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?



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 $?
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.


