Radon-Nikodym Derivative of a Total Variation Measure
$begingroup$
Let $nu$ be a signed measure which is absolutely continuous to a sigma-finite measure $mu$. Show that $frac{d|nu|}{dmu}=|frac{dnu}{dmu}|$, where $|nu|$ is the total variation measure of $nu$.
Now by definition, for any $Ein F$ we have $|nu|(E)=nu^+(E)+nu^-(E)=nu(Ecap P)-nu(Ecap P^c)$, where ${P,P^c}$ is a Hahn decomposition of $nu$ and ${nu^+,nu^-}$ is the corresponding Jordan decomposition. Using these two expressions, I found that $frac{d|nu|}{dmu}=frac{dnu^+}{dmu}+frac{dnu^-}{dmu}=(1_P-1_{P^c})frac{dnu}{dmu}$, but I'm not sure how to equate either of these expressions with $|frac{dnu}{dmu}|$.
measure-theory lebesgue-integral radon-nikodym signed-measures
asked Dec 6 '18 at 15:45
Keshav SrinivasanKeshav Srinivasan
2,05911443
$endgroup$
add a comment |
$begingroup$
Let $nu$ be a signed measure which is absolutely continuous to a sigma-finite measure $mu$. Show that $frac{d|nu|}{dmu}=|frac{dnu}{dmu}|$, where $|nu|$ is the total variation measure of $nu$.
Now by definition, for any $Ein F$ we have $|nu|(E)=nu^+(E)+nu^-(E)=nu(Ecap P)-nu(Ecap P^c)$, where ${P,P^c}$ is a Hahn decomposition of $nu$ and ${nu^+,nu^-}$ is the corresponding Jordan decomposition. Using these two expressions, I found that $frac{d|nu|}{dmu}=frac{dnu^+}{dmu}+frac{dnu^-}{dmu}=(1_P-1_{P^c})frac{dnu}{dmu}$, but I'm not sure how to equate either of these expressions with $|frac{dnu}{dmu}|$.
measure-theory lebesgue-integral radon-nikodym signed-measures
asked Dec 6 '18 at 15:45
Keshav SrinivasanKeshav Srinivasan
2,05911443
$endgroup$
$begingroup$
Are you looking for proof of $frac{d|nu|}{dmu}=|frac{dnu}{dmu}|$.
$endgroup$
– UserS
Dec 6 '18 at 16:03
$begingroup$
@UserS Yes, I am.
$endgroup$
– Keshav Srinivasan
Dec 6 '18 at 16:11
add a comment |
$begingroup$
Let $nu$ be a signed measure which is absolutely continuous to a sigma-finite measure $mu$. Show that $frac{d|nu|}{dmu}=|frac{dnu}{dmu}|$, where $|nu|$ is the total variation measure of $nu$.
Now by definition, for any $Ein F$ we have $|nu|(E)=nu^+(E)+nu^-(E)=nu(Ecap P)-nu(Ecap P^c)$, where ${P,P^c}$ is a Hahn decomposition of $nu$ and ${nu^+,nu^-}$ is the corresponding Jordan decomposition. Using these two expressions, I found that $frac{d|nu|}{dmu}=frac{dnu^+}{dmu}+frac{dnu^-}{dmu}=(1_P-1_{P^c})frac{dnu}{dmu}$, but I'm not sure how to equate either of these expressions with $|frac{dnu}{dmu}|$.
measure-theory lebesgue-integral radon-nikodym signed-measures
asked Dec 6 '18 at 15:45
Keshav SrinivasanKeshav Srinivasan
2,05911443
$endgroup$
Let $nu$ be a signed measure which is absolutely continuous to a sigma-finite measure $mu$. Show that $frac{d|nu|}{dmu}=|frac{dnu}{dmu}|$, where $|nu|$ is the total variation measure of $nu$.
Now by definition, for any $Ein F$ we have $|nu|(E)=nu^+(E)+nu^-(E)=nu(Ecap P)-nu(Ecap P^c)$, where ${P,P^c}$ is a Hahn decomposition of $nu$ and ${nu^+,nu^-}$ is the corresponding Jordan decomposition. Using these two expressions, I found that $frac{d|nu|}{dmu}=frac{dnu^+}{dmu}+frac{dnu^-}{dmu}=(1_P-1_{P^c})frac{dnu}{dmu}$, but I'm not sure how to equate either of these expressions with $|frac{dnu}{dmu}|$.
measure-theory lebesgue-integral radon-nikodym signed-measures
measure-theory lebesgue-integral radon-nikodym signed-measures
asked Dec 6 '18 at 15:45
Keshav SrinivasanKeshav Srinivasan
2,05911443
asked Dec 6 '18 at 15:45
Keshav SrinivasanKeshav Srinivasan
2,05911443
asked Dec 6 '18 at 15:45
Keshav SrinivasanKeshav Srinivasan
2,05911443
asked Dec 6 '18 at 15:45
Keshav SrinivasanKeshav Srinivasan
2,05911443
asked Dec 6 '18 at 15:45
Keshav SrinivasanKeshav Srinivasan
2,05911443
2,05911443
$begingroup$
Are you looking for proof of $frac{d|nu|}{dmu}=|frac{dnu}{dmu}|$.
$endgroup$
– UserS
Dec 6 '18 at 16:03
$begingroup$
@UserS Yes, I am.
$endgroup$
– Keshav Srinivasan
Dec 6 '18 at 16:11
add a comment |
$begingroup$
Are you looking for proof of $frac{d|nu|}{dmu}=|frac{dnu}{dmu}|$.
$endgroup$
– UserS
Dec 6 '18 at 16:03
$begingroup$
@UserS Yes, I am.
$endgroup$
– Keshav Srinivasan
Dec 6 '18 at 16:11
$begingroup$
Are you looking for proof of $frac{d|nu|}{dmu}=|frac{dnu}{dmu}|$.
$endgroup$
– UserS
Dec 6 '18 at 16:03
$begingroup$
Are you looking for proof of $frac{d|nu|}{dmu}=|frac{dnu}{dmu}|$.
$endgroup$
– UserS
Dec 6 '18 at 16:03
$begingroup$
@UserS Yes, I am.
$endgroup$
– Keshav Srinivasan
Dec 6 '18 at 16:11
$begingroup$
@UserS Yes, I am.
$endgroup$
– Keshav Srinivasan
Dec 6 '18 at 16:11
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Note that $|nu|$ is a measure i.e. non-negative valued. Now if you go through the proof of Radon-Nikodym theorem you see that proof has two parts, one is for (non-negative) measure in which case Radon-Nikodym derivative is non-negative and another part is for signed measure which has been done by considering positive and negative parts of signed measure giving extended real valued Radon-Nikodym derivative.
Therefore $frac{d|nu|}{dmu}geq 0$ and so taking modules in the equation $frac{d|nu|}{dmu}=frac{dnu^+}{dmu}+frac{dnu^-}{dmu}=(1_P-1_{P^c})frac{dnu}{dmu}$ we have $frac{d|nu|}{dmu}=|(1_P-1_{P^c})||frac{dnu}{dmu}|=1|frac{dnu}{dmu}|=|
frac{dnu}{dmu}|$ , $mu$ a.e.
answered Dec 6 '18 at 16:33
UserSUserS
1,5391112
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add a comment |
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Let $P, N$ be the Hahn decomposition of $nu$, with the associated Jordan decomposition $nu^+, nu^-$. One sees that $nu << mu implies nu^+ << mu$ and $nu^- << mu$. Let $f = frac{d nu}{d mu}$.
These are both positive measures, and therefore admit Radon-Nikodym derivatives with respect to $mu$, which we denote by $g = frac{d nu^+}{d mu}$ and $h = frac{dnu^-}{d mu}$. Note that $g,h$ are non-negative.
From the above, it follows that $nu^- + nu^+ = |nu| << mu$.
First, compute $g,h$ in terms of $f$. For any function $K$, we have : $$int Kf d mu = int K d nu = int_P K d nu^+ - int_{P^C} K d nu^- = int K left(g1_P - h1_{P^C}right)d mu$$
So, $f = g1_P - h1_{P^c}$ almost everywhere, by uniqueness of the RN derivative.
Task : Use a similar argument to above, to show that $frac{d|nu|}{d mu} = g1_P + h1_{P^c}$. Conclude that $|f| = frac{d |nu|}{d mu}$(since $P^c$ and $P$ are disjoint, the absolute value of $g1_P - h1_{P^c}$ is simply equal to $g1_P + h1_{P^c}$, as desired).
answered Dec 6 '18 at 16:30
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
37.9k33376
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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votes
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votes
$begingroup$
Note that $|nu|$ is a measure i.e. non-negative valued. Now if you go through the proof of Radon-Nikodym theorem you see that proof has two parts, one is for (non-negative) measure in which case Radon-Nikodym derivative is non-negative and another part is for signed measure which has been done by considering positive and negative parts of signed measure giving extended real valued Radon-Nikodym derivative.
Therefore $frac{d|nu|}{dmu}geq 0$ and so taking modules in the equation $frac{d|nu|}{dmu}=frac{dnu^+}{dmu}+frac{dnu^-}{dmu}=(1_P-1_{P^c})frac{dnu}{dmu}$ we have $frac{d|nu|}{dmu}=|(1_P-1_{P^c})||frac{dnu}{dmu}|=1|frac{dnu}{dmu}|=|
frac{dnu}{dmu}|$ , $mu$ a.e.
answered Dec 6 '18 at 16:33
UserSUserS
1,5391112
$endgroup$
add a comment |
$begingroup$
Note that $|nu|$ is a measure i.e. non-negative valued. Now if you go through the proof of Radon-Nikodym theorem you see that proof has two parts, one is for (non-negative) measure in which case Radon-Nikodym derivative is non-negative and another part is for signed measure which has been done by considering positive and negative parts of signed measure giving extended real valued Radon-Nikodym derivative.
Therefore $frac{d|nu|}{dmu}geq 0$ and so taking modules in the equation $frac{d|nu|}{dmu}=frac{dnu^+}{dmu}+frac{dnu^-}{dmu}=(1_P-1_{P^c})frac{dnu}{dmu}$ we have $frac{d|nu|}{dmu}=|(1_P-1_{P^c})||frac{dnu}{dmu}|=1|frac{dnu}{dmu}|=|
frac{dnu}{dmu}|$ , $mu$ a.e.
answered Dec 6 '18 at 16:33
UserSUserS
1,5391112
$endgroup$
add a comment |
$begingroup$
Note that $|nu|$ is a measure i.e. non-negative valued. Now if you go through the proof of Radon-Nikodym theorem you see that proof has two parts, one is for (non-negative) measure in which case Radon-Nikodym derivative is non-negative and another part is for signed measure which has been done by considering positive and negative parts of signed measure giving extended real valued Radon-Nikodym derivative.
Therefore $frac{d|nu|}{dmu}geq 0$ and so taking modules in the equation $frac{d|nu|}{dmu}=frac{dnu^+}{dmu}+frac{dnu^-}{dmu}=(1_P-1_{P^c})frac{dnu}{dmu}$ we have $frac{d|nu|}{dmu}=|(1_P-1_{P^c})||frac{dnu}{dmu}|=1|frac{dnu}{dmu}|=|
frac{dnu}{dmu}|$ , $mu$ a.e.
answered Dec 6 '18 at 16:33
UserSUserS
1,5391112
$endgroup$
Note that $|nu|$ is a measure i.e. non-negative valued. Now if you go through the proof of Radon-Nikodym theorem you see that proof has two parts, one is for (non-negative) measure in which case Radon-Nikodym derivative is non-negative and another part is for signed measure which has been done by considering positive and negative parts of signed measure giving extended real valued Radon-Nikodym derivative.
Therefore $frac{d|nu|}{dmu}geq 0$ and so taking modules in the equation $frac{d|nu|}{dmu}=frac{dnu^+}{dmu}+frac{dnu^-}{dmu}=(1_P-1_{P^c})frac{dnu}{dmu}$ we have $frac{d|nu|}{dmu}=|(1_P-1_{P^c})||frac{dnu}{dmu}|=1|frac{dnu}{dmu}|=|
frac{dnu}{dmu}|$ , $mu$ a.e.
answered Dec 6 '18 at 16:33
UserSUserS
1,5391112
edited Dec 6 '18 at 19:14
answered Dec 6 '18 at 16:33
UserSUserS
1,5391112
answered Dec 6 '18 at 16:33
UserSUserS
1,5391112
answered Dec 6 '18 at 16:33
UserSUserS
1,5391112
1,5391112
add a comment |
add a comment |
$begingroup$
Let $P, N$ be the Hahn decomposition of $nu$, with the associated Jordan decomposition $nu^+, nu^-$. One sees that $nu << mu implies nu^+ << mu$ and $nu^- << mu$. Let $f = frac{d nu}{d mu}$.
These are both positive measures, and therefore admit Radon-Nikodym derivatives with respect to $mu$, which we denote by $g = frac{d nu^+}{d mu}$ and $h = frac{dnu^-}{d mu}$. Note that $g,h$ are non-negative.
From the above, it follows that $nu^- + nu^+ = |nu| << mu$.
First, compute $g,h$ in terms of $f$. For any function $K$, we have : $$int Kf d mu = int K d nu = int_P K d nu^+ - int_{P^C} K d nu^- = int K left(g1_P - h1_{P^C}right)d mu$$
So, $f = g1_P - h1_{P^c}$ almost everywhere, by uniqueness of the RN derivative.
Task : Use a similar argument to above, to show that $frac{d|nu|}{d mu} = g1_P + h1_{P^c}$. Conclude that $|f| = frac{d |nu|}{d mu}$(since $P^c$ and $P$ are disjoint, the absolute value of $g1_P - h1_{P^c}$ is simply equal to $g1_P + h1_{P^c}$, as desired).
answered Dec 6 '18 at 16:30
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
37.9k33376
$endgroup$
add a comment |
$begingroup$
Let $P, N$ be the Hahn decomposition of $nu$, with the associated Jordan decomposition $nu^+, nu^-$. One sees that $nu << mu implies nu^+ << mu$ and $nu^- << mu$. Let $f = frac{d nu}{d mu}$.
These are both positive measures, and therefore admit Radon-Nikodym derivatives with respect to $mu$, which we denote by $g = frac{d nu^+}{d mu}$ and $h = frac{dnu^-}{d mu}$. Note that $g,h$ are non-negative.
From the above, it follows that $nu^- + nu^+ = |nu| << mu$.
First, compute $g,h$ in terms of $f$. For any function $K$, we have : $$int Kf d mu = int K d nu = int_P K d nu^+ - int_{P^C} K d nu^- = int K left(g1_P - h1_{P^C}right)d mu$$
So, $f = g1_P - h1_{P^c}$ almost everywhere, by uniqueness of the RN derivative.
Task : Use a similar argument to above, to show that $frac{d|nu|}{d mu} = g1_P + h1_{P^c}$. Conclude that $|f| = frac{d |nu|}{d mu}$(since $P^c$ and $P$ are disjoint, the absolute value of $g1_P - h1_{P^c}$ is simply equal to $g1_P + h1_{P^c}$, as desired).
answered Dec 6 '18 at 16:30
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
37.9k33376
$endgroup$
add a comment |
$begingroup$
Let $P, N$ be the Hahn decomposition of $nu$, with the associated Jordan decomposition $nu^+, nu^-$. One sees that $nu << mu implies nu^+ << mu$ and $nu^- << mu$. Let $f = frac{d nu}{d mu}$.
These are both positive measures, and therefore admit Radon-Nikodym derivatives with respect to $mu$, which we denote by $g = frac{d nu^+}{d mu}$ and $h = frac{dnu^-}{d mu}$. Note that $g,h$ are non-negative.
From the above, it follows that $nu^- + nu^+ = |nu| << mu$.
First, compute $g,h$ in terms of $f$. For any function $K$, we have : $$int Kf d mu = int K d nu = int_P K d nu^+ - int_{P^C} K d nu^- = int K left(g1_P - h1_{P^C}right)d mu$$
So, $f = g1_P - h1_{P^c}$ almost everywhere, by uniqueness of the RN derivative.
Task : Use a similar argument to above, to show that $frac{d|nu|}{d mu} = g1_P + h1_{P^c}$. Conclude that $|f| = frac{d |nu|}{d mu}$(since $P^c$ and $P$ are disjoint, the absolute value of $g1_P - h1_{P^c}$ is simply equal to $g1_P + h1_{P^c}$, as desired).
answered Dec 6 '18 at 16:30
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
37.9k33376
$endgroup$
Let $P, N$ be the Hahn decomposition of $nu$, with the associated Jordan decomposition $nu^+, nu^-$. One sees that $nu << mu implies nu^+ << mu$ and $nu^- << mu$. Let $f = frac{d nu}{d mu}$.
These are both positive measures, and therefore admit Radon-Nikodym derivatives with respect to $mu$, which we denote by $g = frac{d nu^+}{d mu}$ and $h = frac{dnu^-}{d mu}$. Note that $g,h$ are non-negative.
From the above, it follows that $nu^- + nu^+ = |nu| << mu$.
First, compute $g,h$ in terms of $f$. For any function $K$, we have : $$int Kf d mu = int K d nu = int_P K d nu^+ - int_{P^C} K d nu^- = int K left(g1_P - h1_{P^C}right)d mu$$
So, $f = g1_P - h1_{P^c}$ almost everywhere, by uniqueness of the RN derivative.
Task : Use a similar argument to above, to show that $frac{d|nu|}{d mu} = g1_P + h1_{P^c}$. Conclude that $|f| = frac{d |nu|}{d mu}$(since $P^c$ and $P$ are disjoint, the absolute value of $g1_P - h1_{P^c}$ is simply equal to $g1_P + h1_{P^c}$, as desired).
answered Dec 6 '18 at 16:30
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
37.9k33376
answered Dec 6 '18 at 16:30
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
37.9k33376
answered Dec 6 '18 at 16:30
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
37.9k33376
answered Dec 6 '18 at 16:30
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
37.9k33376
37.9k33376
add a comment |
add a comment |
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$begingroup$
Are you looking for proof of $frac{d|nu|}{dmu}=|frac{dnu}{dmu}|$.
$endgroup$
– UserS
Dec 6 '18 at 16:03
$begingroup$
@UserS Yes, I am.
$endgroup$
– Keshav Srinivasan
Dec 6 '18 at 16:11