Why $frac{dx}{|x|}$ is a Haar measure on $mathbb{R}setminus{0}$
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1
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This is in "A course in the abstract harmonic analysis by G.B. Follan on page 45"
Why $frac{dx}{|x|}$ is a Haar measure on multiplicative group $mathbb{R}setminus{0}$
I've started as following
$frac{dax}{|ax|}=frac{adx}{|ax|}$
but I couldn't figure out the result
Any help will be greatly appreciated
measure-theory haar-measure
asked yesterday
user62498
1,878613
add a comment |
up vote
1
down vote
favorite
This is in "A course in the abstract harmonic analysis by G.B. Follan on page 45"
Why $frac{dx}{|x|}$ is a Haar measure on multiplicative group $mathbb{R}setminus{0}$
I've started as following
$frac{dax}{|ax|}=frac{adx}{|ax|}$
but I couldn't figure out the result
Any help will be greatly appreciated
measure-theory haar-measure
asked yesterday
user62498
1,878613
be careful, we using a change of variables the multiplicative factor you obtain is not the determinant of the jacobian (in this case $a$) but its absolute value (in this case $|a|$). In others words $d(ax)=|a| dx$.
– Delta-u
21 hours ago
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
This is in "A course in the abstract harmonic analysis by G.B. Follan on page 45"
Why $frac{dx}{|x|}$ is a Haar measure on multiplicative group $mathbb{R}setminus{0}$
I've started as following
$frac{dax}{|ax|}=frac{adx}{|ax|}$
but I couldn't figure out the result
Any help will be greatly appreciated
measure-theory haar-measure
asked yesterday
user62498
1,878613
This is in "A course in the abstract harmonic analysis by G.B. Follan on page 45"
Why $frac{dx}{|x|}$ is a Haar measure on multiplicative group $mathbb{R}setminus{0}$
I've started as following
$frac{dax}{|ax|}=frac{adx}{|ax|}$
but I couldn't figure out the result
Any help will be greatly appreciated
measure-theory haar-measure
measure-theory haar-measure
asked yesterday
user62498
1,878613
asked yesterday
user62498
1,878613
asked yesterday
user62498
1,878613
asked yesterday
user62498
1,878613
asked yesterday
user62498
1,878613
1,878613
be careful, we using a change of variables the multiplicative factor you obtain is not the determinant of the jacobian (in this case $a$) but its absolute value (in this case $|a|$). In others words $d(ax)=|a| dx$.
– Delta-u
21 hours ago
add a comment |
be careful, we using a change of variables the multiplicative factor you obtain is not the determinant of the jacobian (in this case $a$) but its absolute value (in this case $|a|$). In others words $d(ax)=|a| dx$.
– Delta-u
21 hours ago
be careful, we using a change of variables the multiplicative factor you obtain is not the determinant of the jacobian (in this case $a$) but its absolute value (in this case $|a|$). In others words $d(ax)=|a| dx$.
– Delta-u
21 hours ago
be careful, we using a change of variables the multiplicative factor you obtain is not the determinant of the jacobian (in this case $a$) but its absolute value (in this case $|a|$). In others words $d(ax)=|a| dx$.
– Delta-u
21 hours ago
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
$int_{aE} frac 1 {|x|} , dx =int_{E} frac 1 {|x|} , dx$ for all Borel sets $E$ in $mathbb Rsetminus {0}$ and all $a neq 0$ (by an obvious change of variable). This is the definition of Haar measure.
answered yesterday
Kavi Rama Murthy
39.3k31748
@Dear Kavi Rama Murthy, I have a problem with $frac{1}{|x|}$
– user62498
yesterday
What exactly is the problem?
– Kavi Rama Murthy
yesterday
$frac{ads}{|a||x|}=frac{ds}{|x|}$
– user62498
yesterday
You can understand what is happening by taking a special case. Take $E=(1,2)$ and $a=-1$. See what $int_{aE} frac 1 {|x|} , dx$ is.
– Kavi Rama Murthy
yesterday
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
$int_{aE} frac 1 {|x|} , dx =int_{E} frac 1 {|x|} , dx$ for all Borel sets $E$ in $mathbb Rsetminus {0}$ and all $a neq 0$ (by an obvious change of variable). This is the definition of Haar measure.
answered yesterday
Kavi Rama Murthy
39.3k31748
@Dear Kavi Rama Murthy, I have a problem with $frac{1}{|x|}$
– user62498
yesterday
What exactly is the problem?
– Kavi Rama Murthy
yesterday
$frac{ads}{|a||x|}=frac{ds}{|x|}$
– user62498
yesterday
You can understand what is happening by taking a special case. Take $E=(1,2)$ and $a=-1$. See what $int_{aE} frac 1 {|x|} , dx$ is.
– Kavi Rama Murthy
yesterday
add a comment |
up vote
1
down vote
$int_{aE} frac 1 {|x|} , dx =int_{E} frac 1 {|x|} , dx$ for all Borel sets $E$ in $mathbb Rsetminus {0}$ and all $a neq 0$ (by an obvious change of variable). This is the definition of Haar measure.
answered yesterday
Kavi Rama Murthy
39.3k31748
@Dear Kavi Rama Murthy, I have a problem with $frac{1}{|x|}$
– user62498
yesterday
What exactly is the problem?
– Kavi Rama Murthy
yesterday
$frac{ads}{|a||x|}=frac{ds}{|x|}$
– user62498
yesterday
You can understand what is happening by taking a special case. Take $E=(1,2)$ and $a=-1$. See what $int_{aE} frac 1 {|x|} , dx$ is.
– Kavi Rama Murthy
yesterday
add a comment |
up vote
1
down vote
up vote
1
down vote
$int_{aE} frac 1 {|x|} , dx =int_{E} frac 1 {|x|} , dx$ for all Borel sets $E$ in $mathbb Rsetminus {0}$ and all $a neq 0$ (by an obvious change of variable). This is the definition of Haar measure.
answered yesterday
Kavi Rama Murthy
39.3k31748
$int_{aE} frac 1 {|x|} , dx =int_{E} frac 1 {|x|} , dx$ for all Borel sets $E$ in $mathbb Rsetminus {0}$ and all $a neq 0$ (by an obvious change of variable). This is the definition of Haar measure.
answered yesterday
Kavi Rama Murthy
39.3k31748
answered yesterday
Kavi Rama Murthy
39.3k31748
answered yesterday
Kavi Rama Murthy
39.3k31748
answered yesterday
Kavi Rama Murthy
39.3k31748
39.3k31748
@Dear Kavi Rama Murthy, I have a problem with $frac{1}{|x|}$
– user62498
yesterday
What exactly is the problem?
– Kavi Rama Murthy
yesterday
$frac{ads}{|a||x|}=frac{ds}{|x|}$
– user62498
yesterday
You can understand what is happening by taking a special case. Take $E=(1,2)$ and $a=-1$. See what $int_{aE} frac 1 {|x|} , dx$ is.
– Kavi Rama Murthy
yesterday
add a comment |
@Dear Kavi Rama Murthy, I have a problem with $frac{1}{|x|}$
– user62498
yesterday
What exactly is the problem?
– Kavi Rama Murthy
yesterday
$frac{ads}{|a||x|}=frac{ds}{|x|}$
– user62498
yesterday
You can understand what is happening by taking a special case. Take $E=(1,2)$ and $a=-1$. See what $int_{aE} frac 1 {|x|} , dx$ is.
– Kavi Rama Murthy
yesterday
@Dear Kavi Rama Murthy, I have a problem with $frac{1}{|x|}$
– user62498
yesterday
@Dear Kavi Rama Murthy, I have a problem with $frac{1}{|x|}$
– user62498
yesterday
What exactly is the problem?
– Kavi Rama Murthy
yesterday
What exactly is the problem?
– Kavi Rama Murthy
yesterday
$frac{ads}{|a||x|}=frac{ds}{|x|}$
– user62498
yesterday
$frac{ads}{|a||x|}=frac{ds}{|x|}$
– user62498
yesterday
You can understand what is happening by taking a special case. Take $E=(1,2)$ and $a=-1$. See what $int_{aE} frac 1 {|x|} , dx$ is.
– Kavi Rama Murthy
yesterday
You can understand what is happening by taking a special case. Take $E=(1,2)$ and $a=-1$. See what $int_{aE} frac 1 {|x|} , dx$ is.
– Kavi Rama Murthy
yesterday
add a comment |
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be careful, we using a change of variables the multiplicative factor you obtain is not the determinant of the jacobian (in this case $a$) but its absolute value (in this case $|a|$). In others words $d(ax)=|a| dx$.
– Delta-u
21 hours ago