Estimate/inequality releated to the Hausdorff measure
$begingroup$
Define the diameter of a subset $Y subseteq mathbb R^n$ of the metric space $(mathbb R^n, d)$ with the standard metric $d$ to be
$$operatorname{diam}(Y) := sup_{mathbf x,mathbf y in Y} d(mathbf x, mathbf y).$$
Now define the ball $B_r(mathbf y) := { mathbf x in mathbb R^n : d(mathbf x, mathbf y) leq r}$ and consider the ball
$A := B_r(mathbf 0)$ and a cover ${A_k}_{kgeq 1} supset A$, which don't have to be optimal (a cover here only need to satisfy $Asubset bigcup_{kgeq 1} A_k$).
The actual claim we proved in class was that the (outer) Hausdorff $n$-dimensional measure (here: $ninmathbb N$) is defined on the same $sigma$-algebra as the $n$-dimensional Lebesgue measure, and that they agreed up to a constant $C(n)$ depending on $n$.
Doing so, we used the inequality
$$displaystylesum_{kgeq 1} operatorname{diam}(A_k)^n geq (2r)^n$$
valid for any cover ${A_k}_{kgeq 1}$ to $A$ satisfying
$$operatorname{diam}(A_k) < 2r ; ; forall k$$
How can one prove this inequality?
measure-theory fractals geometric-measure-theory hausdorff-measure
edited Nov 27 '18 at 19:48
Xander Henderson
14.3k103554
asked Nov 27 '18 at 18:55
Markus KlyverMarkus Klyver
390314
$endgroup$
add a comment |
$begingroup$
Define the diameter of a subset $Y subseteq mathbb R^n$ of the metric space $(mathbb R^n, d)$ with the standard metric $d$ to be
$$operatorname{diam}(Y) := sup_{mathbf x,mathbf y in Y} d(mathbf x, mathbf y).$$
Now define the ball $B_r(mathbf y) := { mathbf x in mathbb R^n : d(mathbf x, mathbf y) leq r}$ and consider the ball
$A := B_r(mathbf 0)$ and a cover ${A_k}_{kgeq 1} supset A$, which don't have to be optimal (a cover here only need to satisfy $Asubset bigcup_{kgeq 1} A_k$).
The actual claim we proved in class was that the (outer) Hausdorff $n$-dimensional measure (here: $ninmathbb N$) is defined on the same $sigma$-algebra as the $n$-dimensional Lebesgue measure, and that they agreed up to a constant $C(n)$ depending on $n$.
Doing so, we used the inequality
$$displaystylesum_{kgeq 1} operatorname{diam}(A_k)^n geq (2r)^n$$
valid for any cover ${A_k}_{kgeq 1}$ to $A$ satisfying
$$operatorname{diam}(A_k) < 2r ; ; forall k$$
How can one prove this inequality?
measure-theory fractals geometric-measure-theory hausdorff-measure
edited Nov 27 '18 at 19:48
Xander Henderson
14.3k103554
asked Nov 27 '18 at 18:55
Markus KlyverMarkus Klyver
390314
$endgroup$
$begingroup$
Can't you just use that $|A_k| le c_n text{diam}(A_k)^n$, where $c_n$ is s.t. $|A| = c_n (2r)^n$?
$endgroup$
– mathworker21
Nov 27 '18 at 19:01
$begingroup$
@mathworker21 What do you mean by |A| here?
$endgroup$
– Markus Klyver
Nov 27 '18 at 19:15
$begingroup$
Lebesgue measure of $A$
$endgroup$
– mathworker21
Nov 27 '18 at 19:16
add a comment |
$begingroup$
Define the diameter of a subset $Y subseteq mathbb R^n$ of the metric space $(mathbb R^n, d)$ with the standard metric $d$ to be
$$operatorname{diam}(Y) := sup_{mathbf x,mathbf y in Y} d(mathbf x, mathbf y).$$
Now define the ball $B_r(mathbf y) := { mathbf x in mathbb R^n : d(mathbf x, mathbf y) leq r}$ and consider the ball
$A := B_r(mathbf 0)$ and a cover ${A_k}_{kgeq 1} supset A$, which don't have to be optimal (a cover here only need to satisfy $Asubset bigcup_{kgeq 1} A_k$).
The actual claim we proved in class was that the (outer) Hausdorff $n$-dimensional measure (here: $ninmathbb N$) is defined on the same $sigma$-algebra as the $n$-dimensional Lebesgue measure, and that they agreed up to a constant $C(n)$ depending on $n$.
Doing so, we used the inequality
$$displaystylesum_{kgeq 1} operatorname{diam}(A_k)^n geq (2r)^n$$
valid for any cover ${A_k}_{kgeq 1}$ to $A$ satisfying
$$operatorname{diam}(A_k) < 2r ; ; forall k$$
How can one prove this inequality?
measure-theory fractals geometric-measure-theory hausdorff-measure
edited Nov 27 '18 at 19:48
Xander Henderson
14.3k103554
asked Nov 27 '18 at 18:55
Markus KlyverMarkus Klyver
390314
$endgroup$
Define the diameter of a subset $Y subseteq mathbb R^n$ of the metric space $(mathbb R^n, d)$ with the standard metric $d$ to be
$$operatorname{diam}(Y) := sup_{mathbf x,mathbf y in Y} d(mathbf x, mathbf y).$$
Now define the ball $B_r(mathbf y) := { mathbf x in mathbb R^n : d(mathbf x, mathbf y) leq r}$ and consider the ball
$A := B_r(mathbf 0)$ and a cover ${A_k}_{kgeq 1} supset A$, which don't have to be optimal (a cover here only need to satisfy $Asubset bigcup_{kgeq 1} A_k$).
The actual claim we proved in class was that the (outer) Hausdorff $n$-dimensional measure (here: $ninmathbb N$) is defined on the same $sigma$-algebra as the $n$-dimensional Lebesgue measure, and that they agreed up to a constant $C(n)$ depending on $n$.
Doing so, we used the inequality
$$displaystylesum_{kgeq 1} operatorname{diam}(A_k)^n geq (2r)^n$$
valid for any cover ${A_k}_{kgeq 1}$ to $A$ satisfying
$$operatorname{diam}(A_k) < 2r ; ; forall k$$
How can one prove this inequality?
measure-theory fractals geometric-measure-theory hausdorff-measure
measure-theory fractals geometric-measure-theory hausdorff-measure
edited Nov 27 '18 at 19:48
Xander Henderson
14.3k103554
asked Nov 27 '18 at 18:55
Markus KlyverMarkus Klyver
390314
edited Nov 27 '18 at 19:48
Xander Henderson
14.3k103554
asked Nov 27 '18 at 18:55
Markus KlyverMarkus Klyver
390314
edited Nov 27 '18 at 19:48
Xander Henderson
14.3k103554
edited Nov 27 '18 at 19:48
Xander Henderson
14.3k103554
edited Nov 27 '18 at 19:48
Xander Henderson
14.3k103554
14.3k103554
asked Nov 27 '18 at 18:55
Markus KlyverMarkus Klyver
390314
asked Nov 27 '18 at 18:55
Markus KlyverMarkus Klyver
390314
asked Nov 27 '18 at 18:55
Markus KlyverMarkus Klyver
390314
390314
$begingroup$
Can't you just use that $|A_k| le c_n text{diam}(A_k)^n$, where $c_n$ is s.t. $|A| = c_n (2r)^n$?
$endgroup$
– mathworker21
Nov 27 '18 at 19:01
$begingroup$
@mathworker21 What do you mean by |A| here?
$endgroup$
– Markus Klyver
Nov 27 '18 at 19:15
$begingroup$
Lebesgue measure of $A$
$endgroup$
– mathworker21
Nov 27 '18 at 19:16
add a comment |
$begingroup$
Can't you just use that $|A_k| le c_n text{diam}(A_k)^n$, where $c_n$ is s.t. $|A| = c_n (2r)^n$?
$endgroup$
– mathworker21
Nov 27 '18 at 19:01
$begingroup$
@mathworker21 What do you mean by |A| here?
$endgroup$
– Markus Klyver
Nov 27 '18 at 19:15
$begingroup$
Lebesgue measure of $A$
$endgroup$
– mathworker21
Nov 27 '18 at 19:16
$begingroup$
Can't you just use that $|A_k| le c_n text{diam}(A_k)^n$, where $c_n$ is s.t. $|A| = c_n (2r)^n$?
$endgroup$
– mathworker21
Nov 27 '18 at 19:01
$begingroup$
Can't you just use that $|A_k| le c_n text{diam}(A_k)^n$, where $c_n$ is s.t. $|A| = c_n (2r)^n$?
$endgroup$
– mathworker21
Nov 27 '18 at 19:01
$begingroup$
@mathworker21 What do you mean by |A| here?
$endgroup$
– Markus Klyver
Nov 27 '18 at 19:15
$begingroup$
@mathworker21 What do you mean by |A| here?
$endgroup$
– Markus Klyver
Nov 27 '18 at 19:15
$begingroup$
Lebesgue measure of $A$
$endgroup$
– mathworker21
Nov 27 '18 at 19:16
$begingroup$
Lebesgue measure of $A$
$endgroup$
– mathworker21
Nov 27 '18 at 19:16
add a comment |
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$begingroup$
Can't you just use that $|A_k| le c_n text{diam}(A_k)^n$, where $c_n$ is s.t. $|A| = c_n (2r)^n$?
$endgroup$
– mathworker21
Nov 27 '18 at 19:01
$begingroup$
@mathworker21 What do you mean by |A| here?
$endgroup$
– Markus Klyver
Nov 27 '18 at 19:15
$begingroup$
Lebesgue measure of $A$
$endgroup$
– mathworker21
Nov 27 '18 at 19:16