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Show that The Hausdorff Measure is a Measure
I am trying to prove that the Hausdorff measure is, in fact, a measure.
Definition. $mu$ is a measure on $mathbb{R}^{m}$ if $mu$ assigns a non-negative number (possibly $infty$), to each subset of $mathbb{R}^{m}$ such that
$mu(emptyset)=0$;
$mu(S)leqmu(T)$ if $Ssubseteq T$;- if $S_{1},S_{2},ldots$ is a countable or finite sequence of sets, then $muleft(bigcup_{i=1}^{infty}S_{i}right)leqsum_{i=1}^{infty}mu(S_{i})$ with equality if $S_{i}$ are disjoint Borel sets.
We call $mu(S)$ the measure of $S$.
Definition. Suppose $Fsubsetmathbb{R}^{n}$ and $sinmathbb{R}_{geq0}$. For $delta>0$, consider all $delta$-covers of $F$ and minimise the sum of the $s$th powers of the diameters. As $deltato0$, the class of permissible covers of $F$ is reduced. So $$mathcal{H}_{delta}^{s}(F)=infleft{sum_{i=1}^{infty}|U_{i}|^{s}:{U_{i}} text{ is a } delta text{-cover of } Fright}$$ increases and so approaches a limit. The $s$-dimensional Hausdorff measure $$mathcal{H}^{s}(F)=lim_{deltato0}mathcal{H}_{delta}^{s}(F)$$ exists for any $Fsubsetmathbb{R}^{n}$ although it can be, and often is, either $0$ of $infty$.
What I've done: Obviously $mathcal{H}^{s}(emptyset)=0$, and it is intuitive that if $Esubseteq F$, then $mathcal{H}^{s}(E)leqslantmathcal{H}^{s}(F)$. What I am not sure how to do it show that $$mathcal{H}^{s}left(bigcup_{i=1}^{infty}F_{i}right)leqslantsum_{i=1}^{infty}mathcal{H}^{s}(F_{i})$$ with equality if ${F{i}}$ are disjoint Borel sets.
measure-theory
edited Nov 20 '18 at 4:57
Lord Shark the Unknown
101k958132
asked Jul 9 '17 at 14:01
JSharpee
29518
|
show 3 more comments
I am trying to prove that the Hausdorff measure is, in fact, a measure.
Definition. $mu$ is a measure on $mathbb{R}^{m}$ if $mu$ assigns a non-negative number (possibly $infty$), to each subset of $mathbb{R}^{m}$ such that
$mu(emptyset)=0$;
$mu(S)leqmu(T)$ if $Ssubseteq T$;- if $S_{1},S_{2},ldots$ is a countable or finite sequence of sets, then $muleft(bigcup_{i=1}^{infty}S_{i}right)leqsum_{i=1}^{infty}mu(S_{i})$ with equality if $S_{i}$ are disjoint Borel sets.
We call $mu(S)$ the measure of $S$.
Definition. Suppose $Fsubsetmathbb{R}^{n}$ and $sinmathbb{R}_{geq0}$. For $delta>0$, consider all $delta$-covers of $F$ and minimise the sum of the $s$th powers of the diameters. As $deltato0$, the class of permissible covers of $F$ is reduced. So $$mathcal{H}_{delta}^{s}(F)=infleft{sum_{i=1}^{infty}|U_{i}|^{s}:{U_{i}} text{ is a } delta text{-cover of } Fright}$$ increases and so approaches a limit. The $s$-dimensional Hausdorff measure $$mathcal{H}^{s}(F)=lim_{deltato0}mathcal{H}_{delta}^{s}(F)$$ exists for any $Fsubsetmathbb{R}^{n}$ although it can be, and often is, either $0$ of $infty$.
What I've done: Obviously $mathcal{H}^{s}(emptyset)=0$, and it is intuitive that if $Esubseteq F$, then $mathcal{H}^{s}(E)leqslantmathcal{H}^{s}(F)$. What I am not sure how to do it show that $$mathcal{H}^{s}left(bigcup_{i=1}^{infty}F_{i}right)leqslantsum_{i=1}^{infty}mathcal{H}^{s}(F_{i})$$ with equality if ${F{i}}$ are disjoint Borel sets.
measure-theory
edited Nov 20 '18 at 4:57
Lord Shark the Unknown
101k958132
asked Jul 9 '17 at 14:01
JSharpee
29518
In your definition of a measure, do you really want to define $mu$ on each subset of $mathbb{R}^m$? Or perhaps only on Borel subsets?
– Lee Mosher
Jul 9 '17 at 14:04
Theorem 2.19 here (books.google.de/…) might be helpful.
– PhoemueX
Jul 9 '17 at 14:09
@LeeMosher only Borel subsets are necessary, but it is my understanding that they are well-defined on all subsets.
– JSharpee
Jul 9 '17 at 14:15
@PhoemueX I will have a read through. Thank you.
– JSharpee
Jul 9 '17 at 14:15
Generally speaking, it is not the case that measures are well-defined on all subsets. That's not even true for the standard Lebesgue measure on the real line (assuming the axiom of choice).
– Lee Mosher
Jul 9 '17 at 14:16
|
show 3 more comments
I am trying to prove that the Hausdorff measure is, in fact, a measure.
Definition. $mu$ is a measure on $mathbb{R}^{m}$ if $mu$ assigns a non-negative number (possibly $infty$), to each subset of $mathbb{R}^{m}$ such that
$mu(emptyset)=0$;
$mu(S)leqmu(T)$ if $Ssubseteq T$;- if $S_{1},S_{2},ldots$ is a countable or finite sequence of sets, then $muleft(bigcup_{i=1}^{infty}S_{i}right)leqsum_{i=1}^{infty}mu(S_{i})$ with equality if $S_{i}$ are disjoint Borel sets.
We call $mu(S)$ the measure of $S$.
Definition. Suppose $Fsubsetmathbb{R}^{n}$ and $sinmathbb{R}_{geq0}$. For $delta>0$, consider all $delta$-covers of $F$ and minimise the sum of the $s$th powers of the diameters. As $deltato0$, the class of permissible covers of $F$ is reduced. So $$mathcal{H}_{delta}^{s}(F)=infleft{sum_{i=1}^{infty}|U_{i}|^{s}:{U_{i}} text{ is a } delta text{-cover of } Fright}$$ increases and so approaches a limit. The $s$-dimensional Hausdorff measure $$mathcal{H}^{s}(F)=lim_{deltato0}mathcal{H}_{delta}^{s}(F)$$ exists for any $Fsubsetmathbb{R}^{n}$ although it can be, and often is, either $0$ of $infty$.
What I've done: Obviously $mathcal{H}^{s}(emptyset)=0$, and it is intuitive that if $Esubseteq F$, then $mathcal{H}^{s}(E)leqslantmathcal{H}^{s}(F)$. What I am not sure how to do it show that $$mathcal{H}^{s}left(bigcup_{i=1}^{infty}F_{i}right)leqslantsum_{i=1}^{infty}mathcal{H}^{s}(F_{i})$$ with equality if ${F{i}}$ are disjoint Borel sets.
measure-theory
edited Nov 20 '18 at 4:57
Lord Shark the Unknown
101k958132
asked Jul 9 '17 at 14:01
JSharpee
29518
I am trying to prove that the Hausdorff measure is, in fact, a measure.
Definition. $mu$ is a measure on $mathbb{R}^{m}$ if $mu$ assigns a non-negative number (possibly $infty$), to each subset of $mathbb{R}^{m}$ such that
$mu(emptyset)=0$;
$mu(S)leqmu(T)$ if $Ssubseteq T$;- if $S_{1},S_{2},ldots$ is a countable or finite sequence of sets, then $muleft(bigcup_{i=1}^{infty}S_{i}right)leqsum_{i=1}^{infty}mu(S_{i})$ with equality if $S_{i}$ are disjoint Borel sets.
We call $mu(S)$ the measure of $S$.
Definition. Suppose $Fsubsetmathbb{R}^{n}$ and $sinmathbb{R}_{geq0}$. For $delta>0$, consider all $delta$-covers of $F$ and minimise the sum of the $s$th powers of the diameters. As $deltato0$, the class of permissible covers of $F$ is reduced. So $$mathcal{H}_{delta}^{s}(F)=infleft{sum_{i=1}^{infty}|U_{i}|^{s}:{U_{i}} text{ is a } delta text{-cover of } Fright}$$ increases and so approaches a limit. The $s$-dimensional Hausdorff measure $$mathcal{H}^{s}(F)=lim_{deltato0}mathcal{H}_{delta}^{s}(F)$$ exists for any $Fsubsetmathbb{R}^{n}$ although it can be, and often is, either $0$ of $infty$.
What I've done: Obviously $mathcal{H}^{s}(emptyset)=0$, and it is intuitive that if $Esubseteq F$, then $mathcal{H}^{s}(E)leqslantmathcal{H}^{s}(F)$. What I am not sure how to do it show that $$mathcal{H}^{s}left(bigcup_{i=1}^{infty}F_{i}right)leqslantsum_{i=1}^{infty}mathcal{H}^{s}(F_{i})$$ with equality if ${F{i}}$ are disjoint Borel sets.
measure-theory
measure-theory
edited Nov 20 '18 at 4:57
Lord Shark the Unknown
101k958132
asked Jul 9 '17 at 14:01
JSharpee
29518
edited Nov 20 '18 at 4:57
Lord Shark the Unknown
101k958132
asked Jul 9 '17 at 14:01
JSharpee
29518
edited Nov 20 '18 at 4:57
Lord Shark the Unknown
101k958132
edited Nov 20 '18 at 4:57
Lord Shark the Unknown
101k958132
edited Nov 20 '18 at 4:57
Lord Shark the Unknown
101k958132
101k958132
asked Jul 9 '17 at 14:01
JSharpee
29518
asked Jul 9 '17 at 14:01
JSharpee
29518
asked Jul 9 '17 at 14:01
JSharpee
29518
29518
In your definition of a measure, do you really want to define $mu$ on each subset of $mathbb{R}^m$? Or perhaps only on Borel subsets?
– Lee Mosher
Jul 9 '17 at 14:04
Theorem 2.19 here (books.google.de/…) might be helpful.
– PhoemueX
Jul 9 '17 at 14:09
@LeeMosher only Borel subsets are necessary, but it is my understanding that they are well-defined on all subsets.
– JSharpee
Jul 9 '17 at 14:15
@PhoemueX I will have a read through. Thank you.
– JSharpee
Jul 9 '17 at 14:15
Generally speaking, it is not the case that measures are well-defined on all subsets. That's not even true for the standard Lebesgue measure on the real line (assuming the axiom of choice).
– Lee Mosher
Jul 9 '17 at 14:16
|
show 3 more comments
In your definition of a measure, do you really want to define $mu$ on each subset of $mathbb{R}^m$? Or perhaps only on Borel subsets?
– Lee Mosher
Jul 9 '17 at 14:04
Theorem 2.19 here (books.google.de/…) might be helpful.
– PhoemueX
Jul 9 '17 at 14:09
@LeeMosher only Borel subsets are necessary, but it is my understanding that they are well-defined on all subsets.
– JSharpee
Jul 9 '17 at 14:15
@PhoemueX I will have a read through. Thank you.
– JSharpee
Jul 9 '17 at 14:15
Generally speaking, it is not the case that measures are well-defined on all subsets. That's not even true for the standard Lebesgue measure on the real line (assuming the axiom of choice).
– Lee Mosher
Jul 9 '17 at 14:16
In your definition of a measure, do you really want to define $mu$ on each subset of $mathbb{R}^m$? Or perhaps only on Borel subsets?
– Lee Mosher
Jul 9 '17 at 14:04
In your definition of a measure, do you really want to define $mu$ on each subset of $mathbb{R}^m$? Or perhaps only on Borel subsets?
– Lee Mosher
Jul 9 '17 at 14:04
Theorem 2.19 here (books.google.de/…) might be helpful.
– PhoemueX
Jul 9 '17 at 14:09
Theorem 2.19 here (books.google.de/…) might be helpful.
– PhoemueX
Jul 9 '17 at 14:09
@LeeMosher only Borel subsets are necessary, but it is my understanding that they are well-defined on all subsets.
– JSharpee
Jul 9 '17 at 14:15
@LeeMosher only Borel subsets are necessary, but it is my understanding that they are well-defined on all subsets.
– JSharpee
Jul 9 '17 at 14:15
@PhoemueX I will have a read through. Thank you.
– JSharpee
Jul 9 '17 at 14:15
@PhoemueX I will have a read through. Thank you.
– JSharpee
Jul 9 '17 at 14:15
Generally speaking, it is not the case that measures are well-defined on all subsets. That's not even true for the standard Lebesgue measure on the real line (assuming the axiom of choice).
– Lee Mosher
Jul 9 '17 at 14:16
Generally speaking, it is not the case that measures are well-defined on all subsets. That's not even true for the standard Lebesgue measure on the real line (assuming the axiom of choice).
– Lee Mosher
Jul 9 '17 at 14:16
|
show 3 more comments
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In your definition of a measure, do you really want to define $mu$ on each subset of $mathbb{R}^m$? Or perhaps only on Borel subsets?
– Lee Mosher
Jul 9 '17 at 14:04
Theorem 2.19 here (books.google.de/…) might be helpful.
– PhoemueX
Jul 9 '17 at 14:09
@LeeMosher only Borel subsets are necessary, but it is my understanding that they are well-defined on all subsets.
– JSharpee
Jul 9 '17 at 14:15
@PhoemueX I will have a read through. Thank you.
– JSharpee
Jul 9 '17 at 14:15
Generally speaking, it is not the case that measures are well-defined on all subsets. That's not even true for the standard Lebesgue measure on the real line (assuming the axiom of choice).
– Lee Mosher
Jul 9 '17 at 14:16