Why $frac{dx}{|x|}$ is a Haar measure on $mathbb{R}setminus{0}$
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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 2 days ago
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 2 days ago
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
2 days 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 2 days ago
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 2 days ago
user62498
1,878613
asked 2 days ago
user62498
1,878613
asked 2 days ago
user62498
1,878613
asked 2 days ago
user62498
1,878613
asked 2 days ago
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
2 days 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
2 days 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
2 days 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
2 days 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 2 days ago
Kavi Rama Murthy
39.3k31748
@Dear Kavi Rama Murthy, I have a problem with $frac{1}{|x|}$
– user62498
2 days ago
What exactly is the problem?
– Kavi Rama Murthy
2 days ago
$frac{ads}{|a||x|}=frac{ds}{|x|}$
– user62498
2 days ago
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
2 days ago
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 2 days ago
Kavi Rama Murthy
39.3k31748
@Dear Kavi Rama Murthy, I have a problem with $frac{1}{|x|}$
– user62498
2 days ago
What exactly is the problem?
– Kavi Rama Murthy
2 days ago
$frac{ads}{|a||x|}=frac{ds}{|x|}$
– user62498
2 days ago
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
2 days ago
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 2 days ago
Kavi Rama Murthy
39.3k31748
@Dear Kavi Rama Murthy, I have a problem with $frac{1}{|x|}$
– user62498
2 days ago
What exactly is the problem?
– Kavi Rama Murthy
2 days ago
$frac{ads}{|a||x|}=frac{ds}{|x|}$
– user62498
2 days ago
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
2 days ago
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 2 days ago
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 2 days ago
Kavi Rama Murthy
39.3k31748
answered 2 days ago
Kavi Rama Murthy
39.3k31748
answered 2 days ago
Kavi Rama Murthy
39.3k31748
answered 2 days ago
Kavi Rama Murthy
39.3k31748
39.3k31748
@Dear Kavi Rama Murthy, I have a problem with $frac{1}{|x|}$
– user62498
2 days ago
What exactly is the problem?
– Kavi Rama Murthy
2 days ago
$frac{ads}{|a||x|}=frac{ds}{|x|}$
– user62498
2 days ago
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
2 days ago
add a comment |
@Dear Kavi Rama Murthy, I have a problem with $frac{1}{|x|}$
– user62498
2 days ago
What exactly is the problem?
– Kavi Rama Murthy
2 days ago
$frac{ads}{|a||x|}=frac{ds}{|x|}$
– user62498
2 days ago
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
2 days ago
@Dear Kavi Rama Murthy, I have a problem with $frac{1}{|x|}$
– user62498
2 days ago
@Dear Kavi Rama Murthy, I have a problem with $frac{1}{|x|}$
– user62498
2 days ago
What exactly is the problem?
– Kavi Rama Murthy
2 days ago
What exactly is the problem?
– Kavi Rama Murthy
2 days ago
$frac{ads}{|a||x|}=frac{ds}{|x|}$
– user62498
2 days ago
$frac{ads}{|a||x|}=frac{ds}{|x|}$
– user62498
2 days ago
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
2 days ago
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
2 days ago
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
2 days ago