Let $f: \mathbb{R}^m\rightarrow \mathbb{R}^{m-k}$ be a Lipschitz map.



> 
> Can we get a uniform estimate on the Hausdorff dimension of the boundaries of fibres of $f$? I.e. do we have an upper bound for
>  $$ \sup\_{y\in \mathbb{R}^{n-k}} \dim\_H(\partial f^{-1}(\{y\})) ?$$ 
> 
> 
> 


Theorem 2.5 in [1] tells us, that for almost every $y\in \mathbb{R}^{n-k}$ we have that $\dim\_H(f^{-1}(y))\leq k$. This tells us
$$ \text{essup}\_{y\in \mathbb{R}^{n-k}} \dim\_H(\partial f^{-1}(\{y\})) \leq k.$$
Can we pass to the supremum? And are there even better bounds? I mean, I used $\partial f^{-1}(\{y\})\subseteq f^{-1}(\{y\})$ as $f$ is continuous and the monotonicity of the Hausdorff dimension, but I guess that one can do better than this.


**[1] G. Alberti, S. Bianchini, G. Crippa,** Structure of level sets and Sard-type properties of Lipschitz maps: results and counterexamples. 
*Ann. Sc. Norm. Super. Pisa Cl. Sci.* (5) 12 (2013), no. 4, 863–902. 



Hausdorff dimension of the boundary of fibres of Lipschitz maps
Unfortunately, you can always find a Lipschitz map
$$
f:\mathbb{R}^m\to\mathbb{R}^{m-k}
\quad
\text{and}
\quad
y\in\mathbb{R}^{m-k}
$$
such that $\partial f^{-1}(y)$ has positive $m$-dimensional measure so 
$\dim\_H \partial f^{-1}(y)=m$. 


Here is an example. Let $K\subset\mathbb{R}^m$ be a Cantor set (i.e. a set homeomorphic to the ternary Cantor set) of positive $m$-dimensional measure. Existence of such a set $K$ is standard. Let $f(x)=\operatorname{dist}(x,K)$. Then $f:\mathbb{R}^m\to\mathbb{R}$ is $1$-Lipschitz and it vanishes precisely on $K$. That is $f^{-1}(0)=K=\partial K$ (the boundary of a Cantor set is the Cantor set itself) has positive $m$-dimensional measure. Now, assuming that $\mathbb{R}\subset\mathbb{R}^{m-k}$ we can regard $f$ as a mapping $f:\mathbb{R}^m\to\mathbb{R}^{m-k}$. 


