Is there only one complete measure extending Borel measure?
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Is Lebesgue measure the only complete mesure extending Borel mesure on $mathbb R$?
If yes, why? If no, what is an example of a complete measure extending Borel measure that is different from Lebesgue measure?
I raise this question following Completion of measure spaces - uniqueness.
measure-theory lebesgue-measure
asked Aug 17 at 9:20
mathcounterexamples.net
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up vote
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Is Lebesgue measure the only complete mesure extending Borel mesure on $mathbb R$?
If yes, why? If no, what is an example of a complete measure extending Borel measure that is different from Lebesgue measure?
I raise this question following Completion of measure spaces - uniqueness.
measure-theory lebesgue-measure
asked Aug 17 at 9:20
mathcounterexamples.net
25.3k21755
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Is Lebesgue measure the only complete mesure extending Borel mesure on $mathbb R$?
If yes, why? If no, what is an example of a complete measure extending Borel measure that is different from Lebesgue measure?
I raise this question following Completion of measure spaces - uniqueness.
measure-theory lebesgue-measure
asked Aug 17 at 9:20
mathcounterexamples.net
25.3k21755
Is Lebesgue measure the only complete mesure extending Borel mesure on $mathbb R$?
If yes, why? If no, what is an example of a complete measure extending Borel measure that is different from Lebesgue measure?
I raise this question following Completion of measure spaces - uniqueness.
measure-theory lebesgue-measure
asked Aug 17 at 9:20
mathcounterexamples.net
25.3k21755
asked Aug 17 at 9:20
mathcounterexamples.net
25.3k21755
asked Aug 17 at 9:20
mathcounterexamples.net
25.3k21755
asked Aug 17 at 9:20
mathcounterexamples.net
25.3k21755
25.3k21755
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add a comment |Â
1 Answer
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And the answer is the same as in the other post. You need to add more conditions to have uniqueness.
It is possible to define extensions of the Lebesgue measure, using Carathéodory's theorem, for example.
https://en.wikipedia.org/wiki/Carathéodory%27s_extension_theorem
Then you can take the completion of any of these extended measures, and there you go.
answered Aug 17 at 9:39
A. Pongrácz
3,877625
I don’t see how Carathéodory theorem answers the question. Reading Wikipedia article Consequently, any measure on a space containing all intervals of real numbers can be extended to the Borel algebra of the set of real numbers.... But I suppose that I start from the Borel measure.
– mathcounterexamples.net
Aug 17 at 10:58
That is not Carathéodory's theorem, just an application of it. Read the general theorem, and think about it, please.
– A. Pongrácz
Aug 17 at 11:21
Carathéodory's extension theorem (named after the mathematician Constantin Carathéodory) states that any measure defined on a given ring R of subsets of a given set Ω can be extended to the Ã-algebra generated by R, and this extension is unique if the measure is Ã-finite. Here, Borel measure is already equipped with a $sigma$-algebra. Hence Carathéodory extension may just be the Borel $sigma$-algebra itself. Can you precise a complete extension of Borel measure that is different from Lebesgue measure?
– mathcounterexamples.net
Aug 17 at 11:43
Look, I am not going to spoonfeed you the solution. You need to put your own effort into it. Hint: pick a set that is not Lebesgue-measurable, and consider the (semi)ring generated by that set and the family of Lebesgue measurable sets. Show that you can assign a value to the new set so that it extends as a measure to the ring. Then you can apply the theorem. I am not going to help you any more until you show your own work.
– A. Pongrácz
Aug 17 at 11:46
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
And the answer is the same as in the other post. You need to add more conditions to have uniqueness.
It is possible to define extensions of the Lebesgue measure, using Carathéodory's theorem, for example.
https://en.wikipedia.org/wiki/Carathéodory%27s_extension_theorem
Then you can take the completion of any of these extended measures, and there you go.
answered Aug 17 at 9:39
A. Pongrácz
3,877625
I don’t see how Carathéodory theorem answers the question. Reading Wikipedia article Consequently, any measure on a space containing all intervals of real numbers can be extended to the Borel algebra of the set of real numbers.... But I suppose that I start from the Borel measure.
– mathcounterexamples.net
Aug 17 at 10:58
That is not Carathéodory's theorem, just an application of it. Read the general theorem, and think about it, please.
– A. Pongrácz
Aug 17 at 11:21
Carathéodory's extension theorem (named after the mathematician Constantin Carathéodory) states that any measure defined on a given ring R of subsets of a given set Ω can be extended to the Ã-algebra generated by R, and this extension is unique if the measure is Ã-finite. Here, Borel measure is already equipped with a $sigma$-algebra. Hence Carathéodory extension may just be the Borel $sigma$-algebra itself. Can you precise a complete extension of Borel measure that is different from Lebesgue measure?
– mathcounterexamples.net
Aug 17 at 11:43
Look, I am not going to spoonfeed you the solution. You need to put your own effort into it. Hint: pick a set that is not Lebesgue-measurable, and consider the (semi)ring generated by that set and the family of Lebesgue measurable sets. Show that you can assign a value to the new set so that it extends as a measure to the ring. Then you can apply the theorem. I am not going to help you any more until you show your own work.
– A. Pongrácz
Aug 17 at 11:46
add a comment |Â
up vote
1
down vote
And the answer is the same as in the other post. You need to add more conditions to have uniqueness.
It is possible to define extensions of the Lebesgue measure, using Carathéodory's theorem, for example.
https://en.wikipedia.org/wiki/Carathéodory%27s_extension_theorem
Then you can take the completion of any of these extended measures, and there you go.
answered Aug 17 at 9:39
A. Pongrácz
3,877625
I don’t see how Carathéodory theorem answers the question. Reading Wikipedia article Consequently, any measure on a space containing all intervals of real numbers can be extended to the Borel algebra of the set of real numbers.... But I suppose that I start from the Borel measure.
– mathcounterexamples.net
Aug 17 at 10:58
That is not Carathéodory's theorem, just an application of it. Read the general theorem, and think about it, please.
– A. Pongrácz
Aug 17 at 11:21
Carathéodory's extension theorem (named after the mathematician Constantin Carathéodory) states that any measure defined on a given ring R of subsets of a given set Ω can be extended to the Ã-algebra generated by R, and this extension is unique if the measure is Ã-finite. Here, Borel measure is already equipped with a $sigma$-algebra. Hence Carathéodory extension may just be the Borel $sigma$-algebra itself. Can you precise a complete extension of Borel measure that is different from Lebesgue measure?
– mathcounterexamples.net
Aug 17 at 11:43
Look, I am not going to spoonfeed you the solution. You need to put your own effort into it. Hint: pick a set that is not Lebesgue-measurable, and consider the (semi)ring generated by that set and the family of Lebesgue measurable sets. Show that you can assign a value to the new set so that it extends as a measure to the ring. Then you can apply the theorem. I am not going to help you any more until you show your own work.
– A. Pongrácz
Aug 17 at 11:46
add a comment |Â
up vote
1
down vote
up vote
1
down vote
And the answer is the same as in the other post. You need to add more conditions to have uniqueness.
It is possible to define extensions of the Lebesgue measure, using Carathéodory's theorem, for example.
https://en.wikipedia.org/wiki/Carathéodory%27s_extension_theorem
Then you can take the completion of any of these extended measures, and there you go.
answered Aug 17 at 9:39
A. Pongrácz
3,877625
And the answer is the same as in the other post. You need to add more conditions to have uniqueness.
It is possible to define extensions of the Lebesgue measure, using Carathéodory's theorem, for example.
https://en.wikipedia.org/wiki/Carathéodory%27s_extension_theorem
Then you can take the completion of any of these extended measures, and there you go.
answered Aug 17 at 9:39
A. Pongrácz
3,877625
answered Aug 17 at 9:39
A. Pongrácz
3,877625
answered Aug 17 at 9:39
A. Pongrácz
3,877625
answered Aug 17 at 9:39
A. Pongrácz
3,877625
3,877625
I don’t see how Carathéodory theorem answers the question. Reading Wikipedia article Consequently, any measure on a space containing all intervals of real numbers can be extended to the Borel algebra of the set of real numbers.... But I suppose that I start from the Borel measure.
– mathcounterexamples.net
Aug 17 at 10:58
That is not Carathéodory's theorem, just an application of it. Read the general theorem, and think about it, please.
– A. Pongrácz
Aug 17 at 11:21
Carathéodory's extension theorem (named after the mathematician Constantin Carathéodory) states that any measure defined on a given ring R of subsets of a given set Ω can be extended to the Ã-algebra generated by R, and this extension is unique if the measure is Ã-finite. Here, Borel measure is already equipped with a $sigma$-algebra. Hence Carathéodory extension may just be the Borel $sigma$-algebra itself. Can you precise a complete extension of Borel measure that is different from Lebesgue measure?
– mathcounterexamples.net
Aug 17 at 11:43
Look, I am not going to spoonfeed you the solution. You need to put your own effort into it. Hint: pick a set that is not Lebesgue-measurable, and consider the (semi)ring generated by that set and the family of Lebesgue measurable sets. Show that you can assign a value to the new set so that it extends as a measure to the ring. Then you can apply the theorem. I am not going to help you any more until you show your own work.
– A. Pongrácz
Aug 17 at 11:46
add a comment |Â
I don’t see how Carathéodory theorem answers the question. Reading Wikipedia article Consequently, any measure on a space containing all intervals of real numbers can be extended to the Borel algebra of the set of real numbers.... But I suppose that I start from the Borel measure.
– mathcounterexamples.net
Aug 17 at 10:58
That is not Carathéodory's theorem, just an application of it. Read the general theorem, and think about it, please.
– A. Pongrácz
Aug 17 at 11:21
Carathéodory's extension theorem (named after the mathematician Constantin Carathéodory) states that any measure defined on a given ring R of subsets of a given set Ω can be extended to the Ã-algebra generated by R, and this extension is unique if the measure is Ã-finite. Here, Borel measure is already equipped with a $sigma$-algebra. Hence Carathéodory extension may just be the Borel $sigma$-algebra itself. Can you precise a complete extension of Borel measure that is different from Lebesgue measure?
– mathcounterexamples.net
Aug 17 at 11:43
Look, I am not going to spoonfeed you the solution. You need to put your own effort into it. Hint: pick a set that is not Lebesgue-measurable, and consider the (semi)ring generated by that set and the family of Lebesgue measurable sets. Show that you can assign a value to the new set so that it extends as a measure to the ring. Then you can apply the theorem. I am not going to help you any more until you show your own work.
– A. Pongrácz
Aug 17 at 11:46
I don’t see how Carathéodory theorem answers the question. Reading Wikipedia article Consequently, any measure on a space containing all intervals of real numbers can be extended to the Borel algebra of the set of real numbers.... But I suppose that I start from the Borel measure.
– mathcounterexamples.net
Aug 17 at 10:58
I don’t see how Carathéodory theorem answers the question. Reading Wikipedia article Consequently, any measure on a space containing all intervals of real numbers can be extended to the Borel algebra of the set of real numbers.... But I suppose that I start from the Borel measure.
– mathcounterexamples.net
Aug 17 at 10:58
That is not Carathéodory's theorem, just an application of it. Read the general theorem, and think about it, please.
– A. Pongrácz
Aug 17 at 11:21
That is not Carathéodory's theorem, just an application of it. Read the general theorem, and think about it, please.
– A. Pongrácz
Aug 17 at 11:21
Carathéodory's extension theorem (named after the mathematician Constantin Carathéodory) states that any measure defined on a given ring R of subsets of a given set Ω can be extended to the Ã-algebra generated by R, and this extension is unique if the measure is Ã-finite. Here, Borel measure is already equipped with a $sigma$-algebra. Hence Carathéodory extension may just be the Borel $sigma$-algebra itself. Can you precise a complete extension of Borel measure that is different from Lebesgue measure?
– mathcounterexamples.net
Aug 17 at 11:43
Carathéodory's extension theorem (named after the mathematician Constantin Carathéodory) states that any measure defined on a given ring R of subsets of a given set Ω can be extended to the Ã-algebra generated by R, and this extension is unique if the measure is Ã-finite. Here, Borel measure is already equipped with a $sigma$-algebra. Hence Carathéodory extension may just be the Borel $sigma$-algebra itself. Can you precise a complete extension of Borel measure that is different from Lebesgue measure?
– mathcounterexamples.net
Aug 17 at 11:43
Look, I am not going to spoonfeed you the solution. You need to put your own effort into it. Hint: pick a set that is not Lebesgue-measurable, and consider the (semi)ring generated by that set and the family of Lebesgue measurable sets. Show that you can assign a value to the new set so that it extends as a measure to the ring. Then you can apply the theorem. I am not going to help you any more until you show your own work.
– A. Pongrácz
Aug 17 at 11:46
Look, I am not going to spoonfeed you the solution. You need to put your own effort into it. Hint: pick a set that is not Lebesgue-measurable, and consider the (semi)ring generated by that set and the family of Lebesgue measurable sets. Show that you can assign a value to the new set so that it extends as a measure to the ring. Then you can apply the theorem. I am not going to help you any more until you show your own work.
– A. Pongrácz
Aug 17 at 11:46
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