Verify that $R^d$ with the usual Lebesgue measure is separable
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A measure space $(X,mu)$ is sepearable if there is a countable family of measurable subsets ${E_k }_{k=1}^infty $ so that if $E$ is any measurable set of finite measure , then $$mu(E triangle E_{n_k}) to 0 ,,,,,,,as ,kto0$$ for an appropriate subsequence ${n_k }$ which depends on $E$ .
Verify that $R^d$ with the usual Lebesgue measure is separable.
I want to show that $E_k$ is a collection of open subset centered at rational numbers with rational radius. However , let $R^d =R$ , and $E=(0,1) cup (2,3)$ , the collection of $E_k$ defines above fails. Then I want to let $M$ denote the collection of all the subset of countable union of $E_k$ , But it is obvious $M$ is not countable.
real-analysis functional-analysis measure-theory
asked Dec 5 '18 at 8:35
J.GuoJ.Guo
2599
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add a comment |
$begingroup$
A measure space $(X,mu)$ is sepearable if there is a countable family of measurable subsets ${E_k }_{k=1}^infty $ so that if $E$ is any measurable set of finite measure , then $$mu(E triangle E_{n_k}) to 0 ,,,,,,,as ,kto0$$ for an appropriate subsequence ${n_k }$ which depends on $E$ .
Verify that $R^d$ with the usual Lebesgue measure is separable.
I want to show that $E_k$ is a collection of open subset centered at rational numbers with rational radius. However , let $R^d =R$ , and $E=(0,1) cup (2,3)$ , the collection of $E_k$ defines above fails. Then I want to let $M$ denote the collection of all the subset of countable union of $E_k$ , But it is obvious $M$ is not countable.
real-analysis functional-analysis measure-theory
asked Dec 5 '18 at 8:35
J.GuoJ.Guo
2599
$endgroup$
add a comment |
$begingroup$
A measure space $(X,mu)$ is sepearable if there is a countable family of measurable subsets ${E_k }_{k=1}^infty $ so that if $E$ is any measurable set of finite measure , then $$mu(E triangle E_{n_k}) to 0 ,,,,,,,as ,kto0$$ for an appropriate subsequence ${n_k }$ which depends on $E$ .
Verify that $R^d$ with the usual Lebesgue measure is separable.
I want to show that $E_k$ is a collection of open subset centered at rational numbers with rational radius. However , let $R^d =R$ , and $E=(0,1) cup (2,3)$ , the collection of $E_k$ defines above fails. Then I want to let $M$ denote the collection of all the subset of countable union of $E_k$ , But it is obvious $M$ is not countable.
real-analysis functional-analysis measure-theory
asked Dec 5 '18 at 8:35
J.GuoJ.Guo
2599
$endgroup$
A measure space $(X,mu)$ is sepearable if there is a countable family of measurable subsets ${E_k }_{k=1}^infty $ so that if $E$ is any measurable set of finite measure , then $$mu(E triangle E_{n_k}) to 0 ,,,,,,,as ,kto0$$ for an appropriate subsequence ${n_k }$ which depends on $E$ .
Verify that $R^d$ with the usual Lebesgue measure is separable.
I want to show that $E_k$ is a collection of open subset centered at rational numbers with rational radius. However , let $R^d =R$ , and $E=(0,1) cup (2,3)$ , the collection of $E_k$ defines above fails. Then I want to let $M$ denote the collection of all the subset of countable union of $E_k$ , But it is obvious $M$ is not countable.
real-analysis functional-analysis measure-theory
real-analysis functional-analysis measure-theory
asked Dec 5 '18 at 8:35
J.GuoJ.Guo
2599
asked Dec 5 '18 at 8:35
J.GuoJ.Guo
2599
asked Dec 5 '18 at 8:35
J.GuoJ.Guo
2599
asked Dec 5 '18 at 8:35
J.GuoJ.Guo
2599
asked Dec 5 '18 at 8:35
J.GuoJ.Guo
2599
2599
add a comment |
add a comment |
1 Answer
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The following basic approximation results will give the answer.
1) Any set $E$ of finite measure in $mathbb R^{d}$ can be approximated by a finite disjoint union of measurable rectangles.
2) Any measurable set of finite measure in $mathbb R$ can be approximated by a finite disjoint union of half- closed intervals.
3) Any half closed interval can be approximated by a half closed interval with rational end points.
Here approximating $A$ by $B$ means making $m(ADelta B)$ small.
answered Dec 5 '18 at 8:41
Kavi Rama MurthyKavi Rama Murthy
53k32055
$endgroup$
$begingroup$
Any measurable set of finite measure in $R$ can be approximated by a finite disjoint union of half closed intervals. But we need to approximate $E$ by just one subset , we can not take the "union"
$endgroup$
– J.Guo
Dec 5 '18 at 8:59
$begingroup$
The countable dense set I am suggesting is the collection of all sets which can be written as finite unions of sets of the type $[a_1,b_n)times [a_2,b_2)timescdots times [a_d,b_d)$ with $a_i$'s and $b_i$'s rational.
$endgroup$
– Kavi Rama Murthy
Dec 5 '18 at 9:20
$begingroup$
@J.Guo In you argument for one-dimensional case you are considering just single intervals with rational end points. Why not consider finite unions of these. In that case $(0,1)cup (2,3)$ is already in the countable family.
$endgroup$
– Kavi Rama Murthy
Dec 5 '18 at 9:23
add a comment |
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1 Answer
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1 Answer
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active
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active
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active
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votes
$begingroup$
The following basic approximation results will give the answer.
1) Any set $E$ of finite measure in $mathbb R^{d}$ can be approximated by a finite disjoint union of measurable rectangles.
2) Any measurable set of finite measure in $mathbb R$ can be approximated by a finite disjoint union of half- closed intervals.
3) Any half closed interval can be approximated by a half closed interval with rational end points.
Here approximating $A$ by $B$ means making $m(ADelta B)$ small.
answered Dec 5 '18 at 8:41
Kavi Rama MurthyKavi Rama Murthy
53k32055
$endgroup$
$begingroup$
Any measurable set of finite measure in $R$ can be approximated by a finite disjoint union of half closed intervals. But we need to approximate $E$ by just one subset , we can not take the "union"
$endgroup$
– J.Guo
Dec 5 '18 at 8:59
$begingroup$
The countable dense set I am suggesting is the collection of all sets which can be written as finite unions of sets of the type $[a_1,b_n)times [a_2,b_2)timescdots times [a_d,b_d)$ with $a_i$'s and $b_i$'s rational.
$endgroup$
– Kavi Rama Murthy
Dec 5 '18 at 9:20
$begingroup$
@J.Guo In you argument for one-dimensional case you are considering just single intervals with rational end points. Why not consider finite unions of these. In that case $(0,1)cup (2,3)$ is already in the countable family.
$endgroup$
– Kavi Rama Murthy
Dec 5 '18 at 9:23
add a comment |
$begingroup$
The following basic approximation results will give the answer.
1) Any set $E$ of finite measure in $mathbb R^{d}$ can be approximated by a finite disjoint union of measurable rectangles.
2) Any measurable set of finite measure in $mathbb R$ can be approximated by a finite disjoint union of half- closed intervals.
3) Any half closed interval can be approximated by a half closed interval with rational end points.
Here approximating $A$ by $B$ means making $m(ADelta B)$ small.
answered Dec 5 '18 at 8:41
Kavi Rama MurthyKavi Rama Murthy
53k32055
$endgroup$
$begingroup$
Any measurable set of finite measure in $R$ can be approximated by a finite disjoint union of half closed intervals. But we need to approximate $E$ by just one subset , we can not take the "union"
$endgroup$
– J.Guo
Dec 5 '18 at 8:59
$begingroup$
The countable dense set I am suggesting is the collection of all sets which can be written as finite unions of sets of the type $[a_1,b_n)times [a_2,b_2)timescdots times [a_d,b_d)$ with $a_i$'s and $b_i$'s rational.
$endgroup$
– Kavi Rama Murthy
Dec 5 '18 at 9:20
$begingroup$
@J.Guo In you argument for one-dimensional case you are considering just single intervals with rational end points. Why not consider finite unions of these. In that case $(0,1)cup (2,3)$ is already in the countable family.
$endgroup$
– Kavi Rama Murthy
Dec 5 '18 at 9:23
add a comment |
$begingroup$
The following basic approximation results will give the answer.
1) Any set $E$ of finite measure in $mathbb R^{d}$ can be approximated by a finite disjoint union of measurable rectangles.
2) Any measurable set of finite measure in $mathbb R$ can be approximated by a finite disjoint union of half- closed intervals.
3) Any half closed interval can be approximated by a half closed interval with rational end points.
Here approximating $A$ by $B$ means making $m(ADelta B)$ small.
answered Dec 5 '18 at 8:41
Kavi Rama MurthyKavi Rama Murthy
53k32055
$endgroup$
The following basic approximation results will give the answer.
1) Any set $E$ of finite measure in $mathbb R^{d}$ can be approximated by a finite disjoint union of measurable rectangles.
2) Any measurable set of finite measure in $mathbb R$ can be approximated by a finite disjoint union of half- closed intervals.
3) Any half closed interval can be approximated by a half closed interval with rational end points.
Here approximating $A$ by $B$ means making $m(ADelta B)$ small.
answered Dec 5 '18 at 8:41
Kavi Rama MurthyKavi Rama Murthy
53k32055
answered Dec 5 '18 at 8:41
Kavi Rama MurthyKavi Rama Murthy
53k32055
answered Dec 5 '18 at 8:41
Kavi Rama MurthyKavi Rama Murthy
53k32055
answered Dec 5 '18 at 8:41
Kavi Rama MurthyKavi Rama Murthy
53k32055
53k32055
$begingroup$
Any measurable set of finite measure in $R$ can be approximated by a finite disjoint union of half closed intervals. But we need to approximate $E$ by just one subset , we can not take the "union"
$endgroup$
– J.Guo
Dec 5 '18 at 8:59
$begingroup$
The countable dense set I am suggesting is the collection of all sets which can be written as finite unions of sets of the type $[a_1,b_n)times [a_2,b_2)timescdots times [a_d,b_d)$ with $a_i$'s and $b_i$'s rational.
$endgroup$
– Kavi Rama Murthy
Dec 5 '18 at 9:20
$begingroup$
@J.Guo In you argument for one-dimensional case you are considering just single intervals with rational end points. Why not consider finite unions of these. In that case $(0,1)cup (2,3)$ is already in the countable family.
$endgroup$
– Kavi Rama Murthy
Dec 5 '18 at 9:23
add a comment |
$begingroup$
Any measurable set of finite measure in $R$ can be approximated by a finite disjoint union of half closed intervals. But we need to approximate $E$ by just one subset , we can not take the "union"
$endgroup$
– J.Guo
Dec 5 '18 at 8:59
$begingroup$
The countable dense set I am suggesting is the collection of all sets which can be written as finite unions of sets of the type $[a_1,b_n)times [a_2,b_2)timescdots times [a_d,b_d)$ with $a_i$'s and $b_i$'s rational.
$endgroup$
– Kavi Rama Murthy
Dec 5 '18 at 9:20
$begingroup$
@J.Guo In you argument for one-dimensional case you are considering just single intervals with rational end points. Why not consider finite unions of these. In that case $(0,1)cup (2,3)$ is already in the countable family.
$endgroup$
– Kavi Rama Murthy
Dec 5 '18 at 9:23
$begingroup$
Any measurable set of finite measure in $R$ can be approximated by a finite disjoint union of half closed intervals. But we need to approximate $E$ by just one subset , we can not take the "union"
$endgroup$
– J.Guo
Dec 5 '18 at 8:59
$begingroup$
Any measurable set of finite measure in $R$ can be approximated by a finite disjoint union of half closed intervals. But we need to approximate $E$ by just one subset , we can not take the "union"
$endgroup$
– J.Guo
Dec 5 '18 at 8:59
$begingroup$
The countable dense set I am suggesting is the collection of all sets which can be written as finite unions of sets of the type $[a_1,b_n)times [a_2,b_2)timescdots times [a_d,b_d)$ with $a_i$'s and $b_i$'s rational.
$endgroup$
– Kavi Rama Murthy
Dec 5 '18 at 9:20
$begingroup$
The countable dense set I am suggesting is the collection of all sets which can be written as finite unions of sets of the type $[a_1,b_n)times [a_2,b_2)timescdots times [a_d,b_d)$ with $a_i$'s and $b_i$'s rational.
$endgroup$
– Kavi Rama Murthy
Dec 5 '18 at 9:20
$begingroup$
@J.Guo In you argument for one-dimensional case you are considering just single intervals with rational end points. Why not consider finite unions of these. In that case $(0,1)cup (2,3)$ is already in the countable family.
$endgroup$
– Kavi Rama Murthy
Dec 5 '18 at 9:23
$begingroup$
@J.Guo In you argument for one-dimensional case you are considering just single intervals with rational end points. Why not consider finite unions of these. In that case $(0,1)cup (2,3)$ is already in the countable family.
$endgroup$
– Kavi Rama Murthy
Dec 5 '18 at 9:23
add a comment |
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