I am reading a [blog](https://matheuscmss.wordpress.com/2015/05/10/the-hausdorff-measure-at-adequate-scale-of-simply-connected-planar-domains) 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](https://terrytao.wordpress.com/2009/05/19/245c-notes-5-hausdorff-dimension-optional/).  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 $\infty$?

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Related [**What is the Point of a Horseshoe Map ?**](https://math.stackexchange.com/questions/357034/whats-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](http://w3.impa.br/~cmateus/files/mpy2012.pdf)

$\sqrt{2}$ 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.