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Let $A\subseteq [-1,1]^d$ be a measurable set and $\mu$ be the Lebesgue measure. For any $\delta>0$, define $A_\delta := \{x: d(x,A)\leq \delta\}$, where $d(x,A) := \inf_{y\in A}\|x-y\|_2$.

The upper and lower Minkowski's content (of order $d-1$) is defined as $$ \overline M(A) := \limsup_{\delta\to 0^+}\frac{\mu(A_\delta)}{\delta} \;\;\;\;\;\text{and}\;\;\;\;\; \overline M(A) := \liminf_{\delta\to 0^+}\frac{\mu(A_\delta)}{\delta}. $$ In particular, if $A$ is an isotropic $\ell_2$ ball $\{x: \|x\|_2\leq a\}$, then $0<\overline M(A)=\underline M(A)<\infty$.

My question: I wonder whether an uniform version of the Minkowski's content is available. In particular, let $\mathcal A$ be a collection of measurable sets in $[-1,1]^d$, and for example purposes take $\mathcal A$ to be $\ell_2$ balls of all radius (i.e., $A\in\mathcal A$ if $A=\{x:\|x|_2\leq a\}$ for some $0<a< 1$). Can we define something like $$ \limsup_{\delta\to 0^+}\sup_{A\in\mathcal A}\frac{\mu(A_\delta)}{\delta} $$ and show that it is finite?

My preliminary calculation shows that the above definition cannot work, but if we consider $$ \limsup_{\delta\to 0^+}\sup_{A\in\mathcal A}\frac{\mu(A_\delta)}{\delta\cdot \mu(A)^{\frac{d-1}{d}} + \delta^d} $$ then the quantity is finite. I wonder whether there is an existing concept/work that could characterize this uniform behavior of Minkowski's content?

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