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0

$begingroup$

Let $pi: (X,mathcalM,nu) to (Y,mathcalN,eta)$ be a measurable map i.e. $pi^-1(E) in mathcalM$ for all $E in mathcalN$. I want to define a map from $L^infty(Y,eta)$ to $L^infty(X,nu)$. The natural thing to do is the following:

Define a map $tildepi:L^infty(Y,eta) to L^infty(X,nu)$ by $$tildepi(f)=fcircpi,f in L^infty(Y,eta)$$
Recall that
$$|f|_infty=infleft{alpha ge 0: muleft(f(x)$$

Claim-1: $tildepi(f)$ is measurable.

Proof: Let's assume that $tildepi(f)$ is real-valued, without any loss of generality. Observe that $tildepi(f)^-1([a,infty))=pi^-1left(f^-1([a,infty)right)$. Since $f$ is a measurable, $f^-1([a,infty)) in mathcalN$. Since $pi$ is a measurable map, $$tildepi(f)^-1left([a,infty)right)=pi^-1left(f^-1([a,infty))right) in mathcalM$$

Claim-2: $tildepi(f) in L^infty(X,nu)$

Proof: Observe that for a.e $x in X$ , we have $|f(pi(x))| le |f|_infty < infty$.

At this point of time, I need one assumption that $pi_*(nu)=eta$. I don't see how else to move forward without this assumption.

Let $E_alpha=f(pi(x))$. Let $F_alpha=f(y)$. Then, $E_alphasubseteqpi^-1left(F_alpharight)$. For $alpha > |f|_infty$, observe that $eta(F_alpha)=0$. Since $pi^-1(F_alpha) in mathcalN$, we have that $nu(pi^-1(F_alpha))=0$ and hence $nu(E_alpha)=0$. Thus, $|tildepi(f)|_infty le |f|_infty< infty$.

Please let me know if there is something wrong with this proof. Thanks for the help!!

analysis measure-theory lp-spaces measurable-functions

share|cite|improve this question

asked 2 days ago

tattwamasi amrutamtattwamasi amrutam

8,25721643

$endgroup$

add a comment | 

0

$begingroup$

Let $pi: (X,mathcalM,nu) to (Y,mathcalN,eta)$ be a measurable map i.e. $pi^-1(E) in mathcalM$ for all $E in mathcalN$. I want to define a map from $L^infty(Y,eta)$ to $L^infty(X,nu)$. The natural thing to do is the following:

Define a map $tildepi:L^infty(Y,eta) to L^infty(X,nu)$ by $$tildepi(f)=fcircpi,f in L^infty(Y,eta)$$
Recall that
$$|f|_infty=infleft{alpha ge 0: muleft(f(x)$$

Claim-1: $tildepi(f)$ is measurable.

Proof: Let's assume that $tildepi(f)$ is real-valued, without any loss of generality. Observe that $tildepi(f)^-1([a,infty))=pi^-1left(f^-1([a,infty)right)$. Since $f$ is a measurable, $f^-1([a,infty)) in mathcalN$. Since $pi$ is a measurable map, $$tildepi(f)^-1left([a,infty)right)=pi^-1left(f^-1([a,infty))right) in mathcalM$$

Claim-2: $tildepi(f) in L^infty(X,nu)$

Proof: Observe that for a.e $x in X$ , we have $|f(pi(x))| le |f|_infty < infty$.

At this point of time, I need one assumption that $pi_*(nu)=eta$. I don't see how else to move forward without this assumption.

Let $E_alpha=f(pi(x))$. Let $F_alpha=f(y)$. Then, $E_alphasubseteqpi^-1left(F_alpharight)$. For $alpha > |f|_infty$, observe that $eta(F_alpha)=0$. Since $pi^-1(F_alpha) in mathcalN$, we have that $nu(pi^-1(F_alpha))=0$ and hence $nu(E_alpha)=0$. Thus, $|tildepi(f)|_infty le |f|_infty< infty$.

Please let me know if there is something wrong with this proof. Thanks for the help!!

analysis measure-theory lp-spaces measurable-functions

share|cite|improve this question

asked 2 days ago

tattwamasi amrutamtattwamasi amrutam

8,25721643

$endgroup$

add a comment | 

$begingroup$

Let $pi: (X,mathcalM,nu) to (Y,mathcalN,eta)$ be a measurable map i.e. $pi^-1(E) in mathcalM$ for all $E in mathcalN$. I want to define a map from $L^infty(Y,eta)$ to $L^infty(X,nu)$. The natural thing to do is the following:

Define a map $tildepi:L^infty(Y,eta) to L^infty(X,nu)$ by $$tildepi(f)=fcircpi,f in L^infty(Y,eta)$$
Recall that
$$|f|_infty=infleft{alpha ge 0: muleft(f(x)$$

Claim-1: $tildepi(f)$ is measurable.

Proof: Let's assume that $tildepi(f)$ is real-valued, without any loss of generality. Observe that $tildepi(f)^-1([a,infty))=pi^-1left(f^-1([a,infty)right)$. Since $f$ is a measurable, $f^-1([a,infty)) in mathcalN$. Since $pi$ is a measurable map, $$tildepi(f)^-1left([a,infty)right)=pi^-1left(f^-1([a,infty))right) in mathcalM$$

Claim-2: $tildepi(f) in L^infty(X,nu)$

Proof: Observe that for a.e $x in X$ , we have $|f(pi(x))| le |f|_infty < infty$.

At this point of time, I need one assumption that $pi_*(nu)=eta$. I don't see how else to move forward without this assumption.

Let $E_alpha=f(pi(x))$. Let $F_alpha=f(y)$. Then, $E_alphasubseteqpi^-1left(F_alpharight)$. For $alpha > |f|_infty$, observe that $eta(F_alpha)=0$. Since $pi^-1(F_alpha) in mathcalN$, we have that $nu(pi^-1(F_alpha))=0$ and hence $nu(E_alpha)=0$. Thus, $|tildepi(f)|_infty le |f|_infty< infty$.

Please let me know if there is something wrong with this proof. Thanks for the help!!

analysis measure-theory lp-spaces measurable-functions

share|cite|improve this question

asked 2 days ago

tattwamasi amrutamtattwamasi amrutam

8,25721643

$endgroup$

share|cite|improve this question

asked 2 days ago

tattwamasi amrutamtattwamasi amrutam

8,25721643

share|cite|improve this question

asked 2 days ago

tattwamasi amrutamtattwamasi amrutam

8,25721643

asked 2 days ago

tattwamasi amrutamtattwamasi amrutam

8,25721643

asked 2 days ago

tattwamasi amrutamtattwamasi amrutam

8,25721643

1 Answer
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The function $tilde pi $ is not well defined function unless you make an extra assumption on $pi$. You need the condition $nu (pi ^-1(E))=0$ whenever $E in mathcal N$ and $eta(E)=0$. You don't need the stronger assumption that $eta =nu circ pi^-1$.

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

Kavi Rama MurthyKavi Rama Murthy

66.3k52867

$endgroup$

add a comment | 

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1

$begingroup$

The function $tilde pi $ is not well defined function unless you make an extra assumption on $pi$. You need the condition $nu (pi ^-1(E))=0$ whenever $E in mathcal N$ and $eta(E)=0$. You don't need the stronger assumption that $eta =nu circ pi^-1$.

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

Kavi Rama MurthyKavi Rama Murthy

66.3k52867

$endgroup$

add a comment | 

1

$begingroup$

The function $tilde pi $ is not well defined function unless you make an extra assumption on $pi$. You need the condition $nu (pi ^-1(E))=0$ whenever $E in mathcal N$ and $eta(E)=0$. You don't need the stronger assumption that $eta =nu circ pi^-1$.

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

Kavi Rama MurthyKavi Rama Murthy

66.3k52867

$endgroup$

add a comment | 

$begingroup$

The function $tilde pi $ is not well defined function unless you make an extra assumption on $pi$. You need the condition $nu (pi ^-1(E))=0$ whenever $E in mathcal N$ and $eta(E)=0$. You don't need the stronger assumption that $eta =nu circ pi^-1$.

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

Kavi Rama MurthyKavi Rama Murthy

66.3k52867

$endgroup$

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

Kavi Rama MurthyKavi Rama Murthy

66.3k52867

answered 2 days ago

Kavi Rama MurthyKavi Rama Murthy

66.3k52867

answered 2 days ago

Kavi Rama MurthyKavi Rama Murthy

66.3k52867