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







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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















up vote
1
down vote

favorite
1












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







share|cite|improve this question






















  • 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













up vote
1
down vote

favorite
1









up vote
1
down vote

favorite
1






1





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







share|cite|improve this question













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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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


















  • 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










1 Answer
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$int_{aE} frac 1 {|x|} , dx =int_{E} frac 1 {|x|} , dx$ 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.






share|cite|improve this answer





















  • @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











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1 Answer
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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.






share|cite|improve this answer





















  • @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















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.






share|cite|improve this answer





















  • @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













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.






share|cite|improve this answer












$int_{aE} frac 1 {|x|} , dx =int_{E} frac 1 {|x|} , dx$ 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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • @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


















 

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