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I'm currently reading a paper. Let $F\subset\mathbb R^n$ and $\epsilon\gt0$, the paper defined $m^s(F):=\liminf_{\epsilon \to 0}\epsilon^{s-n}\lambda(F_\epsilon)$ and $M^s(F):=\limsup_{\epsilon \to 0}\epsilon^{s-n}\lambda(F_\epsilon)$, where $\lambda$ is Lebesgue measure on $\mathbb R^n$ and $F_\epsilon$ is $\epsilon$-parallel set of $F$, i.e. $F_\epsilon:=\{x:d(x,F)\le\epsilon\}$. The author called $$\underline{D}(F):=\inf\{t\ge0:m^t(F)=0\}=\sup\{t\ge0:m^t(F)=\infty\},$$ $$\overline{D}(F):=\inf\{t\ge0:M^t(F)=0\}=\sup\{t\ge0:M^t(F)=\infty\}$$ as lower and upper Minkowski dimension of $F$ respectively. If $\underline{D}(F)=\overline{D}(F)$ then the common value $D=D(F)$ is called the Minkowski dimension of $F$. In the other hand, all textbooks i know defined lower and upper Minkowski dimension (or box-counting dimension) respectively as $$\underline{B}(F):=\liminf_{\delta \to 0}\frac{\ln N_\delta(F)}{-\ln\delta}$$ and $$\overline{B}(F):=\limsup_{\delta \to 0}\frac{\ln N_\delta(F)}{-\ln\delta}$$ where $N_\delta(F)$ is the smallest number of sets of diameter at most $\delta$ which can cover $F$. Likewise, if $\underline{B}(F)=\overline{B}(F)$ then the common value $B=B(F)$ is called the Minkowski dimension (or box-counting dimension) of $F$.

My questions: Are these definitions equal or their names just coincide? If $D=B$, where can i find the proof? Thank you.

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These definitions are equivalent. This is proved on pages 78-79 of Geometry of Sets and Measures in Euclidean Spaces by Mattila. (The book that should be constantly on your desk if you are working in this area.) The key point is $$ \alpha(n) \epsilon^{n} P_\epsilon(F) \le \lambda(F_\epsilon) \le \alpha(n) (2\epsilon)^{n} N_\epsilon(F) \tag1 $$ Here $\alpha(n)$ is the volume of unit ball in $n$ dimensions, $N_\epsilon(F)$ is the $\epsilon$-covering number of $F$ defined in your post, while $P_\epsilon(F)$ is the $\epsilon$-packing number of $F$: the maximal number of disjoint balls of radius $\epsilon$ with centers lying at $F$.

The proof of (1) is straightforward: the balls from the definition of $P_\epsilon(F)$ are disjoint and are contained in $F_\epsilon$, hence bounds its measure from below. Also, doubling the size of balls in an $\epsilon$-cover of $F$, we cover $F_\epsilon$ as well.

It remains to relate packing and covering:
$$N_{2\epsilon}(F)\le P_\epsilon(F)\le N_{\epsilon/2}(F)\tag2$$ Indeed, a maximal $\epsilon$-packing produces $2\epsilon$-covering by doubling the size of every ball. Conversely, the existence of an $\epsilon/2$-covering obstructs $\epsilon$-packing.

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  • $\begingroup$ Thanks, i'll get the book :) $\endgroup$ – Deco Jul 23 '13 at 20:56

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