Mixing
The relation between $L^p$-norm and distribution function
$begingroup$
In my textbook, there is a theorem
Let $(X,mu)$ be a measurable theorem.
For $f$ in $L^p(X,mu )$, $0 < p < infty$, we have $$ ||f||_{L^p}^p = p intlimits_{0}^{infty} alpha^{p-1} d_f(alpha ) mathrm{d}alpha . $$
Where $||.||_{L^p}^p$ is the ordinary $L^p$ norm and $d_f$ is the ditribution function.
The first step, they did
$$ p intlimits_{0}^{infty} alpha^{p-1} d_f(alpha ) mathrm{d}alpha = p intlimits_{0}^{infty} alpha^{p-1} int_{X} chi_{[|f| > alpha]} mathrm{d} mu (x) mathrm{d} alpha$$
which is normal, as we just apply the definition of a distribution function.
But in the next step, they said they applied thẻ Fubini's theorem and obtain the following
$$int_X intlimits_0^{|f(x)|} p alpha ^{p-1} mathrm{d} alpha mathrm{d} mu (x) $$
which doesn't make sense to me.
As far as I know, the Fubini's theorem states that if $f$ is integrable on the product measure space then the order of integrating can be changed.
But how did it change the interval $[0;+infty )$ into $[0; |f(x)|]$ and not $[|f(x)|, +infty )$?
Please give me a clarification. Thank you.
measure-theory lebesgue-integral
asked Jan 9 at 21:06
ElementXElementX
404111
$endgroup$
add a comment |
$begingroup$
In my textbook, there is a theorem
Let $(X,mu)$ be a measurable theorem.
For $f$ in $L^p(X,mu )$, $0 < p < infty$, we have $$ ||f||_{L^p}^p = p intlimits_{0}^{infty} alpha^{p-1} d_f(alpha ) mathrm{d}alpha . $$
Where $||.||_{L^p}^p$ is the ordinary $L^p$ norm and $d_f$ is the ditribution function.
The first step, they did
$$ p intlimits_{0}^{infty} alpha^{p-1} d_f(alpha ) mathrm{d}alpha = p intlimits_{0}^{infty} alpha^{p-1} int_{X} chi_{[|f| > alpha]} mathrm{d} mu (x) mathrm{d} alpha$$
which is normal, as we just apply the definition of a distribution function.
But in the next step, they said they applied thẻ Fubini's theorem and obtain the following
$$int_X intlimits_0^{|f(x)|} p alpha ^{p-1} mathrm{d} alpha mathrm{d} mu (x) $$
which doesn't make sense to me.
As far as I know, the Fubini's theorem states that if $f$ is integrable on the product measure space then the order of integrating can be changed.
But how did it change the interval $[0;+infty )$ into $[0; |f(x)|]$ and not $[|f(x)|, +infty )$?
Please give me a clarification. Thank you.
measure-theory lebesgue-integral
asked Jan 9 at 21:06
ElementXElementX
404111
$endgroup$
2
$begingroup$
Notice that you have $ chi_{|f(x)| >alpha} $, which means when you switch the order of the integration that function gives you that $ alpha < |f(x)| $. I won't recommend switching the order as you would do it in a calculus course.
$endgroup$
– HarshCurious
Jan 9 at 21:16
add a comment |
$begingroup$
In my textbook, there is a theorem
Let $(X,mu)$ be a measurable theorem.
For $f$ in $L^p(X,mu )$, $0 < p < infty$, we have $$ ||f||_{L^p}^p = p intlimits_{0}^{infty} alpha^{p-1} d_f(alpha ) mathrm{d}alpha . $$
Where $||.||_{L^p}^p$ is the ordinary $L^p$ norm and $d_f$ is the ditribution function.
The first step, they did
$$ p intlimits_{0}^{infty} alpha^{p-1} d_f(alpha ) mathrm{d}alpha = p intlimits_{0}^{infty} alpha^{p-1} int_{X} chi_{[|f| > alpha]} mathrm{d} mu (x) mathrm{d} alpha$$
which is normal, as we just apply the definition of a distribution function.
But in the next step, they said they applied thẻ Fubini's theorem and obtain the following
$$int_X intlimits_0^{|f(x)|} p alpha ^{p-1} mathrm{d} alpha mathrm{d} mu (x) $$
which doesn't make sense to me.
As far as I know, the Fubini's theorem states that if $f$ is integrable on the product measure space then the order of integrating can be changed.
But how did it change the interval $[0;+infty )$ into $[0; |f(x)|]$ and not $[|f(x)|, +infty )$?
Please give me a clarification. Thank you.
measure-theory lebesgue-integral
asked Jan 9 at 21:06
ElementXElementX
404111
$endgroup$
In my textbook, there is a theorem
Let $(X,mu)$ be a measurable theorem.
For $f$ in $L^p(X,mu )$, $0 < p < infty$, we have $$ ||f||_{L^p}^p = p intlimits_{0}^{infty} alpha^{p-1} d_f(alpha ) mathrm{d}alpha . $$
Where $||.||_{L^p}^p$ is the ordinary $L^p$ norm and $d_f$ is the ditribution function.
The first step, they did
$$ p intlimits_{0}^{infty} alpha^{p-1} d_f(alpha ) mathrm{d}alpha = p intlimits_{0}^{infty} alpha^{p-1} int_{X} chi_{[|f| > alpha]} mathrm{d} mu (x) mathrm{d} alpha$$
which is normal, as we just apply the definition of a distribution function.
But in the next step, they said they applied thẻ Fubini's theorem and obtain the following
$$int_X intlimits_0^{|f(x)|} p alpha ^{p-1} mathrm{d} alpha mathrm{d} mu (x) $$
which doesn't make sense to me.
As far as I know, the Fubini's theorem states that if $f$ is integrable on the product measure space then the order of integrating can be changed.
But how did it change the interval $[0;+infty )$ into $[0; |f(x)|]$ and not $[|f(x)|, +infty )$?
Please give me a clarification. Thank you.
measure-theory lebesgue-integral
measure-theory lebesgue-integral
asked Jan 9 at 21:06
ElementXElementX
404111
asked Jan 9 at 21:06
ElementXElementX
404111
asked Jan 9 at 21:06
ElementXElementX
404111
asked Jan 9 at 21:06
ElementXElementX
404111
asked Jan 9 at 21:06
ElementXElementX
404111
404111
2
$begingroup$
Notice that you have $ chi_{|f(x)| >alpha} $, which means when you switch the order of the integration that function gives you that $ alpha < |f(x)| $. I won't recommend switching the order as you would do it in a calculus course.
$endgroup$
– HarshCurious
Jan 9 at 21:16
add a comment |
2
$begingroup$
Notice that you have $ chi_{|f(x)| >alpha} $, which means when you switch the order of the integration that function gives you that $ alpha < |f(x)| $. I won't recommend switching the order as you would do it in a calculus course.
$endgroup$
– HarshCurious
Jan 9 at 21:16
2
2
$begingroup$
Notice that you have $ chi_{|f(x)| >alpha} $, which means when you switch the order of the integration that function gives you that $ alpha < |f(x)| $. I won't recommend switching the order as you would do it in a calculus course.
$endgroup$
– HarshCurious
Jan 9 at 21:16
$begingroup$
Notice that you have $ chi_{|f(x)| >alpha} $, which means when you switch the order of the integration that function gives you that $ alpha < |f(x)| $. I won't recommend switching the order as you would do it in a calculus course.
$endgroup$
– HarshCurious
Jan 9 at 21:16
add a comment |
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$begingroup$
Notice that you have $ chi_{|f(x)| >alpha} $, which means when you switch the order of the integration that function gives you that $ alpha < |f(x)| $. I won't recommend switching the order as you would do it in a calculus course.
$endgroup$
– HarshCurious
Jan 9 at 21:16