Singularity of two measures is equivalent to the only measure being dominated by both measures is the zero...
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I am looking at the proof of the Lebesgue decomposition theorem. Let $mu,nu$ be finite measures.
Here, they first define a measure $nu^o$ which is absolutely continuous with respect to $mu$. Now they set $nu^perp := nu - nu^o$, and show that $nu^o$ is maximal among all measures $rho$ such that $rho le nu$ and $rho ll mu$.
Using this they show Orthogonality. Let $tau$ be a measure such that $tau le mu$ and $tau le nu^perp$. Clearly, this implies that $nu^o + tau le nu$ and $nu^o + tau ll mu$. By the maximality, $nu^o + tau le nu^o$ and we conclude that $tau = 0$ and $nu^perp perp mu$.
I don't follow the final line. How does $tau=0$ conclude that $nu^perp perp mu$?
real-analysis analysis measure-theory
asked Nov 23 at 22:02
takecare
2,27721435
add a comment |
up vote
0
down vote
favorite
I am looking at the proof of the Lebesgue decomposition theorem. Let $mu,nu$ be finite measures.
Here, they first define a measure $nu^o$ which is absolutely continuous with respect to $mu$. Now they set $nu^perp := nu - nu^o$, and show that $nu^o$ is maximal among all measures $rho$ such that $rho le nu$ and $rho ll mu$.
Using this they show Orthogonality. Let $tau$ be a measure such that $tau le mu$ and $tau le nu^perp$. Clearly, this implies that $nu^o + tau le nu$ and $nu^o + tau ll mu$. By the maximality, $nu^o + tau le nu^o$ and we conclude that $tau = 0$ and $nu^perp perp mu$.
I don't follow the final line. How does $tau=0$ conclude that $nu^perp perp mu$?
real-analysis analysis measure-theory
asked Nov 23 at 22:02
takecare
2,27721435
Do you mean $ll$ instead of $leq$? I'm not sure if I understand the question. If $tau=0$ the whole situation is trivial and $nu$ and $mu$ need not be related whatsoever.
– Ivo Terek
Nov 23 at 22:09
@IvoTerek I fixed the question.
– takecare
Nov 23 at 22:10
@IvoTerek it is common to denote $leq$ in lieu of $ll$ because, well, absolute continuity is a transitive and reflexive.
– Will M.
Nov 23 at 23:35
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am looking at the proof of the Lebesgue decomposition theorem. Let $mu,nu$ be finite measures.
Here, they first define a measure $nu^o$ which is absolutely continuous with respect to $mu$. Now they set $nu^perp := nu - nu^o$, and show that $nu^o$ is maximal among all measures $rho$ such that $rho le nu$ and $rho ll mu$.
Using this they show Orthogonality. Let $tau$ be a measure such that $tau le mu$ and $tau le nu^perp$. Clearly, this implies that $nu^o + tau le nu$ and $nu^o + tau ll mu$. By the maximality, $nu^o + tau le nu^o$ and we conclude that $tau = 0$ and $nu^perp perp mu$.
I don't follow the final line. How does $tau=0$ conclude that $nu^perp perp mu$?
real-analysis analysis measure-theory
asked Nov 23 at 22:02
takecare
2,27721435
I am looking at the proof of the Lebesgue decomposition theorem. Let $mu,nu$ be finite measures.
Here, they first define a measure $nu^o$ which is absolutely continuous with respect to $mu$. Now they set $nu^perp := nu - nu^o$, and show that $nu^o$ is maximal among all measures $rho$ such that $rho le nu$ and $rho ll mu$.
Using this they show Orthogonality. Let $tau$ be a measure such that $tau le mu$ and $tau le nu^perp$. Clearly, this implies that $nu^o + tau le nu$ and $nu^o + tau ll mu$. By the maximality, $nu^o + tau le nu^o$ and we conclude that $tau = 0$ and $nu^perp perp mu$.
I don't follow the final line. How does $tau=0$ conclude that $nu^perp perp mu$?
real-analysis analysis measure-theory
real-analysis analysis measure-theory
asked Nov 23 at 22:02
takecare
2,27721435
asked Nov 23 at 22:02
takecare
2,27721435
edited Nov 23 at 22:22
asked Nov 23 at 22:02
takecare
2,27721435
asked Nov 23 at 22:02
takecare
2,27721435
asked Nov 23 at 22:02
takecare
2,27721435
2,27721435
Do you mean $ll$ instead of $leq$? I'm not sure if I understand the question. If $tau=0$ the whole situation is trivial and $nu$ and $mu$ need not be related whatsoever.
– Ivo Terek
Nov 23 at 22:09
@IvoTerek I fixed the question.
– takecare
Nov 23 at 22:10
@IvoTerek it is common to denote $leq$ in lieu of $ll$ because, well, absolute continuity is a transitive and reflexive.
– Will M.
Nov 23 at 23:35
add a comment |
Do you mean $ll$ instead of $leq$? I'm not sure if I understand the question. If $tau=0$ the whole situation is trivial and $nu$ and $mu$ need not be related whatsoever.
– Ivo Terek
Nov 23 at 22:09
@IvoTerek I fixed the question.
– takecare
Nov 23 at 22:10
@IvoTerek it is common to denote $leq$ in lieu of $ll$ because, well, absolute continuity is a transitive and reflexive.
– Will M.
Nov 23 at 23:35
Do you mean $ll$ instead of $leq$? I'm not sure if I understand the question. If $tau=0$ the whole situation is trivial and $nu$ and $mu$ need not be related whatsoever.
– Ivo Terek
Nov 23 at 22:09
Do you mean $ll$ instead of $leq$? I'm not sure if I understand the question. If $tau=0$ the whole situation is trivial and $nu$ and $mu$ need not be related whatsoever.
– Ivo Terek
Nov 23 at 22:09
@IvoTerek I fixed the question.
– takecare
Nov 23 at 22:10
@IvoTerek I fixed the question.
– takecare
Nov 23 at 22:10
@IvoTerek it is common to denote $leq$ in lieu of $ll$ because, well, absolute continuity is a transitive and reflexive.
– Will M.
Nov 23 at 23:35
@IvoTerek it is common to denote $leq$ in lieu of $ll$ because, well, absolute continuity is a transitive and reflexive.
– Will M.
Nov 23 at 23:35
add a comment |
1 Answer
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I suppose $mu$ and $nu$ positive measures. Let $f$ be the Radon-Nikodym derivative of $mu$ w.r.t. $mu+nu^{perp}$ and $g$ be the Radon-Nikodym derivative of $nu^{perp}$ w.r.t. $mu+nu^{perp}$. Let $tau (E)=int_E min {f,g} d (mu+nu^{perp})$. Then $tau leq mu$ and $tau leq nu^{perp}$ so $tau =0$. Hence $int_E min {f,g}d(mu+nu^{perp})=0$ for every measurable set $E$ which implies $min {f,g}=0$ almost everywhere w.r.t. $mu+nu^{perp}$ (hence w.r.t. each of $mu$ and $nu^{perp})$. This means one of $f,g$ is $0$ at every point excluding a null set w.r.t. each of $mu$ and $nu^{perp}$. Can you now prove that $mu perp nu^{perp}$?
answered Nov 23 at 23:34
Kavi Rama Murthy
43.5k31751
So we may have $f=0$ on a set $N^c$ for which $mu(N)=nu^perp (N)=0$. I don't see how to guarantee $mu(N) = nu^perp (N^c)=0$ from this.
– takecare
Nov 23 at 23:40
1
@takecare $mu {f=0}=0$ and $nu^{perp} {g=0}=0$. And $(mu+nu^{perp}) (Acup B)^{c}=0$ where $A={f=0}$ and $B={g=0}$. $mu$ is concentrated on $A^{c}$, $nu^{perp}$ on ${gneq 0, fg=0}$ and these two are disjoint sets.
– Kavi Rama Murthy
Nov 23 at 23:43
I can see we should always have $(mu + nu^perp) (A cup B)^c = 0$ but why is $mu (A) = 0$ and $nu^perp(B)=0$? And how do these give $mu perp nu^perp$?
– takecare
Nov 23 at 23:49
1
@takecare $mu (A) =int_{A} f d(mu+nu^{perp})$ by definition of RND so $mu(A)=0$.
– Kavi Rama Murthy
Nov 23 at 23:53
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
accepted
I suppose $mu$ and $nu$ positive measures. Let $f$ be the Radon-Nikodym derivative of $mu$ w.r.t. $mu+nu^{perp}$ and $g$ be the Radon-Nikodym derivative of $nu^{perp}$ w.r.t. $mu+nu^{perp}$. Let $tau (E)=int_E min {f,g} d (mu+nu^{perp})$. Then $tau leq mu$ and $tau leq nu^{perp}$ so $tau =0$. Hence $int_E min {f,g}d(mu+nu^{perp})=0$ for every measurable set $E$ which implies $min {f,g}=0$ almost everywhere w.r.t. $mu+nu^{perp}$ (hence w.r.t. each of $mu$ and $nu^{perp})$. This means one of $f,g$ is $0$ at every point excluding a null set w.r.t. each of $mu$ and $nu^{perp}$. Can you now prove that $mu perp nu^{perp}$?
answered Nov 23 at 23:34
Kavi Rama Murthy
43.5k31751
So we may have $f=0$ on a set $N^c$ for which $mu(N)=nu^perp (N)=0$. I don't see how to guarantee $mu(N) = nu^perp (N^c)=0$ from this.
– takecare
Nov 23 at 23:40
1
@takecare $mu {f=0}=0$ and $nu^{perp} {g=0}=0$. And $(mu+nu^{perp}) (Acup B)^{c}=0$ where $A={f=0}$ and $B={g=0}$. $mu$ is concentrated on $A^{c}$, $nu^{perp}$ on ${gneq 0, fg=0}$ and these two are disjoint sets.
– Kavi Rama Murthy
Nov 23 at 23:43
I can see we should always have $(mu + nu^perp) (A cup B)^c = 0$ but why is $mu (A) = 0$ and $nu^perp(B)=0$? And how do these give $mu perp nu^perp$?
– takecare
Nov 23 at 23:49
1
@takecare $mu (A) =int_{A} f d(mu+nu^{perp})$ by definition of RND so $mu(A)=0$.
– Kavi Rama Murthy
Nov 23 at 23:53
add a comment |
up vote
1
down vote
accepted
I suppose $mu$ and $nu$ positive measures. Let $f$ be the Radon-Nikodym derivative of $mu$ w.r.t. $mu+nu^{perp}$ and $g$ be the Radon-Nikodym derivative of $nu^{perp}$ w.r.t. $mu+nu^{perp}$. Let $tau (E)=int_E min {f,g} d (mu+nu^{perp})$. Then $tau leq mu$ and $tau leq nu^{perp}$ so $tau =0$. Hence $int_E min {f,g}d(mu+nu^{perp})=0$ for every measurable set $E$ which implies $min {f,g}=0$ almost everywhere w.r.t. $mu+nu^{perp}$ (hence w.r.t. each of $mu$ and $nu^{perp})$. This means one of $f,g$ is $0$ at every point excluding a null set w.r.t. each of $mu$ and $nu^{perp}$. Can you now prove that $mu perp nu^{perp}$?
answered Nov 23 at 23:34
Kavi Rama Murthy
43.5k31751
So we may have $f=0$ on a set $N^c$ for which $mu(N)=nu^perp (N)=0$. I don't see how to guarantee $mu(N) = nu^perp (N^c)=0$ from this.
– takecare
Nov 23 at 23:40
1
@takecare $mu {f=0}=0$ and $nu^{perp} {g=0}=0$. And $(mu+nu^{perp}) (Acup B)^{c}=0$ where $A={f=0}$ and $B={g=0}$. $mu$ is concentrated on $A^{c}$, $nu^{perp}$ on ${gneq 0, fg=0}$ and these two are disjoint sets.
– Kavi Rama Murthy
Nov 23 at 23:43
I can see we should always have $(mu + nu^perp) (A cup B)^c = 0$ but why is $mu (A) = 0$ and $nu^perp(B)=0$? And how do these give $mu perp nu^perp$?
– takecare
Nov 23 at 23:49
1
@takecare $mu (A) =int_{A} f d(mu+nu^{perp})$ by definition of RND so $mu(A)=0$.
– Kavi Rama Murthy
Nov 23 at 23:53
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
I suppose $mu$ and $nu$ positive measures. Let $f$ be the Radon-Nikodym derivative of $mu$ w.r.t. $mu+nu^{perp}$ and $g$ be the Radon-Nikodym derivative of $nu^{perp}$ w.r.t. $mu+nu^{perp}$. Let $tau (E)=int_E min {f,g} d (mu+nu^{perp})$. Then $tau leq mu$ and $tau leq nu^{perp}$ so $tau =0$. Hence $int_E min {f,g}d(mu+nu^{perp})=0$ for every measurable set $E$ which implies $min {f,g}=0$ almost everywhere w.r.t. $mu+nu^{perp}$ (hence w.r.t. each of $mu$ and $nu^{perp})$. This means one of $f,g$ is $0$ at every point excluding a null set w.r.t. each of $mu$ and $nu^{perp}$. Can you now prove that $mu perp nu^{perp}$?
answered Nov 23 at 23:34
Kavi Rama Murthy
43.5k31751
I suppose $mu$ and $nu$ positive measures. Let $f$ be the Radon-Nikodym derivative of $mu$ w.r.t. $mu+nu^{perp}$ and $g$ be the Radon-Nikodym derivative of $nu^{perp}$ w.r.t. $mu+nu^{perp}$. Let $tau (E)=int_E min {f,g} d (mu+nu^{perp})$. Then $tau leq mu$ and $tau leq nu^{perp}$ so $tau =0$. Hence $int_E min {f,g}d(mu+nu^{perp})=0$ for every measurable set $E$ which implies $min {f,g}=0$ almost everywhere w.r.t. $mu+nu^{perp}$ (hence w.r.t. each of $mu$ and $nu^{perp})$. This means one of $f,g$ is $0$ at every point excluding a null set w.r.t. each of $mu$ and $nu^{perp}$. Can you now prove that $mu perp nu^{perp}$?
answered Nov 23 at 23:34
Kavi Rama Murthy
43.5k31751
answered Nov 23 at 23:34
Kavi Rama Murthy
43.5k31751
answered Nov 23 at 23:34
Kavi Rama Murthy
43.5k31751
answered Nov 23 at 23:34
Kavi Rama Murthy
43.5k31751
43.5k31751
So we may have $f=0$ on a set $N^c$ for which $mu(N)=nu^perp (N)=0$. I don't see how to guarantee $mu(N) = nu^perp (N^c)=0$ from this.
– takecare
Nov 23 at 23:40
1
@takecare $mu {f=0}=0$ and $nu^{perp} {g=0}=0$. And $(mu+nu^{perp}) (Acup B)^{c}=0$ where $A={f=0}$ and $B={g=0}$. $mu$ is concentrated on $A^{c}$, $nu^{perp}$ on ${gneq 0, fg=0}$ and these two are disjoint sets.
– Kavi Rama Murthy
Nov 23 at 23:43
I can see we should always have $(mu + nu^perp) (A cup B)^c = 0$ but why is $mu (A) = 0$ and $nu^perp(B)=0$? And how do these give $mu perp nu^perp$?
– takecare
Nov 23 at 23:49
1
@takecare $mu (A) =int_{A} f d(mu+nu^{perp})$ by definition of RND so $mu(A)=0$.
– Kavi Rama Murthy
Nov 23 at 23:53
add a comment |
So we may have $f=0$ on a set $N^c$ for which $mu(N)=nu^perp (N)=0$. I don't see how to guarantee $mu(N) = nu^perp (N^c)=0$ from this.
– takecare
Nov 23 at 23:40
1
@takecare $mu {f=0}=0$ and $nu^{perp} {g=0}=0$. And $(mu+nu^{perp}) (Acup B)^{c}=0$ where $A={f=0}$ and $B={g=0}$. $mu$ is concentrated on $A^{c}$, $nu^{perp}$ on ${gneq 0, fg=0}$ and these two are disjoint sets.
– Kavi Rama Murthy
Nov 23 at 23:43
I can see we should always have $(mu + nu^perp) (A cup B)^c = 0$ but why is $mu (A) = 0$ and $nu^perp(B)=0$? And how do these give $mu perp nu^perp$?
– takecare
Nov 23 at 23:49
1
@takecare $mu (A) =int_{A} f d(mu+nu^{perp})$ by definition of RND so $mu(A)=0$.
– Kavi Rama Murthy
Nov 23 at 23:53
So we may have $f=0$ on a set $N^c$ for which $mu(N)=nu^perp (N)=0$. I don't see how to guarantee $mu(N) = nu^perp (N^c)=0$ from this.
– takecare
Nov 23 at 23:40
So we may have $f=0$ on a set $N^c$ for which $mu(N)=nu^perp (N)=0$. I don't see how to guarantee $mu(N) = nu^perp (N^c)=0$ from this.
– takecare
Nov 23 at 23:40
1
1
@takecare $mu {f=0}=0$ and $nu^{perp} {g=0}=0$. And $(mu+nu^{perp}) (Acup B)^{c}=0$ where $A={f=0}$ and $B={g=0}$. $mu$ is concentrated on $A^{c}$, $nu^{perp}$ on ${gneq 0, fg=0}$ and these two are disjoint sets.
– Kavi Rama Murthy
Nov 23 at 23:43
@takecare $mu {f=0}=0$ and $nu^{perp} {g=0}=0$. And $(mu+nu^{perp}) (Acup B)^{c}=0$ where $A={f=0}$ and $B={g=0}$. $mu$ is concentrated on $A^{c}$, $nu^{perp}$ on ${gneq 0, fg=0}$ and these two are disjoint sets.
– Kavi Rama Murthy
Nov 23 at 23:43
I can see we should always have $(mu + nu^perp) (A cup B)^c = 0$ but why is $mu (A) = 0$ and $nu^perp(B)=0$? And how do these give $mu perp nu^perp$?
– takecare
Nov 23 at 23:49
I can see we should always have $(mu + nu^perp) (A cup B)^c = 0$ but why is $mu (A) = 0$ and $nu^perp(B)=0$? And how do these give $mu perp nu^perp$?
– takecare
Nov 23 at 23:49
1
1
@takecare $mu (A) =int_{A} f d(mu+nu^{perp})$ by definition of RND so $mu(A)=0$.
– Kavi Rama Murthy
Nov 23 at 23:53
@takecare $mu (A) =int_{A} f d(mu+nu^{perp})$ by definition of RND so $mu(A)=0$.
– Kavi Rama Murthy
Nov 23 at 23:53
add a comment |
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Do you mean $ll$ instead of $leq$? I'm not sure if I understand the question. If $tau=0$ the whole situation is trivial and $nu$ and $mu$ need not be related whatsoever.
– Ivo Terek
Nov 23 at 22:09
@IvoTerek I fixed the question.
– takecare
Nov 23 at 22:10
@IvoTerek it is common to denote $leq$ in lieu of $ll$ because, well, absolute continuity is a transitive and reflexive.
– Will M.
Nov 23 at 23:35