter $$\sqrt 2/3^k$$.

For measurable sets the Hausdorff measure is defined by the limit $$\mathcal H^s(A)=\lim_{\epsilon\to 0^+}\mathcal H^s_\epsilon(A)$$, where

$$\mathcal H_\epsilon^s(A):=\inf\left\{\sum_{k=0}^\infty\operatorname{diam}(O_k)^s,\, \operatorname{diam}(O_k)\le \epsilon,\, A\subset\bigcup_{k=0}^\infty O_k,\, O_k\text{ is open}\right\}\tag2$$

Now we can try to approximate the value of $$\mathcal H^s(A)$$ from above. Suppose we are trying to cover $$A$$ with open squares of diameter $$\gamma\sqrt 2/ 3^k$$, then for enough small $$\gamma>1$$ it seems easy to prove (by the inner structure of $$C\times C$$ that can be seen in it construction) that the most efficient way to do it is covering $$A$$ as if we were covering $$A_k$$, so we will need $$4^k$$ of this squares. The same happen for all $$k\in\Bbb N$$, hence we will find that

$$\mathcal H^s(A)\le\lim_{k\to\infty} 4^k\left(\gamma\frac{\sqrt 2}{3^k}\right)^s,\quad\forall \gamma>1\tag3$$

where the inequality comes from the fact that a box-covering is less efficient than a circle-covering using the same diameter for squares and circles (because a square cover less area than a circle with the same diameter). Hence we find that $$\mathcal H^s(A)\le(\sqrt 2)^s$$ when we choose $$s:=\ln 4/\ln 3$$.

From all of this I see that if we shows that, for each $$\gamma_k\sqrt 2/3^k$$ (for enough small $$\gamma_k>1$$), the covering of circles cannot be better than the covering of boxes, that is that $$\mathcal H^s_{\gamma_k\sqrt 2/3^k}(A)=4^k(\gamma_k\sqrt 2/3^k)^s$$ then we will had that $$\mathcal H^s(A)=\lim_{k\to\infty}\mathcal H^s_{\gamma_k\sqrt 2/3^k}(A)=(\sqrt 2)^s$$ for $$s:=\ln 4/\ln 3$$.

About possible references: I didn't read it but, the most complete textbook that cover this questions, seems to by Fractal geometry of Falconer.