Why is my proof that $L^1(nu) = L^1(|nu|)$ for signed measure $nu$ incorrect?
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The following exercise is in Folland:
Let $nu$ be a signed measure on $(X, mathcal{M})$ show that $L^1(nu) = L^1(|nu|)$.
My attempt:
Consider $f in L^1(nu)$ then we know that
$$int |f| dnu = int |f| dnu^+ - int |f| dnu^- < infty$$
This can only happen iff $int |f| dnu^+ < infty$ and $int |f| dnu^- < infty$. Thus, we see that
$$int |f| d|nu| = int |f| dnu^+ + int |f| dnu^- < infty$$
So we must have $f in L^1(|nu|)$. Since all these steps were two ways, we have also that $f in L^1(nu)$.
This "proof" seems too simple and others seem to be giving me the same idea that this is not correct. Where is the flaw and how can I fix it?
real-analysis analysis measure-theory
asked Nov 18 at 2:08
carsandpulsars
35117
|
show 5 more comments
up vote
2
down vote
favorite
The following exercise is in Folland:
Let $nu$ be a signed measure on $(X, mathcal{M})$ show that $L^1(nu) = L^1(|nu|)$.
My attempt:
Consider $f in L^1(nu)$ then we know that
$$int |f| dnu = int |f| dnu^+ - int |f| dnu^- < infty$$
This can only happen iff $int |f| dnu^+ < infty$ and $int |f| dnu^- < infty$. Thus, we see that
$$int |f| d|nu| = int |f| dnu^+ + int |f| dnu^- < infty$$
So we must have $f in L^1(|nu|)$. Since all these steps were two ways, we have also that $f in L^1(nu)$.
This "proof" seems too simple and others seem to be giving me the same idea that this is not correct. Where is the flaw and how can I fix it?
real-analysis analysis measure-theory
asked Nov 18 at 2:08
carsandpulsars
35117
What's the exact definition of $L^1(nu)$ for a signed measure $nu$?
– Nate Eldredge
Nov 18 at 2:19
Remembering that $L^1$ is really a set of equivalence classes, presumably part of what's being asserted is that the equivalence classes are the same, so you would have to show that $f = g$ $nu$-a.e. (however that's defined) iff $f = g$ $|nu|$-a.e.
– Nate Eldredge
Nov 18 at 2:21
@NateEldredge In Folland it has been defined as $L^1(nu) = L^1(nu^+) cap L^1(nu^-)$ where $nu = nu^+ - nu^-$ is the Jordan decomposition of $nu$.
– carsandpulsars
Nov 18 at 2:22
Okay, so then your statement that $int |f| dnu^+ < infty$ and $int |f| dnu^- < infty$ is already the definition. The line about $int f dnu^+ - int f dnu^-$ is not the definition and would have to be justified.
– Nate Eldredge
Nov 18 at 2:25
But there is still another statement to be shown: that $f = g$ $|nu|$-a.e. iff $f = g$ $nu^+$-a.e. and $f=g$ $nu^-$-a.e.
– Nate Eldredge
Nov 18 at 2:25
|
show 5 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
The following exercise is in Folland:
Let $nu$ be a signed measure on $(X, mathcal{M})$ show that $L^1(nu) = L^1(|nu|)$.
My attempt:
Consider $f in L^1(nu)$ then we know that
$$int |f| dnu = int |f| dnu^+ - int |f| dnu^- < infty$$
This can only happen iff $int |f| dnu^+ < infty$ and $int |f| dnu^- < infty$. Thus, we see that
$$int |f| d|nu| = int |f| dnu^+ + int |f| dnu^- < infty$$
So we must have $f in L^1(|nu|)$. Since all these steps were two ways, we have also that $f in L^1(nu)$.
This "proof" seems too simple and others seem to be giving me the same idea that this is not correct. Where is the flaw and how can I fix it?
real-analysis analysis measure-theory
asked Nov 18 at 2:08
carsandpulsars
35117
The following exercise is in Folland:
Let $nu$ be a signed measure on $(X, mathcal{M})$ show that $L^1(nu) = L^1(|nu|)$.
My attempt:
Consider $f in L^1(nu)$ then we know that
$$int |f| dnu = int |f| dnu^+ - int |f| dnu^- < infty$$
This can only happen iff $int |f| dnu^+ < infty$ and $int |f| dnu^- < infty$. Thus, we see that
$$int |f| d|nu| = int |f| dnu^+ + int |f| dnu^- < infty$$
So we must have $f in L^1(|nu|)$. Since all these steps were two ways, we have also that $f in L^1(nu)$.
This "proof" seems too simple and others seem to be giving me the same idea that this is not correct. Where is the flaw and how can I fix it?
real-analysis analysis measure-theory
real-analysis analysis measure-theory
asked Nov 18 at 2:08
carsandpulsars
35117
asked Nov 18 at 2:08
carsandpulsars
35117
asked Nov 18 at 2:08
carsandpulsars
35117
asked Nov 18 at 2:08
carsandpulsars
35117
asked Nov 18 at 2:08
carsandpulsars
35117
35117
What's the exact definition of $L^1(nu)$ for a signed measure $nu$?
– Nate Eldredge
Nov 18 at 2:19
Remembering that $L^1$ is really a set of equivalence classes, presumably part of what's being asserted is that the equivalence classes are the same, so you would have to show that $f = g$ $nu$-a.e. (however that's defined) iff $f = g$ $|nu|$-a.e.
– Nate Eldredge
Nov 18 at 2:21
@NateEldredge In Folland it has been defined as $L^1(nu) = L^1(nu^+) cap L^1(nu^-)$ where $nu = nu^+ - nu^-$ is the Jordan decomposition of $nu$.
– carsandpulsars
Nov 18 at 2:22
Okay, so then your statement that $int |f| dnu^+ < infty$ and $int |f| dnu^- < infty$ is already the definition. The line about $int f dnu^+ - int f dnu^-$ is not the definition and would have to be justified.
– Nate Eldredge
Nov 18 at 2:25
But there is still another statement to be shown: that $f = g$ $|nu|$-a.e. iff $f = g$ $nu^+$-a.e. and $f=g$ $nu^-$-a.e.
– Nate Eldredge
Nov 18 at 2:25
|
show 5 more comments
What's the exact definition of $L^1(nu)$ for a signed measure $nu$?
– Nate Eldredge
Nov 18 at 2:19
Remembering that $L^1$ is really a set of equivalence classes, presumably part of what's being asserted is that the equivalence classes are the same, so you would have to show that $f = g$ $nu$-a.e. (however that's defined) iff $f = g$ $|nu|$-a.e.
– Nate Eldredge
Nov 18 at 2:21
@NateEldredge In Folland it has been defined as $L^1(nu) = L^1(nu^+) cap L^1(nu^-)$ where $nu = nu^+ - nu^-$ is the Jordan decomposition of $nu$.
– carsandpulsars
Nov 18 at 2:22
Okay, so then your statement that $int |f| dnu^+ < infty$ and $int |f| dnu^- < infty$ is already the definition. The line about $int f dnu^+ - int f dnu^-$ is not the definition and would have to be justified.
– Nate Eldredge
Nov 18 at 2:25
But there is still another statement to be shown: that $f = g$ $|nu|$-a.e. iff $f = g$ $nu^+$-a.e. and $f=g$ $nu^-$-a.e.
– Nate Eldredge
Nov 18 at 2:25
What's the exact definition of $L^1(nu)$ for a signed measure $nu$?
– Nate Eldredge
Nov 18 at 2:19
What's the exact definition of $L^1(nu)$ for a signed measure $nu$?
– Nate Eldredge
Nov 18 at 2:19
Remembering that $L^1$ is really a set of equivalence classes, presumably part of what's being asserted is that the equivalence classes are the same, so you would have to show that $f = g$ $nu$-a.e. (however that's defined) iff $f = g$ $|nu|$-a.e.
– Nate Eldredge
Nov 18 at 2:21
Remembering that $L^1$ is really a set of equivalence classes, presumably part of what's being asserted is that the equivalence classes are the same, so you would have to show that $f = g$ $nu$-a.e. (however that's defined) iff $f = g$ $|nu|$-a.e.
– Nate Eldredge
Nov 18 at 2:21
@NateEldredge In Folland it has been defined as $L^1(nu) = L^1(nu^+) cap L^1(nu^-)$ where $nu = nu^+ - nu^-$ is the Jordan decomposition of $nu$.
– carsandpulsars
Nov 18 at 2:22
@NateEldredge In Folland it has been defined as $L^1(nu) = L^1(nu^+) cap L^1(nu^-)$ where $nu = nu^+ - nu^-$ is the Jordan decomposition of $nu$.
– carsandpulsars
Nov 18 at 2:22
Okay, so then your statement that $int |f| dnu^+ < infty$ and $int |f| dnu^- < infty$ is already the definition. The line about $int f dnu^+ - int f dnu^-$ is not the definition and would have to be justified.
– Nate Eldredge
Nov 18 at 2:25
Okay, so then your statement that $int |f| dnu^+ < infty$ and $int |f| dnu^- < infty$ is already the definition. The line about $int f dnu^+ - int f dnu^-$ is not the definition and would have to be justified.
– Nate Eldredge
Nov 18 at 2:25
But there is still another statement to be shown: that $f = g$ $|nu|$-a.e. iff $f = g$ $nu^+$-a.e. and $f=g$ $nu^-$-a.e.
– Nate Eldredge
Nov 18 at 2:25
But there is still another statement to be shown: that $f = g$ $|nu|$-a.e. iff $f = g$ $nu^+$-a.e. and $f=g$ $nu^-$-a.e.
– Nate Eldredge
Nov 18 at 2:25
|
show 5 more comments
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What's the exact definition of $L^1(nu)$ for a signed measure $nu$?
– Nate Eldredge
Nov 18 at 2:19
Remembering that $L^1$ is really a set of equivalence classes, presumably part of what's being asserted is that the equivalence classes are the same, so you would have to show that $f = g$ $nu$-a.e. (however that's defined) iff $f = g$ $|nu|$-a.e.
– Nate Eldredge
Nov 18 at 2:21
@NateEldredge In Folland it has been defined as $L^1(nu) = L^1(nu^+) cap L^1(nu^-)$ where $nu = nu^+ - nu^-$ is the Jordan decomposition of $nu$.
– carsandpulsars
Nov 18 at 2:22
Okay, so then your statement that $int |f| dnu^+ < infty$ and $int |f| dnu^- < infty$ is already the definition. The line about $int f dnu^+ - int f dnu^-$ is not the definition and would have to be justified.
– Nate Eldredge
Nov 18 at 2:25
But there is still another statement to be shown: that $f = g$ $|nu|$-a.e. iff $f = g$ $nu^+$-a.e. and $f=g$ $nu^-$-a.e.
– Nate Eldredge
Nov 18 at 2:25