I am reading a blog which talks about a $C^1$ diffeomorphism $f: \mathbb{D}\{ x^2 + y^2 < 1\} \to \mathbb{R}^2$ and estimates the Hausdorff dimension of its image $\mathcal{H}_\sqrt{2}^d (f(\mathbb{D}))$. In the blog it is shown:

$$ \mathcal{H}_\sqrt{2}^d (f(\mathbb{D})) \leq 170 \pi \cdot \max \{ K, L\}^{2-d}\cdot L^{d-1}$$

My question is how can the diffeomorphic image of a disk have Hausdorff dimension between 1 and 2?

To remind myself, I read up on Hausdorff measure. The Hausdorff measure of $\mathbb{D}$ is zero in dimensions bigger than 2 and infinite in dimensions less than 2.

So how can the diffeomorphic image of a disk have $\mathcal{H}_\sqrt{2}^d (f(\mathbb{D}))$ which is neither $0$ nor $\infty$?

Related What is the Point of a Horseshoe Map ? and some more discussion where the authors announce finding Hausdorff dimension of the (un)stable set to be less than 2. Fractal Geometry of non-Uniformly Hyperbolic Horseshoes

$\sqrt{2}$ seems to have to do with the diameter of the unit square, so these Hausdorff measures are counting squares. These result is only meaningful with dervatives $||Df(p)|| < K$ and $|\det Df(p)| < L$ are both large. Therefore, $f$ must exhibit quite a bit of distortion, even thought it is $C^1$.

This measure rewards you for stretching. A very long thin rectangle with unit Euclidean area must be covered by (very inefficiently) by a lot of unit squares.