Reference for: a Lebesgue integrable function and its rearrangements have the same Lebesgue integral.
In http://kunklet.people.cofc.edu/condint.pdf there's the sentence at the beginning of the Introduction:
"An elementary fact from measure theory states that a Lebesgue integrable function
and its rearrangements have the same Lebesgue integral."
Can someone please refer a book that proves this statement? Thanks.
measure-theory reference-request lebesgue-integral lebesgue-measure
asked Nov 24 at 18:01
anonymous
12810
|
show 4 more comments
In http://kunklet.people.cofc.edu/condint.pdf there's the sentence at the beginning of the Introduction:
"An elementary fact from measure theory states that a Lebesgue integrable function
and its rearrangements have the same Lebesgue integral."
Can someone please refer a book that proves this statement? Thanks.
measure-theory reference-request lebesgue-integral lebesgue-measure
asked Nov 24 at 18:01
anonymous
12810
I suspect that you could prove it yourself.
– herb steinberg
Nov 24 at 18:47
@herb steinberg : Haha, yes, but I need a book reference :D
– anonymous
Nov 24 at 19:31
Is Exercise 1.4.38 on page 103 of Terry Tao's book "An introduction to measure theory" at terrytao.files.wordpress.com/2012/12/… a more general result that has the statement "An elementary fact from measure theory states that a Lebesgue integrable function and its rearrangements have the same Lebesgue integral." as its specialized corollary?
– anonymous
Nov 25 at 11:15
@herb steinberg how would you prove the statement please?
– anonymous
Nov 29 at 14:47
It depends on how the Lebesgue integral was first developed. I first encountered it using the horizontal strip approach. In this approach, the horizontal strips would give the same value for the rearranged function.
– herb steinberg
Nov 29 at 22:54
|
show 4 more comments
In http://kunklet.people.cofc.edu/condint.pdf there's the sentence at the beginning of the Introduction:
"An elementary fact from measure theory states that a Lebesgue integrable function
and its rearrangements have the same Lebesgue integral."
Can someone please refer a book that proves this statement? Thanks.
measure-theory reference-request lebesgue-integral lebesgue-measure
asked Nov 24 at 18:01
anonymous
12810
In http://kunklet.people.cofc.edu/condint.pdf there's the sentence at the beginning of the Introduction:
"An elementary fact from measure theory states that a Lebesgue integrable function
and its rearrangements have the same Lebesgue integral."
Can someone please refer a book that proves this statement? Thanks.
measure-theory reference-request lebesgue-integral lebesgue-measure
measure-theory reference-request lebesgue-integral lebesgue-measure
asked Nov 24 at 18:01
anonymous
12810
asked Nov 24 at 18:01
anonymous
12810
asked Nov 24 at 18:01
anonymous
12810
asked Nov 24 at 18:01
anonymous
12810
asked Nov 24 at 18:01
anonymous
12810
12810
I suspect that you could prove it yourself.
– herb steinberg
Nov 24 at 18:47
@herb steinberg : Haha, yes, but I need a book reference :D
– anonymous
Nov 24 at 19:31
Is Exercise 1.4.38 on page 103 of Terry Tao's book "An introduction to measure theory" at terrytao.files.wordpress.com/2012/12/… a more general result that has the statement "An elementary fact from measure theory states that a Lebesgue integrable function and its rearrangements have the same Lebesgue integral." as its specialized corollary?
– anonymous
Nov 25 at 11:15
@herb steinberg how would you prove the statement please?
– anonymous
Nov 29 at 14:47
It depends on how the Lebesgue integral was first developed. I first encountered it using the horizontal strip approach. In this approach, the horizontal strips would give the same value for the rearranged function.
– herb steinberg
Nov 29 at 22:54
|
show 4 more comments
I suspect that you could prove it yourself.
– herb steinberg
Nov 24 at 18:47
@herb steinberg : Haha, yes, but I need a book reference :D
– anonymous
Nov 24 at 19:31
Is Exercise 1.4.38 on page 103 of Terry Tao's book "An introduction to measure theory" at terrytao.files.wordpress.com/2012/12/… a more general result that has the statement "An elementary fact from measure theory states that a Lebesgue integrable function and its rearrangements have the same Lebesgue integral." as its specialized corollary?
– anonymous
Nov 25 at 11:15
@herb steinberg how would you prove the statement please?
– anonymous
Nov 29 at 14:47
It depends on how the Lebesgue integral was first developed. I first encountered it using the horizontal strip approach. In this approach, the horizontal strips would give the same value for the rearranged function.
– herb steinberg
Nov 29 at 22:54
I suspect that you could prove it yourself.
– herb steinberg
Nov 24 at 18:47
I suspect that you could prove it yourself.
– herb steinberg
Nov 24 at 18:47
@herb steinberg : Haha, yes, but I need a book reference :D
– anonymous
Nov 24 at 19:31
@herb steinberg : Haha, yes, but I need a book reference :D
– anonymous
Nov 24 at 19:31
Is Exercise 1.4.38 on page 103 of Terry Tao's book "An introduction to measure theory" at terrytao.files.wordpress.com/2012/12/… a more general result that has the statement "An elementary fact from measure theory states that a Lebesgue integrable function and its rearrangements have the same Lebesgue integral." as its specialized corollary?
– anonymous
Nov 25 at 11:15
Is Exercise 1.4.38 on page 103 of Terry Tao's book "An introduction to measure theory" at terrytao.files.wordpress.com/2012/12/… a more general result that has the statement "An elementary fact from measure theory states that a Lebesgue integrable function and its rearrangements have the same Lebesgue integral." as its specialized corollary?
– anonymous
Nov 25 at 11:15
@herb steinberg how would you prove the statement please?
– anonymous
Nov 29 at 14:47
@herb steinberg how would you prove the statement please?
– anonymous
Nov 29 at 14:47
It depends on how the Lebesgue integral was first developed. I first encountered it using the horizontal strip approach. In this approach, the horizontal strips would give the same value for the rearranged function.
– herb steinberg
Nov 29 at 22:54
It depends on how the Lebesgue integral was first developed. I first encountered it using the horizontal strip approach. In this approach, the horizontal strips would give the same value for the rearranged function.
– herb steinberg
Nov 29 at 22:54
|
show 4 more comments
2 Answers
2
active
oldest
votes
Denote $f_#mu=mucirc f^{-1}$, namely the law of $f$ under the Lebesgue measure. Since $f$ and $g$ have the same distribution it follows that $f_#mu=g_#mu$ (uniqueness of measure). Moreover one has by the change of measure/variables formula
$$int_mathbb R f(x) , dmu(x) =int_mathbb R x, df_#mu(x) $$
Using $f_#mu=g_#mu$ we conclude that
$$int f, dmu=int g, dmu$$
Surely, you have to verify why the uniqueness of measure theorem applies.
answered Nov 24 at 18:49
Shashi
7,0361528
Yes, but I need a book reference. Do you happen to know of any book that has this proven, please?
– anonymous
Nov 24 at 19:35
@anonymous everything here is actually basic measure theory. What did we use? Change of measure formula and uniqueness of measure, both of them could be found by just googling.
– Shashi
Nov 24 at 19:39
I was expecting some respectable book about measure theory, especially about the rearrangements... Googling gives... lots of very varied results :D A renown book on it would help tremendously!
– anonymous
Nov 24 at 19:43
@anonymous I have put some links to the theorems I have used. To be honest, I don't have a book suggestion about rearrangements.
– Shashi
Nov 24 at 22:20
How would you verify why the uniqueness of measure theorem applies please?
– anonymous
Nov 29 at 14:46
|
show 2 more comments
Use the definition in the link. Let a strip be defined by $[a,b)$. Then from the definition of rearrangement the measure of the set $(x:ale f(x)lt b)$ is the same as the measure of the set $(x:ale g(x)lt b)$. Therefore the strip approach leads to the integrals being equal. In case it isn't obvious, use the fact that the sets $(x:f(x)lt b)-(x:f(x)lt a)=(:ale f(x)lt b)$ so the rearrangement definition applies to strips.
answered Nov 30 at 22:47
herb steinberg
2,4832310
Yes, it seems to be trivial... I can't believe it's this easy. I have to convince myself still... Will give it few days for thinking about it. Thanks, will chime in!
– anonymous
Nov 30 at 23:15
The problem I'm having with convincing myself is about the statement being very general, about all Lebesgue integrable functions, so what if $f(x)$ is not bounded and what if integration is over $xin [1,infty)$...
– anonymous
Nov 30 at 23:27
The domain of x is irrelevant, since we are dealing with measures of the function within strips. To avoid complications, omit the strip containing the x axis, which will shrink to $0$ width eventually. If $f(x)$ is not bounded, then the sums for the strip approximation will have an infinite number of terms, but will converge as long as the function is integrable in the first place. When you rearrange, the terms of the strip approximations for $f$ and $g$ are identical term by term as long as the summation order is determined by strip order.
– herb steinberg
Dec 1 at 1:57
add a comment |
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2 Answers
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Denote $f_#mu=mucirc f^{-1}$, namely the law of $f$ under the Lebesgue measure. Since $f$ and $g$ have the same distribution it follows that $f_#mu=g_#mu$ (uniqueness of measure). Moreover one has by the change of measure/variables formula
$$int_mathbb R f(x) , dmu(x) =int_mathbb R x, df_#mu(x) $$
Using $f_#mu=g_#mu$ we conclude that
$$int f, dmu=int g, dmu$$
Surely, you have to verify why the uniqueness of measure theorem applies.
answered Nov 24 at 18:49
Shashi
7,0361528
Yes, but I need a book reference. Do you happen to know of any book that has this proven, please?
– anonymous
Nov 24 at 19:35
@anonymous everything here is actually basic measure theory. What did we use? Change of measure formula and uniqueness of measure, both of them could be found by just googling.
– Shashi
Nov 24 at 19:39
I was expecting some respectable book about measure theory, especially about the rearrangements... Googling gives... lots of very varied results :D A renown book on it would help tremendously!
– anonymous
Nov 24 at 19:43
@anonymous I have put some links to the theorems I have used. To be honest, I don't have a book suggestion about rearrangements.
– Shashi
Nov 24 at 22:20
How would you verify why the uniqueness of measure theorem applies please?
– anonymous
Nov 29 at 14:46
|
show 2 more comments
Denote $f_#mu=mucirc f^{-1}$, namely the law of $f$ under the Lebesgue measure. Since $f$ and $g$ have the same distribution it follows that $f_#mu=g_#mu$ (uniqueness of measure). Moreover one has by the change of measure/variables formula
$$int_mathbb R f(x) , dmu(x) =int_mathbb R x, df_#mu(x) $$
Using $f_#mu=g_#mu$ we conclude that
$$int f, dmu=int g, dmu$$
Surely, you have to verify why the uniqueness of measure theorem applies.
answered Nov 24 at 18:49
Shashi
7,0361528
Yes, but I need a book reference. Do you happen to know of any book that has this proven, please?
– anonymous
Nov 24 at 19:35
@anonymous everything here is actually basic measure theory. What did we use? Change of measure formula and uniqueness of measure, both of them could be found by just googling.
– Shashi
Nov 24 at 19:39
I was expecting some respectable book about measure theory, especially about the rearrangements... Googling gives... lots of very varied results :D A renown book on it would help tremendously!
– anonymous
Nov 24 at 19:43
@anonymous I have put some links to the theorems I have used. To be honest, I don't have a book suggestion about rearrangements.
– Shashi
Nov 24 at 22:20
How would you verify why the uniqueness of measure theorem applies please?
– anonymous
Nov 29 at 14:46
|
show 2 more comments
Denote $f_#mu=mucirc f^{-1}$, namely the law of $f$ under the Lebesgue measure. Since $f$ and $g$ have the same distribution it follows that $f_#mu=g_#mu$ (uniqueness of measure). Moreover one has by the change of measure/variables formula
$$int_mathbb R f(x) , dmu(x) =int_mathbb R x, df_#mu(x) $$
Using $f_#mu=g_#mu$ we conclude that
$$int f, dmu=int g, dmu$$
Surely, you have to verify why the uniqueness of measure theorem applies.
answered Nov 24 at 18:49
Shashi
7,0361528
Denote $f_#mu=mucirc f^{-1}$, namely the law of $f$ under the Lebesgue measure. Since $f$ and $g$ have the same distribution it follows that $f_#mu=g_#mu$ (uniqueness of measure). Moreover one has by the change of measure/variables formula
$$int_mathbb R f(x) , dmu(x) =int_mathbb R x, df_#mu(x) $$
Using $f_#mu=g_#mu$ we conclude that
$$int f, dmu=int g, dmu$$
Surely, you have to verify why the uniqueness of measure theorem applies.
answered Nov 24 at 18:49
Shashi
7,0361528
edited Nov 24 at 22:15
answered Nov 24 at 18:49
Shashi
7,0361528
answered Nov 24 at 18:49
Shashi
7,0361528
answered Nov 24 at 18:49
Shashi
7,0361528
7,0361528
Yes, but I need a book reference. Do you happen to know of any book that has this proven, please?
– anonymous
Nov 24 at 19:35
@anonymous everything here is actually basic measure theory. What did we use? Change of measure formula and uniqueness of measure, both of them could be found by just googling.
– Shashi
Nov 24 at 19:39
I was expecting some respectable book about measure theory, especially about the rearrangements... Googling gives... lots of very varied results :D A renown book on it would help tremendously!
– anonymous
Nov 24 at 19:43
@anonymous I have put some links to the theorems I have used. To be honest, I don't have a book suggestion about rearrangements.
– Shashi
Nov 24 at 22:20
How would you verify why the uniqueness of measure theorem applies please?
– anonymous
Nov 29 at 14:46
|
show 2 more comments
Yes, but I need a book reference. Do you happen to know of any book that has this proven, please?
– anonymous
Nov 24 at 19:35
@anonymous everything here is actually basic measure theory. What did we use? Change of measure formula and uniqueness of measure, both of them could be found by just googling.
– Shashi
Nov 24 at 19:39
I was expecting some respectable book about measure theory, especially about the rearrangements... Googling gives... lots of very varied results :D A renown book on it would help tremendously!
– anonymous
Nov 24 at 19:43
@anonymous I have put some links to the theorems I have used. To be honest, I don't have a book suggestion about rearrangements.
– Shashi
Nov 24 at 22:20
How would you verify why the uniqueness of measure theorem applies please?
– anonymous
Nov 29 at 14:46
Yes, but I need a book reference. Do you happen to know of any book that has this proven, please?
– anonymous
Nov 24 at 19:35
Yes, but I need a book reference. Do you happen to know of any book that has this proven, please?
– anonymous
Nov 24 at 19:35
@anonymous everything here is actually basic measure theory. What did we use? Change of measure formula and uniqueness of measure, both of them could be found by just googling.
– Shashi
Nov 24 at 19:39
@anonymous everything here is actually basic measure theory. What did we use? Change of measure formula and uniqueness of measure, both of them could be found by just googling.
– Shashi
Nov 24 at 19:39
I was expecting some respectable book about measure theory, especially about the rearrangements... Googling gives... lots of very varied results :D A renown book on it would help tremendously!
– anonymous
Nov 24 at 19:43
I was expecting some respectable book about measure theory, especially about the rearrangements... Googling gives... lots of very varied results :D A renown book on it would help tremendously!
– anonymous
Nov 24 at 19:43
@anonymous I have put some links to the theorems I have used. To be honest, I don't have a book suggestion about rearrangements.
– Shashi
Nov 24 at 22:20
@anonymous I have put some links to the theorems I have used. To be honest, I don't have a book suggestion about rearrangements.
– Shashi
Nov 24 at 22:20
How would you verify why the uniqueness of measure theorem applies please?
– anonymous
Nov 29 at 14:46
How would you verify why the uniqueness of measure theorem applies please?
– anonymous
Nov 29 at 14:46
|
show 2 more comments
Use the definition in the link. Let a strip be defined by $[a,b)$. Then from the definition of rearrangement the measure of the set $(x:ale f(x)lt b)$ is the same as the measure of the set $(x:ale g(x)lt b)$. Therefore the strip approach leads to the integrals being equal. In case it isn't obvious, use the fact that the sets $(x:f(x)lt b)-(x:f(x)lt a)=(:ale f(x)lt b)$ so the rearrangement definition applies to strips.
answered Nov 30 at 22:47
herb steinberg
2,4832310
Yes, it seems to be trivial... I can't believe it's this easy. I have to convince myself still... Will give it few days for thinking about it. Thanks, will chime in!
– anonymous
Nov 30 at 23:15
The problem I'm having with convincing myself is about the statement being very general, about all Lebesgue integrable functions, so what if $f(x)$ is not bounded and what if integration is over $xin [1,infty)$...
– anonymous
Nov 30 at 23:27
The domain of x is irrelevant, since we are dealing with measures of the function within strips. To avoid complications, omit the strip containing the x axis, which will shrink to $0$ width eventually. If $f(x)$ is not bounded, then the sums for the strip approximation will have an infinite number of terms, but will converge as long as the function is integrable in the first place. When you rearrange, the terms of the strip approximations for $f$ and $g$ are identical term by term as long as the summation order is determined by strip order.
– herb steinberg
Dec 1 at 1:57
add a comment |
Use the definition in the link. Let a strip be defined by $[a,b)$. Then from the definition of rearrangement the measure of the set $(x:ale f(x)lt b)$ is the same as the measure of the set $(x:ale g(x)lt b)$. Therefore the strip approach leads to the integrals being equal. In case it isn't obvious, use the fact that the sets $(x:f(x)lt b)-(x:f(x)lt a)=(:ale f(x)lt b)$ so the rearrangement definition applies to strips.
answered Nov 30 at 22:47
herb steinberg
2,4832310
Yes, it seems to be trivial... I can't believe it's this easy. I have to convince myself still... Will give it few days for thinking about it. Thanks, will chime in!
– anonymous
Nov 30 at 23:15
The problem I'm having with convincing myself is about the statement being very general, about all Lebesgue integrable functions, so what if $f(x)$ is not bounded and what if integration is over $xin [1,infty)$...
– anonymous
Nov 30 at 23:27
The domain of x is irrelevant, since we are dealing with measures of the function within strips. To avoid complications, omit the strip containing the x axis, which will shrink to $0$ width eventually. If $f(x)$ is not bounded, then the sums for the strip approximation will have an infinite number of terms, but will converge as long as the function is integrable in the first place. When you rearrange, the terms of the strip approximations for $f$ and $g$ are identical term by term as long as the summation order is determined by strip order.
– herb steinberg
Dec 1 at 1:57
add a comment |
Use the definition in the link. Let a strip be defined by $[a,b)$. Then from the definition of rearrangement the measure of the set $(x:ale f(x)lt b)$ is the same as the measure of the set $(x:ale g(x)lt b)$. Therefore the strip approach leads to the integrals being equal. In case it isn't obvious, use the fact that the sets $(x:f(x)lt b)-(x:f(x)lt a)=(:ale f(x)lt b)$ so the rearrangement definition applies to strips.
answered Nov 30 at 22:47
herb steinberg
2,4832310
Use the definition in the link. Let a strip be defined by $[a,b)$. Then from the definition of rearrangement the measure of the set $(x:ale f(x)lt b)$ is the same as the measure of the set $(x:ale g(x)lt b)$. Therefore the strip approach leads to the integrals being equal. In case it isn't obvious, use the fact that the sets $(x:f(x)lt b)-(x:f(x)lt a)=(:ale f(x)lt b)$ so the rearrangement definition applies to strips.
answered Nov 30 at 22:47
herb steinberg
2,4832310
answered Nov 30 at 22:47
herb steinberg
2,4832310
answered Nov 30 at 22:47
herb steinberg
2,4832310
answered Nov 30 at 22:47
herb steinberg
2,4832310
2,4832310
Yes, it seems to be trivial... I can't believe it's this easy. I have to convince myself still... Will give it few days for thinking about it. Thanks, will chime in!
– anonymous
Nov 30 at 23:15
The problem I'm having with convincing myself is about the statement being very general, about all Lebesgue integrable functions, so what if $f(x)$ is not bounded and what if integration is over $xin [1,infty)$...
– anonymous
Nov 30 at 23:27
The domain of x is irrelevant, since we are dealing with measures of the function within strips. To avoid complications, omit the strip containing the x axis, which will shrink to $0$ width eventually. If $f(x)$ is not bounded, then the sums for the strip approximation will have an infinite number of terms, but will converge as long as the function is integrable in the first place. When you rearrange, the terms of the strip approximations for $f$ and $g$ are identical term by term as long as the summation order is determined by strip order.
– herb steinberg
Dec 1 at 1:57
add a comment |
Yes, it seems to be trivial... I can't believe it's this easy. I have to convince myself still... Will give it few days for thinking about it. Thanks, will chime in!
– anonymous
Nov 30 at 23:15
The problem I'm having with convincing myself is about the statement being very general, about all Lebesgue integrable functions, so what if $f(x)$ is not bounded and what if integration is over $xin [1,infty)$...
– anonymous
Nov 30 at 23:27
The domain of x is irrelevant, since we are dealing with measures of the function within strips. To avoid complications, omit the strip containing the x axis, which will shrink to $0$ width eventually. If $f(x)$ is not bounded, then the sums for the strip approximation will have an infinite number of terms, but will converge as long as the function is integrable in the first place. When you rearrange, the terms of the strip approximations for $f$ and $g$ are identical term by term as long as the summation order is determined by strip order.
– herb steinberg
Dec 1 at 1:57
Yes, it seems to be trivial... I can't believe it's this easy. I have to convince myself still... Will give it few days for thinking about it. Thanks, will chime in!
– anonymous
Nov 30 at 23:15
Yes, it seems to be trivial... I can't believe it's this easy. I have to convince myself still... Will give it few days for thinking about it. Thanks, will chime in!
– anonymous
Nov 30 at 23:15
The problem I'm having with convincing myself is about the statement being very general, about all Lebesgue integrable functions, so what if $f(x)$ is not bounded and what if integration is over $xin [1,infty)$...
– anonymous
Nov 30 at 23:27
The problem I'm having with convincing myself is about the statement being very general, about all Lebesgue integrable functions, so what if $f(x)$ is not bounded and what if integration is over $xin [1,infty)$...
– anonymous
Nov 30 at 23:27
The domain of x is irrelevant, since we are dealing with measures of the function within strips. To avoid complications, omit the strip containing the x axis, which will shrink to $0$ width eventually. If $f(x)$ is not bounded, then the sums for the strip approximation will have an infinite number of terms, but will converge as long as the function is integrable in the first place. When you rearrange, the terms of the strip approximations for $f$ and $g$ are identical term by term as long as the summation order is determined by strip order.
– herb steinberg
Dec 1 at 1:57
The domain of x is irrelevant, since we are dealing with measures of the function within strips. To avoid complications, omit the strip containing the x axis, which will shrink to $0$ width eventually. If $f(x)$ is not bounded, then the sums for the strip approximation will have an infinite number of terms, but will converge as long as the function is integrable in the first place. When you rearrange, the terms of the strip approximations for $f$ and $g$ are identical term by term as long as the summation order is determined by strip order.
– herb steinberg
Dec 1 at 1:57
add a comment |
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I suspect that you could prove it yourself.
– herb steinberg
Nov 24 at 18:47
@herb steinberg : Haha, yes, but I need a book reference :D
– anonymous
Nov 24 at 19:31
Is Exercise 1.4.38 on page 103 of Terry Tao's book "An introduction to measure theory" at terrytao.files.wordpress.com/2012/12/… a more general result that has the statement "An elementary fact from measure theory states that a Lebesgue integrable function and its rearrangements have the same Lebesgue integral." as its specialized corollary?
– anonymous
Nov 25 at 11:15
@herb steinberg how would you prove the statement please?
– anonymous
Nov 29 at 14:47
It depends on how the Lebesgue integral was first developed. I first encountered it using the horizontal strip approach. In this approach, the horizontal strips would give the same value for the rearranged function.
– herb steinberg
Nov 29 at 22:54