Hausdorff dimension of the boundary of fibres of Lipschitz maps

Question

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.

Answer 1

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 $f:\mathbb{R}^m\to\mathbb{R}^{m-k}$.