Calculate the Minkowski dimension of the Koch curve.
We can already observe that the number of sides of the $n$-th generation in the construction of the Koch curve is $4^{n-1}$.

Moreover, the central part of the Koch curve (roughly an equilateral triangle) fits into a square of side $\frac{1}{3}$.

With $\epsilon_{1}=\frac{1}{3}$, we can then cover the curve with 3 squares of size $\epsilon_{1}: N\left(\frac{1}{3}\right)=3$. Continuing the calculations, we notice that $N\left(\frac{1}{3^{2}}\right)=3 \cdot 4$, and then $N\left(\frac{1}{3^{n}}\right)=3 \cdot 4^{n-1}$.

We then have ( $K$ is the Koch curve):

$$
\begin{gathered}
\operatorname{dim}_{\text {box }} K=\lim _{n \rightarrow \infty} \frac{\log \left(N\left(\frac{1}{3^{n}}\right)\right)}{\log \left(\frac{1}{3^{n}}\right)} \\
\operatorname{dim}_{\text {box }} K=\lim _{n \rightarrow \infty} \frac{(n-1) \log (4)+\log (3)}{n \log (3)} \\
\operatorname{dim}_{\text {box }} K=\frac{\log (4)}{\log (3)}
\end{gathered}
$$

We find the same value as with the dimension defined for self-similar fractals.

We can also prove that the box dimension of $A_{\alpha}=\{0\} \cup\left\{\frac{1}{n^{\alpha}}, n \geq 1\right\}$ for $\alpha>0$ is $\operatorname{dim}_{b o x}\left(A_{\alpha}\right)=\frac{1}{\alpha+1}$

## The Hausdorff-Besicovitch Dimension

Definition 105 (Hausdorff $s$-dimensional measure). Let $F$ be the fractal object to be measured. For $\epsilon>0$, we say that a countable part $\left(A_{i}\right)_{i \in \mathbb{N}}$ is an $\epsilon$-covering of $F$ if for all $i, \operatorname{diam} A_{i} \leq \epsilon$, and

$$
F \subset \bigcup_{i=0}^{+\infty} A_{i}
$$

If $s \geq 0$, we introduce the quantity:

$$
H_{\epsilon}^{s}(F)=\inf \left\{\sum_{i_{0}}^{+\infty}\left(\operatorname{diam} A_{i}\right)^{s},\left(A_{i}\right)_{i \in \mathbb{N}} \epsilon \text{-covering of } F\right\}
$$

For a fixed $s$, the function $\epsilon \in \mathbb{R}_{+}^{*} \mapsto H_{\epsilon}^{s}(F) \in[0,+\infty]$ is a decreasing function of $\epsilon$, so it has a limit as $\epsilon$ tends to 0:

$$
H^{s}(F):=\lim _{\epsilon \rightarrow 0} H_{\epsilon}^{s}(F)
$$

This limit, possibly infinite, is called the $s$-dimensional Hausdorff measure of $F$.

We note that $H^{s}(F)$ is a decreasing function of $F$. Moreover, if $H^{s}(F)$ is finite for a certain $s$, then for all $t>s, H^{t}(F)=+\infty$. There therefore exists a particular value $s$ separating the $s$ for which $H^{s}(F)=0$ and those for which $H^{s}(F)=+\infty$. This number is the Hausdorff dimension.

Definition 106 (Hausdorff dimension). We call the Hausdorff dimension the following quantity:

$$
\operatorname{dim}_{H}(F)=\inf \left\{s, H^{s}(F)=0\right\}=\sup \left\{s, H^{s}(F)=+\infty\right\}
$$

However, we cannot guarantee that, for $s=\operatorname{dim}_{H}(F)$, the quantity $H^{s}(F)$ is non-zero or finite.

Here are some elementary properties: