Hausdorff measure (open subset containing x)
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Let $s geq 0$ be a real number and $E subset mathbb{R^n}$.
Suppose that for all $x in E$ there is an open subset $U subset mathbb{R^n}$ which contains $x$, and $mathcal{H^s}(E cap U)=0 Rightarrow mathcal{H^s}(E)=0$.
How to show this implication?
I tried to use:
Since $E subset mathbb{R^n}$, it exists a number $s_0=$inf$lbrace s in [0, infty):mathcal{H^s}(E)=0 rbrace$.
So $s_0(E)leq s$ and for all $x in E$ there is an open subset $U Rightarrow s_0(E) geq s$.
Here I don't see how to conclude. It should be possible to use that every open cover of $E subset mathbb{R^n}$ has a countable subcover, but how to apply it here?
Is this way correct or can it be proved differently?
measure-theory
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up vote
1
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favorite
Let $s geq 0$ be a real number and $E subset mathbb{R^n}$.
Suppose that for all $x in E$ there is an open subset $U subset mathbb{R^n}$ which contains $x$, and $mathcal{H^s}(E cap U)=0 Rightarrow mathcal{H^s}(E)=0$.
How to show this implication?
I tried to use:
Since $E subset mathbb{R^n}$, it exists a number $s_0=$inf$lbrace s in [0, infty):mathcal{H^s}(E)=0 rbrace$.
So $s_0(E)leq s$ and for all $x in E$ there is an open subset $U Rightarrow s_0(E) geq s$.
Here I don't see how to conclude. It should be possible to use that every open cover of $E subset mathbb{R^n}$ has a countable subcover, but how to apply it here?
Is this way correct or can it be proved differently?
measure-theory
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $s geq 0$ be a real number and $E subset mathbb{R^n}$.
Suppose that for all $x in E$ there is an open subset $U subset mathbb{R^n}$ which contains $x$, and $mathcal{H^s}(E cap U)=0 Rightarrow mathcal{H^s}(E)=0$.
How to show this implication?
I tried to use:
Since $E subset mathbb{R^n}$, it exists a number $s_0=$inf$lbrace s in [0, infty):mathcal{H^s}(E)=0 rbrace$.
So $s_0(E)leq s$ and for all $x in E$ there is an open subset $U Rightarrow s_0(E) geq s$.
Here I don't see how to conclude. It should be possible to use that every open cover of $E subset mathbb{R^n}$ has a countable subcover, but how to apply it here?
Is this way correct or can it be proved differently?
measure-theory
Let $s geq 0$ be a real number and $E subset mathbb{R^n}$.
Suppose that for all $x in E$ there is an open subset $U subset mathbb{R^n}$ which contains $x$, and $mathcal{H^s}(E cap U)=0 Rightarrow mathcal{H^s}(E)=0$.
How to show this implication?
I tried to use:
Since $E subset mathbb{R^n}$, it exists a number $s_0=$inf$lbrace s in [0, infty):mathcal{H^s}(E)=0 rbrace$.
So $s_0(E)leq s$ and for all $x in E$ there is an open subset $U Rightarrow s_0(E) geq s$.
Here I don't see how to conclude. It should be possible to use that every open cover of $E subset mathbb{R^n}$ has a countable subcover, but how to apply it here?
Is this way correct or can it be proved differently?
measure-theory
measure-theory
asked Nov 15 at 8:41
Tartulop
534
534
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1 Answer
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1
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accepted
$H^{s}$ is countably subadditive and every open cover has a countable subcover. If $Esubset cup U_i$ with $H^{s} (Ecap U_i)=0$ for all $i$ then $H^{s}(E) leq sum H^{s} (Ecap U_i)=0$.
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
$H^{s}$ is countably subadditive and every open cover has a countable subcover. If $Esubset cup U_i$ with $H^{s} (Ecap U_i)=0$ for all $i$ then $H^{s}(E) leq sum H^{s} (Ecap U_i)=0$.
add a comment |
up vote
1
down vote
accepted
$H^{s}$ is countably subadditive and every open cover has a countable subcover. If $Esubset cup U_i$ with $H^{s} (Ecap U_i)=0$ for all $i$ then $H^{s}(E) leq sum H^{s} (Ecap U_i)=0$.
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
$H^{s}$ is countably subadditive and every open cover has a countable subcover. If $Esubset cup U_i$ with $H^{s} (Ecap U_i)=0$ for all $i$ then $H^{s}(E) leq sum H^{s} (Ecap U_i)=0$.
$H^{s}$ is countably subadditive and every open cover has a countable subcover. If $Esubset cup U_i$ with $H^{s} (Ecap U_i)=0$ for all $i$ then $H^{s}(E) leq sum H^{s} (Ecap U_i)=0$.
answered Nov 15 at 8:48
Kavi Rama Murthy
42.1k31751
42.1k31751
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