Radon-Nikodym Derivative of a Total Variation Measure












0












$begingroup$


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







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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
















0












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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














0












0








0





$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











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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


















  • $begingroup$
    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










2 Answers
2






active

oldest

votes


















1












$begingroup$

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.






share|cite|improve this answer











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    0












    $begingroup$

    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).






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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

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      active

      oldest

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      active

      oldest

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      1












      $begingroup$

      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.






      share|cite|improve this answer











      $endgroup$


















        1












        $begingroup$

        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.






        share|cite|improve this answer











        $endgroup$
















          1












          1








          1





          $begingroup$

          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.






          share|cite|improve this answer











          $endgroup$



          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.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 6 '18 at 19:14

























          answered Dec 6 '18 at 16:33









          UserSUserS

          1,5391112




          1,5391112























              0












              $begingroup$

              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).






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                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).






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  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).






                  share|cite|improve this answer









                  $endgroup$



                  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).







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 6 '18 at 16:30









                  астон вілла олоф мэллбэргастон вілла олоф мэллбэрг

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