Return to Question

Consider a compact set $E\subset \mathbb{R}^n$ ad assume that ${\cal H}^{n-1}(E)=0$.

If $\epsilon>0$, there is a finite covering of $E$ made of open balls $B_{r_k}(x_k)$ such that $\sum_{k}r_k^{n-1} \leq \epsilon$.

(Indeed, for $\epsilon>0$, there is a countable family of sets $F_k$ of diameters $diam(F_j)$ such that $\sum_j diam(F_j)^{n-1}<\epsilon$. Choose open balls $B_{r_j}(y_j)$ of radius $\rho_j =diam(F_j)$ such that $B_{\rho_j}(y_j)\supset F_j$. From compactness, a finite subcovering $B_{\rho_{j_1}}(y_{j_1}), \ldots, B_{\rho_{j_N}}(y_{j_N})$ of $E$ exists such that $\sum_{k=1}^N \rho^{n-1}_{j_k} \leq \epsilon$ proving the assertion.)

My question is: in the given hypoteses on $E$ and for any given $\epsilon>0$, is it possible to find a finite covering of $E$ made of balls $B_{k}$ with some constant radius $r>0$ such that $\sum_{k} r^{n-1} \leq \epsilon$?

If considering the Lebesgue misure (i.e. ${\cal H}^n$) the assertion should be positive, what about ${\cal H}^{n-1}$ or ${\cal H}^{k}$ with $k< n$?

Or is there an explicit counterexample?

Consider a compact set $E\subset \mathbb{R}^n$ ad assume that ${\cal H}^{n-1}(E)=0$.

If $\epsilon>0$, there is a finite covering of $E$ made of $N_\epsilon$ open balls $B_{r_{\epsilon,k}}(x_{\epsilon,k})$ such that $\sum_{k=1}^{N_\epsilon}r_{\epsilon,k}^{n-1} \leq \epsilon$.

Indeed (I will omit the indices $_\epsilon$) for $\epsilon>0$, there is a countable family of sets $F_k$ of diameters $diam(F_j)$ such that $\sum_j diam(F_j)^{n-1}<\epsilon$. Choose open balls $B_{r_j}(y_j)$ of radius $\rho_j =diam(F_j)$ such that $B_{\rho_j}(y_j)\supset F_j$. From compactness, a finite subcovering $B_{\rho_{j_1}}(y_{j_1}), \ldots, B_{\rho_{j_N}}(y_{j_N})$ of $E$ exists such that $\sum_{k=1}^N \rho^{n-1}_{j_k} \leq \epsilon$ proving the assertion.

My question is: in the given hypoteses on $E$ and for any given $\epsilon>0$, is it possible to find a finite covering of $E$ made of balls $B_{r_\epsilon}(x_{\epsilon,k})$ with the same radius $r_\epsilon>0$, such that $\sum_{k=1}^{N_\epsilon} r_\epsilon^{n-1} \leq \epsilon$?

If considering the Lebesgue measure (i.e. ${\cal H}^n$) the assertion should be positive, what about ${\cal H}^{n-1}$ or ${\cal H}^{k}$ with $k< n$?

Or is there an explicit counterexample?

Consider a compact set $E\subset \mathbb{R}^n$ ad assume that ${\cal H}^{n-1}(E)=0$.

If $\epsilon>0$, there is a finite covering of $E$ made of open balls $B_{r_k}(x_k)$ such that $\sum_{k}r_k^{n-1} \leq \epsilon$.

(Indeed, for $\epsilon>0$, there is a countable family of sets $F_k$ of diameters $diam(F_j)$ such that $\sum_j diam(F_j)^{n-1}<\epsilon$. Choose open balls $B_{r_j}(y_j)$ of radius $\rho_j =diam(F_j)$ such that $B_{\rho_j}(y_j)\supset F_j$. From compactness, a finite subcovering $B_{\rho_{j_1}}(y_{j_1}), \ldots, B_{\rho_{j_N}}(y_{j_N})$ of $E$ exists such that $\sum_{k=1}^N \rho^{n-1}_{j_k} \leq \epsilon$ proving the assertion.)

My question is: in the given hypoteses on $E$ and for a given $\epsilon>0$, is it possible to find a finite covering of $E$ made of balls $B_{k}$ with some constant radius $r>0$ such that $\sum_{k} r^{n-1} \leq \epsilon$?

If considering the Lebesgue misure (i.e. ${\cal H}^n$) the assertion should be positive, what about ${\cal H}^{n-1}$ or ${\cal H}^{k}$ with $k< n$?

Or is there an explicit counterexample?

Consider a compact set $E\subset \mathbb{R}^n$ ad assume that ${\cal H}^{n-1}(E)=0$.

If $\epsilon>0$, there is a finite covering of $E$ made of open balls $B_{r_k}(x_k)$ such that $\sum_{k}r_k^{n-1} \leq \epsilon$.

(Indeed, for $\epsilon>0$, there is a countable family of sets $F_k$ of diameters $diam(F_j)$ such that $\sum_j diam(F_j)^{n-1}<\epsilon$. Choose open balls $B_{r_j}(y_j)$ of radius $\rho_j =diam(F_j)$ such that $B_{\rho_j}(y_j)\supset F_j$. From compactness, a finite subcovering $B_{\rho_{j_1}}(y_{j_1}), \ldots, B_{\rho_{j_N}}(y_{j_N})$ of $E$ exists such that $\sum_{k=1}^N \rho^{n-1}_{j_k} \leq \epsilon$ proving the assertion.)

My question is: in the given hypoteses on $E$ and for any given $\epsilon>0$, is it possible to find a finite covering of $E$ made of balls $B_{k}$ with some constant radius $r>0$ such that $\sum_{k} r^{n-1} \leq \epsilon$?

If considering the Lebesgue misure (i.e. ${\cal H}^n$) the assertion should be positive, what about ${\cal H}^{n-1}$ or ${\cal H}^{k}$ with $k< n$?

Or is there an explicit counterexample?

Consider a compact set $E\subset \mathbb{R}^n$ ad assume that ${\cal H}^{n-1}(E)=0$.

If $\epsilon>0$, there is a finite covering of $E$ made of open balls $B_{r_k}(x_k)$ such that $\sum_{k}r_k^{n-1} \leq \epsilon$.

(Indeed, for $\epsilon>0$, there is a countable family of sets $F_k$ of diameters $diam(F_j)$ such that $\sum_j diam(F_j)^{n-1}<\epsilon$. Choose open balls $B_{r_j}(y_j)$ of radius $\rho_j =diam(F_j)$ such that $B_{\rho_j}(y_j)\supset F_j$. From compactness, a finite subcovering $B_{\rho_{j_1}}(y_{j_1}), \ldots, B_{\rho_{j_N}}(y_{j_N})$ of $E$ exists such that $\sum_{k=1}^N \rho^{n-1}_{j_k} \leq \epsilon$ proving the assertion.)

My question is: in the given hypoteses on $E$ and for a given $\epsilon>0$, is it possible to find a finite covering of $E$ made of balls $B_{k}$ with some constant radius $r>0$ such that $\sum_{k} r^{n-1} \leq \epsilon$?

If considering the Lebesgue misure (i.e. ${\cal H}^n$) the assertion should be positive, what about ${\cal H}^{n-1}$ or ${\cal H}^{k}$ with $k< n$?

Consider a compact set $E\subset \mathbb{R}^n$ ad assume that ${\cal H}^{n-1}(E)=0$.

If $\epsilon>0$, there is a finite covering of $E$ made of open balls $B_{r_k}(x_k)$ such that $\sum_{k}r_k^{n-1} \leq \epsilon$.

(Indeed, for $\epsilon>0$, there is a countable family of sets $F_k$ of diameters $diam(F_j)$ such that $\sum_j diam(F_j)^{n-1}<\epsilon$. Choose open balls $B_{r_j}(y_j)$ of radius $\rho_j =diam(F_j)$ such that $B_{\rho_j}(y_j)\supset F_j$. From compactness, a finite subcovering $B_{\rho_{j_1}}(y_{j_1}), \ldots, B_{\rho_{j_N}}(y_{j_N})$ of $E$ exists such that $\sum_{k=1}^N \rho^{n-1}_{j_k} \leq \epsilon$ proving the assertion.)

My question is: in the given hypoteses on $E$ and for a given $\epsilon>0$, is it possible to find a finite covering of $E$ made of balls $B_{k}$ with some constant radius $r>0$ such that $\sum_{k} r^{n-1} \leq \epsilon$?

If considering the Lebesgue misure (i.e. ${\cal H}^n$) the assertion should be positive, what about ${\cal H}^{n-1}$ or ${\cal H}^{k}$ with $k< n$?

Or is there an explicit counterexample?