Why the product of measure is a tensor product?
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Let $mu$ and $nu$ two measure. What is the justification to say that the measure product is a tensor product ? i.e. why $mutimes nu$ is written $muotimes nu$ ? (Of course, beside the fact that is by definition or it's just a notation). Our teacher told us that the reason that we write it as a tensor product it's because it behave as a tensor product). But I don't understand this (I'm not very confortable with tensor product).
measure-theory
asked Nov 13 at 11:12
lovemath
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Let $mu$ and $nu$ two measure. What is the justification to say that the measure product is a tensor product ? i.e. why $mutimes nu$ is written $muotimes nu$ ? (Of course, beside the fact that is by definition or it's just a notation). Our teacher told us that the reason that we write it as a tensor product it's because it behave as a tensor product). But I don't understand this (I'm not very confortable with tensor product).
measure-theory
asked Nov 13 at 11:12
lovemath
506
New contributor
lovemath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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For example, assuming $sigma$-finiteness, $L^1(muotimesnu)=L^1(mu)otimes L^1(nu)$ as Banach spaces.
– user10354138
Nov 13 at 13:58
add a comment |
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up vote
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down vote
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Let $mu$ and $nu$ two measure. What is the justification to say that the measure product is a tensor product ? i.e. why $mutimes nu$ is written $muotimes nu$ ? (Of course, beside the fact that is by definition or it's just a notation). Our teacher told us that the reason that we write it as a tensor product it's because it behave as a tensor product). But I don't understand this (I'm not very confortable with tensor product).
measure-theory
asked Nov 13 at 11:12
lovemath
506
New contributor
lovemath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Let $mu$ and $nu$ two measure. What is the justification to say that the measure product is a tensor product ? i.e. why $mutimes nu$ is written $muotimes nu$ ? (Of course, beside the fact that is by definition or it's just a notation). Our teacher told us that the reason that we write it as a tensor product it's because it behave as a tensor product). But I don't understand this (I'm not very confortable with tensor product).
measure-theory
measure-theory
asked Nov 13 at 11:12
lovemath
506
New contributor
lovemath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 13 at 11:12
lovemath
506
New contributor
lovemath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 13 at 11:12
lovemath
506
New contributor
lovemath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 13 at 11:12
lovemath
506
asked Nov 13 at 11:12
lovemath
506
506
New contributor
lovemath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
lovemath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
lovemath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
For example, assuming $sigma$-finiteness, $L^1(muotimesnu)=L^1(mu)otimes L^1(nu)$ as Banach spaces.
– user10354138
Nov 13 at 13:58
add a comment |
For example, assuming $sigma$-finiteness, $L^1(muotimesnu)=L^1(mu)otimes L^1(nu)$ as Banach spaces.
– user10354138
Nov 13 at 13:58
For example, assuming $sigma$-finiteness, $L^1(muotimesnu)=L^1(mu)otimes L^1(nu)$ as Banach spaces.
– user10354138
Nov 13 at 13:58
For example, assuming $sigma$-finiteness, $L^1(muotimesnu)=L^1(mu)otimes L^1(nu)$ as Banach spaces.
– user10354138
Nov 13 at 13:58
add a comment |
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lovemath is a new contributor. Be nice, and check out our Code of Conduct.
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For example, assuming $sigma$-finiteness, $L^1(muotimesnu)=L^1(mu)otimes L^1(nu)$ as Banach spaces.
– user10354138
Nov 13 at 13:58