Countable $varepsilon$-net of measurable sets
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Let $I=[0,1]$, $mathcal{A}$ is the family that contains all the Lebesgue measurable sets of $I$, for any $A_1,A_2in mathcal{A}$, we define a metric
$$d(A_1,A_2)=int |1_{A_1}-1_{A_2}|.$$
A $varepsilon$-net of $mathcal{A}$ is a sub-family $mathcal{A}'subset mathcal{A}$ such that
$$forall Ain mathcal{A},exists A'in mathcal{A}',~s.t.~d(A,A')le varepsilon.$$
My question is:
For all $varepsilon>0$, does there exist a countable $varepsilon$-net for $mathcal{A}$?
general-topology measure-theory lebesgue-measure lindelof-spaces
edited Nov 18 at 9:50
Asaf Karagila♦
300k32421751
asked Nov 18 at 9:23
Paul
184111
add a comment |
up vote
0
down vote
favorite
Let $I=[0,1]$, $mathcal{A}$ is the family that contains all the Lebesgue measurable sets of $I$, for any $A_1,A_2in mathcal{A}$, we define a metric
$$d(A_1,A_2)=int |1_{A_1}-1_{A_2}|.$$
A $varepsilon$-net of $mathcal{A}$ is a sub-family $mathcal{A}'subset mathcal{A}$ such that
$$forall Ain mathcal{A},exists A'in mathcal{A}',~s.t.~d(A,A')le varepsilon.$$
My question is:
For all $varepsilon>0$, does there exist a countable $varepsilon$-net for $mathcal{A}$?
general-topology measure-theory lebesgue-measure lindelof-spaces
edited Nov 18 at 9:50
Asaf Karagila♦
300k32421751
asked Nov 18 at 9:23
Paul
184111
Is it a metric ? It seems that the distance between two sets of measure $0$ is also $0$.
– Henno Brandsma
Nov 18 at 9:26
Yes, but we can quotient on the equivalent relation $A_1sim A_2$ iff $d(A_1,A_2)=0$ if needed.
– Paul
Nov 18 at 9:40
1
Note that this is the same as the standard metric on a measure space $d(A,B) = mu(ADelta B)$ and this is a separable pseudometric space.
– Henno Brandsma
Nov 18 at 10:04
@HennoBrandsma Thanks, this is exactly what I'm looking for. I also find this wiki page is useful: en.wikipedia.org/wiki/Sigma-algebra#Separable_%CF%83-algebras
– Paul
Nov 18 at 10:31
And for (pseudo)metric spaces your net property is equivalent to separability of the measure space. And Lebesgue measure is separable (classical fact, that Wikipedia also mentions).
– Henno Brandsma
Nov 18 at 10:35
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $I=[0,1]$, $mathcal{A}$ is the family that contains all the Lebesgue measurable sets of $I$, for any $A_1,A_2in mathcal{A}$, we define a metric
$$d(A_1,A_2)=int |1_{A_1}-1_{A_2}|.$$
A $varepsilon$-net of $mathcal{A}$ is a sub-family $mathcal{A}'subset mathcal{A}$ such that
$$forall Ain mathcal{A},exists A'in mathcal{A}',~s.t.~d(A,A')le varepsilon.$$
My question is:
For all $varepsilon>0$, does there exist a countable $varepsilon$-net for $mathcal{A}$?
general-topology measure-theory lebesgue-measure lindelof-spaces
edited Nov 18 at 9:50
Asaf Karagila♦
300k32421751
asked Nov 18 at 9:23
Paul
184111
Let $I=[0,1]$, $mathcal{A}$ is the family that contains all the Lebesgue measurable sets of $I$, for any $A_1,A_2in mathcal{A}$, we define a metric
$$d(A_1,A_2)=int |1_{A_1}-1_{A_2}|.$$
A $varepsilon$-net of $mathcal{A}$ is a sub-family $mathcal{A}'subset mathcal{A}$ such that
$$forall Ain mathcal{A},exists A'in mathcal{A}',~s.t.~d(A,A')le varepsilon.$$
My question is:
For all $varepsilon>0$, does there exist a countable $varepsilon$-net for $mathcal{A}$?
general-topology measure-theory lebesgue-measure lindelof-spaces
general-topology measure-theory lebesgue-measure lindelof-spaces
edited Nov 18 at 9:50
Asaf Karagila♦
300k32421751
asked Nov 18 at 9:23
Paul
184111
edited Nov 18 at 9:50
Asaf Karagila♦
300k32421751
asked Nov 18 at 9:23
Paul
184111
edited Nov 18 at 9:50
Asaf Karagila♦
300k32421751
edited Nov 18 at 9:50
Asaf Karagila♦
300k32421751
edited Nov 18 at 9:50
Asaf Karagila♦
300k32421751
300k32421751
asked Nov 18 at 9:23
Paul
184111
asked Nov 18 at 9:23
Paul
184111
asked Nov 18 at 9:23
Paul
184111
184111
Is it a metric ? It seems that the distance between two sets of measure $0$ is also $0$.
– Henno Brandsma
Nov 18 at 9:26
Yes, but we can quotient on the equivalent relation $A_1sim A_2$ iff $d(A_1,A_2)=0$ if needed.
– Paul
Nov 18 at 9:40
1
Note that this is the same as the standard metric on a measure space $d(A,B) = mu(ADelta B)$ and this is a separable pseudometric space.
– Henno Brandsma
Nov 18 at 10:04
@HennoBrandsma Thanks, this is exactly what I'm looking for. I also find this wiki page is useful: en.wikipedia.org/wiki/Sigma-algebra#Separable_%CF%83-algebras
– Paul
Nov 18 at 10:31
And for (pseudo)metric spaces your net property is equivalent to separability of the measure space. And Lebesgue measure is separable (classical fact, that Wikipedia also mentions).
– Henno Brandsma
Nov 18 at 10:35
add a comment |
Is it a metric ? It seems that the distance between two sets of measure $0$ is also $0$.
– Henno Brandsma
Nov 18 at 9:26
Yes, but we can quotient on the equivalent relation $A_1sim A_2$ iff $d(A_1,A_2)=0$ if needed.
– Paul
Nov 18 at 9:40
1
Note that this is the same as the standard metric on a measure space $d(A,B) = mu(ADelta B)$ and this is a separable pseudometric space.
– Henno Brandsma
Nov 18 at 10:04
@HennoBrandsma Thanks, this is exactly what I'm looking for. I also find this wiki page is useful: en.wikipedia.org/wiki/Sigma-algebra#Separable_%CF%83-algebras
– Paul
Nov 18 at 10:31
And for (pseudo)metric spaces your net property is equivalent to separability of the measure space. And Lebesgue measure is separable (classical fact, that Wikipedia also mentions).
– Henno Brandsma
Nov 18 at 10:35
Is it a metric ? It seems that the distance between two sets of measure $0$ is also $0$.
– Henno Brandsma
Nov 18 at 9:26
Is it a metric ? It seems that the distance between two sets of measure $0$ is also $0$.
– Henno Brandsma
Nov 18 at 9:26
Yes, but we can quotient on the equivalent relation $A_1sim A_2$ iff $d(A_1,A_2)=0$ if needed.
– Paul
Nov 18 at 9:40
Yes, but we can quotient on the equivalent relation $A_1sim A_2$ iff $d(A_1,A_2)=0$ if needed.
– Paul
Nov 18 at 9:40
1
1
Note that this is the same as the standard metric on a measure space $d(A,B) = mu(ADelta B)$ and this is a separable pseudometric space.
– Henno Brandsma
Nov 18 at 10:04
Note that this is the same as the standard metric on a measure space $d(A,B) = mu(ADelta B)$ and this is a separable pseudometric space.
– Henno Brandsma
Nov 18 at 10:04
@HennoBrandsma Thanks, this is exactly what I'm looking for. I also find this wiki page is useful: en.wikipedia.org/wiki/Sigma-algebra#Separable_%CF%83-algebras
– Paul
Nov 18 at 10:31
@HennoBrandsma Thanks, this is exactly what I'm looking for. I also find this wiki page is useful: en.wikipedia.org/wiki/Sigma-algebra#Separable_%CF%83-algebras
– Paul
Nov 18 at 10:31
And for (pseudo)metric spaces your net property is equivalent to separability of the measure space. And Lebesgue measure is separable (classical fact, that Wikipedia also mentions).
– Henno Brandsma
Nov 18 at 10:35
And for (pseudo)metric spaces your net property is equivalent to separability of the measure space. And Lebesgue measure is separable (classical fact, that Wikipedia also mentions).
– Henno Brandsma
Nov 18 at 10:35
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
To make $d$ a metric you have to think of your set as a subset of $L^{1}$ by identifying sets which differ only by a null set. In that case the assertion is true. Consider finite disjoint unions of intervals with rational end points and their characteristic functions. Characteristic function of any $A inmathcal A$ can be approximated in $L^{1}$ norm by those of finite disjoint unions of intervals and then we can approximate these intervals by those with rational end points which proves the assertion.
answered Nov 18 at 12:26
Kavi Rama Murthy
43.8k31852
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
To make $d$ a metric you have to think of your set as a subset of $L^{1}$ by identifying sets which differ only by a null set. In that case the assertion is true. Consider finite disjoint unions of intervals with rational end points and their characteristic functions. Characteristic function of any $A inmathcal A$ can be approximated in $L^{1}$ norm by those of finite disjoint unions of intervals and then we can approximate these intervals by those with rational end points which proves the assertion.
answered Nov 18 at 12:26
Kavi Rama Murthy
43.8k31852
add a comment |
up vote
1
down vote
accepted
To make $d$ a metric you have to think of your set as a subset of $L^{1}$ by identifying sets which differ only by a null set. In that case the assertion is true. Consider finite disjoint unions of intervals with rational end points and their characteristic functions. Characteristic function of any $A inmathcal A$ can be approximated in $L^{1}$ norm by those of finite disjoint unions of intervals and then we can approximate these intervals by those with rational end points which proves the assertion.
answered Nov 18 at 12:26
Kavi Rama Murthy
43.8k31852
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
To make $d$ a metric you have to think of your set as a subset of $L^{1}$ by identifying sets which differ only by a null set. In that case the assertion is true. Consider finite disjoint unions of intervals with rational end points and their characteristic functions. Characteristic function of any $A inmathcal A$ can be approximated in $L^{1}$ norm by those of finite disjoint unions of intervals and then we can approximate these intervals by those with rational end points which proves the assertion.
answered Nov 18 at 12:26
Kavi Rama Murthy
43.8k31852
To make $d$ a metric you have to think of your set as a subset of $L^{1}$ by identifying sets which differ only by a null set. In that case the assertion is true. Consider finite disjoint unions of intervals with rational end points and their characteristic functions. Characteristic function of any $A inmathcal A$ can be approximated in $L^{1}$ norm by those of finite disjoint unions of intervals and then we can approximate these intervals by those with rational end points which proves the assertion.
answered Nov 18 at 12:26
Kavi Rama Murthy
43.8k31852
answered Nov 18 at 12:26
Kavi Rama Murthy
43.8k31852
answered Nov 18 at 12:26
Kavi Rama Murthy
43.8k31852
answered Nov 18 at 12:26
Kavi Rama Murthy
43.8k31852
43.8k31852
add a comment |
add a comment |
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Is it a metric ? It seems that the distance between two sets of measure $0$ is also $0$.
– Henno Brandsma
Nov 18 at 9:26
Yes, but we can quotient on the equivalent relation $A_1sim A_2$ iff $d(A_1,A_2)=0$ if needed.
– Paul
Nov 18 at 9:40
1
Note that this is the same as the standard metric on a measure space $d(A,B) = mu(ADelta B)$ and this is a separable pseudometric space.
– Henno Brandsma
Nov 18 at 10:04
@HennoBrandsma Thanks, this is exactly what I'm looking for. I also find this wiki page is useful: en.wikipedia.org/wiki/Sigma-algebra#Separable_%CF%83-algebras
– Paul
Nov 18 at 10:31
And for (pseudo)metric spaces your net property is equivalent to separability of the measure space. And Lebesgue measure is separable (classical fact, that Wikipedia also mentions).
– Henno Brandsma
Nov 18 at 10:35