If $ \mathcal{H}^k(B_1(0)\cap S)\leq A\omega_k $ when $ \mathcal{H}^k(B_r(x)\cap S)\leq A\omega_kr^k $ for all $ 0<r<\delta $, $ x\in\mathbb{R}^n $?

Question

Let $ S\subset\mathbb{R}^n $ is of finite $ k $-dimensional Hausdorff and $ 0<\delta<1 $ is a constant. If for any $ x\in\mathbb{R}^n $ and $ r>0 $, we hae $$ \mathcal{H}^k(S\cap B_r(x))\leq A\omega_kr^k. $$ I want to ask if I can get that $$ \mathcal{H}^k(S\cap B_1(0))\leq A\omega_k. $$ Intuitively thinking it is true. By using the definition of Haudorff measure, for any $ \epsilon>0 $, we can obtain a covering $ \{B_{r_i}(x_i)\}_{i=0}^{\infty} $ such that $$ S\cap B_1(0)\subset\bigcup_{i}B_{r_i}(x_i),\,\,r_i<\delta $$ and $$ \sum_i\omega_k r_i^k-\epsilon<\mathcal{H}^k(S\cap B_1(0))\leq \sum_i\omega_k r_i^k. $$ However, I cannot go on. Can you give me some references or hints?

Answer 1

You cannot get this bound. My heuristic explanation for this failure would be that your bound 'does not see folds of $S$ at scales larger than $\delta$'. However, these folds may well make positive contributions to the $k$-dimensional area of $S$.

For a counterexample, you could for instance take any compact, smoothly embedded submanifold $S^k \subset B_1^n$ with $\mathcal{H}^k(S \cap B_1) > \omega_k$, and pick $A > 1$ close enough to one that \begin{equation} A \omega_k < \mathcal{H}^k(S \cap B_1). \end{equation} By compactness and regularity of $S$, there is $\delta > 0$ so that for all points $x \in B_1$ and all radii $r \in (0,\delta)$, \begin{equation} \mathcal{H}^k(S \cap B_r(x)) < A \omega_k r^k. \end{equation} By construction, this defeats the desired estimate.