(Hint Needed) Real-Analysis Exam












1












$begingroup$


So I'm trying to study for a qualifying exam and need a hint on the following problem. If someone knows of some useful lemmas and techniques, I would like to hear it!



Let $(X,mathcal{A},mu)$ be a $sigma$-finite measure space, and let $mathcal{A}_0$ be a sub-$sigma$-algebra of $mathcal{A}$ so that $(X,mathcal{A}_0,mu)$ is $sigma$-finite. Given a nonnegative $mathcal{A}$-measurable function $f$ on $X$, show that there is a nonnegative $mathcal{A}_0$-measurable function $f_0$ on $X$, such that
$$int_Xfg,dmu=int_Xf_0g,dmu $$
for every nonnegative $mathcal{A}_0$-measurable function $g$. In what sense is $f_0$ unique?






real-analysis analysis measure-theory lebesgue-measure







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    So I'm trying to study for a qualifying exam and need a hint on the following problem. If someone knows of some useful lemmas and techniques, I would like to hear it!



    Let $(X,mathcal{A},mu)$ be a $sigma$-finite measure space, and let $mathcal{A}_0$ be a sub-$sigma$-algebra of $mathcal{A}$ so that $(X,mathcal{A}_0,mu)$ is $sigma$-finite. Given a nonnegative $mathcal{A}$-measurable function $f$ on $X$, show that there is a nonnegative $mathcal{A}_0$-measurable function $f_0$ on $X$, such that
    $$int_Xfg,dmu=int_Xf_0g,dmu $$
    for every nonnegative $mathcal{A}_0$-measurable function $g$. In what sense is $f_0$ unique?






    real-analysis analysis measure-theory lebesgue-measure







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      So I'm trying to study for a qualifying exam and need a hint on the following problem. If someone knows of some useful lemmas and techniques, I would like to hear it!



      Let $(X,mathcal{A},mu)$ be a $sigma$-finite measure space, and let $mathcal{A}_0$ be a sub-$sigma$-algebra of $mathcal{A}$ so that $(X,mathcal{A}_0,mu)$ is $sigma$-finite. Given a nonnegative $mathcal{A}$-measurable function $f$ on $X$, show that there is a nonnegative $mathcal{A}_0$-measurable function $f_0$ on $X$, such that
      $$int_Xfg,dmu=int_Xf_0g,dmu $$
      for every nonnegative $mathcal{A}_0$-measurable function $g$. In what sense is $f_0$ unique?






      real-analysis analysis measure-theory lebesgue-measure







      share|cite|improve this question











      $endgroup$




      So I'm trying to study for a qualifying exam and need a hint on the following problem. If someone knows of some useful lemmas and techniques, I would like to hear it!



      Let $(X,mathcal{A},mu)$ be a $sigma$-finite measure space, and let $mathcal{A}_0$ be a sub-$sigma$-algebra of $mathcal{A}$ so that $(X,mathcal{A}_0,mu)$ is $sigma$-finite. Given a nonnegative $mathcal{A}$-measurable function $f$ on $X$, show that there is a nonnegative $mathcal{A}_0$-measurable function $f_0$ on $X$, such that
      $$int_Xfg,dmu=int_Xf_0g,dmu $$
      for every nonnegative $mathcal{A}_0$-measurable function $g$. In what sense is $f_0$ unique?






      real-analysis analysis measure-theory lebesgue-measure




      real-analysis analysis measure-theory lebesgue-measure






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 23 at 4:48









      Key Flex

      8,38261233




      8,38261233










      asked Jan 23 at 4:09









      DarelDarel

      1249




      1249






















          1 Answer
          1






          active

          oldest

          votes


















          3












          $begingroup$

          Notice that the above problem is false if we do not require $(X,mathcal{A}_0,mu)$ to be $sigma$-finite. As a counter-example, take the usual Lebesgue measure on $mathbb{R}$, and $mathcal{A}_0={∅,mathbb{R}}$. Let $g = 1$ and $f =X_{[0,1]}$. Then LHS is $1$ while RHS is either $0$ and $infty ($contradiction $)$. Thus we will assume $(X, mathcal{A}_0, mu)$ is σ-finite in this proof.



          We first construct $f_0$ that holds for $g$ to be simple functions, i.e. we want $f_0$ satisfies
          $$int_Ef dmu=int_E f_0 dmu,mbox{ for any }Einmathcal{A}_0tag{1}$$



          Consider the $sigma$-finite measure space $(X,mathcal{A}_0,mu)$. Let $v$ be a measure defined on this space by $$v(E)=int_Ef dmumbox{ for my }Ein mathcal{A}_0$$



          It is clear that $v$ is a $sigma$-finite measure and $v<<mu$. Radon-Nikodym theorem allows us to
          construct an $mathcal{A}_0$-measurable function $f_0=dfrac{dv}{dmu}ge0$.



          In the case when $gge0$ is $mathcal{A}_0$-measurable, we can find an increasing sequence of $mathcal{A}_0$-simple functions $psi_n$ converges to $g$ pointwise. Then $fpsi _n$ and $f_0psi _n$ monotonically converges to $fg$ and $f_0g$ respectively. By previous case and the Monotone convergence theorem$$int_Xfg d mu=limint_Xfpsi _n dmu=limint_Xf_0psi _n dmu=int_Xf_0g dmu$$



          Therefore, $f_0$ satisfies all the hypotheses. By Radon-Nikodym construction, $f_0$ is unique up to $muupharpoonright mathcal{A}_0$






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            I will add the fact that $mathcal{A}_0$ is $sigma$-finite. So the key was with the Radon-Nikodym Theorem. I'll have to do the reading again, I don't think I remember how to use the Theorem too well.
            $endgroup$
            – Darel
            Jan 23 at 4:45








          • 1




            $begingroup$
            @Darel Yes, the main point is to notice that $mathcal{A}_0$ is $sigma$-finite and to use Radon-Nikodym Theorem.
            $endgroup$
            – Key Flex
            Jan 23 at 4:48












          • $begingroup$
            Just to make this question a little more useful to anyone else reading, do you have any suggested readings?
            $endgroup$
            – Darel
            Jan 23 at 4:49













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          active

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          active

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          3












          $begingroup$

          Notice that the above problem is false if we do not require $(X,mathcal{A}_0,mu)$ to be $sigma$-finite. As a counter-example, take the usual Lebesgue measure on $mathbb{R}$, and $mathcal{A}_0={∅,mathbb{R}}$. Let $g = 1$ and $f =X_{[0,1]}$. Then LHS is $1$ while RHS is either $0$ and $infty ($contradiction $)$. Thus we will assume $(X, mathcal{A}_0, mu)$ is σ-finite in this proof.



          We first construct $f_0$ that holds for $g$ to be simple functions, i.e. we want $f_0$ satisfies
          $$int_Ef dmu=int_E f_0 dmu,mbox{ for any }Einmathcal{A}_0tag{1}$$



          Consider the $sigma$-finite measure space $(X,mathcal{A}_0,mu)$. Let $v$ be a measure defined on this space by $$v(E)=int_Ef dmumbox{ for my }Ein mathcal{A}_0$$



          It is clear that $v$ is a $sigma$-finite measure and $v<<mu$. Radon-Nikodym theorem allows us to
          construct an $mathcal{A}_0$-measurable function $f_0=dfrac{dv}{dmu}ge0$.



          In the case when $gge0$ is $mathcal{A}_0$-measurable, we can find an increasing sequence of $mathcal{A}_0$-simple functions $psi_n$ converges to $g$ pointwise. Then $fpsi _n$ and $f_0psi _n$ monotonically converges to $fg$ and $f_0g$ respectively. By previous case and the Monotone convergence theorem$$int_Xfg d mu=limint_Xfpsi _n dmu=limint_Xf_0psi _n dmu=int_Xf_0g dmu$$



          Therefore, $f_0$ satisfies all the hypotheses. By Radon-Nikodym construction, $f_0$ is unique up to $muupharpoonright mathcal{A}_0$






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            I will add the fact that $mathcal{A}_0$ is $sigma$-finite. So the key was with the Radon-Nikodym Theorem. I'll have to do the reading again, I don't think I remember how to use the Theorem too well.
            $endgroup$
            – Darel
            Jan 23 at 4:45








          • 1




            $begingroup$
            @Darel Yes, the main point is to notice that $mathcal{A}_0$ is $sigma$-finite and to use Radon-Nikodym Theorem.
            $endgroup$
            – Key Flex
            Jan 23 at 4:48












          • $begingroup$
            Just to make this question a little more useful to anyone else reading, do you have any suggested readings?
            $endgroup$
            – Darel
            Jan 23 at 4:49


















          3












          $begingroup$

          Notice that the above problem is false if we do not require $(X,mathcal{A}_0,mu)$ to be $sigma$-finite. As a counter-example, take the usual Lebesgue measure on $mathbb{R}$, and $mathcal{A}_0={∅,mathbb{R}}$. Let $g = 1$ and $f =X_{[0,1]}$. Then LHS is $1$ while RHS is either $0$ and $infty ($contradiction $)$. Thus we will assume $(X, mathcal{A}_0, mu)$ is σ-finite in this proof.



          We first construct $f_0$ that holds for $g$ to be simple functions, i.e. we want $f_0$ satisfies
          $$int_Ef dmu=int_E f_0 dmu,mbox{ for any }Einmathcal{A}_0tag{1}$$



          Consider the $sigma$-finite measure space $(X,mathcal{A}_0,mu)$. Let $v$ be a measure defined on this space by $$v(E)=int_Ef dmumbox{ for my }Ein mathcal{A}_0$$



          It is clear that $v$ is a $sigma$-finite measure and $v<<mu$. Radon-Nikodym theorem allows us to
          construct an $mathcal{A}_0$-measurable function $f_0=dfrac{dv}{dmu}ge0$.



          In the case when $gge0$ is $mathcal{A}_0$-measurable, we can find an increasing sequence of $mathcal{A}_0$-simple functions $psi_n$ converges to $g$ pointwise. Then $fpsi _n$ and $f_0psi _n$ monotonically converges to $fg$ and $f_0g$ respectively. By previous case and the Monotone convergence theorem$$int_Xfg d mu=limint_Xfpsi _n dmu=limint_Xf_0psi _n dmu=int_Xf_0g dmu$$



          Therefore, $f_0$ satisfies all the hypotheses. By Radon-Nikodym construction, $f_0$ is unique up to $muupharpoonright mathcal{A}_0$






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            I will add the fact that $mathcal{A}_0$ is $sigma$-finite. So the key was with the Radon-Nikodym Theorem. I'll have to do the reading again, I don't think I remember how to use the Theorem too well.
            $endgroup$
            – Darel
            Jan 23 at 4:45








          • 1




            $begingroup$
            @Darel Yes, the main point is to notice that $mathcal{A}_0$ is $sigma$-finite and to use Radon-Nikodym Theorem.
            $endgroup$
            – Key Flex
            Jan 23 at 4:48












          • $begingroup$
            Just to make this question a little more useful to anyone else reading, do you have any suggested readings?
            $endgroup$
            – Darel
            Jan 23 at 4:49
















          3












          3








          3





          $begingroup$

          Notice that the above problem is false if we do not require $(X,mathcal{A}_0,mu)$ to be $sigma$-finite. As a counter-example, take the usual Lebesgue measure on $mathbb{R}$, and $mathcal{A}_0={∅,mathbb{R}}$. Let $g = 1$ and $f =X_{[0,1]}$. Then LHS is $1$ while RHS is either $0$ and $infty ($contradiction $)$. Thus we will assume $(X, mathcal{A}_0, mu)$ is σ-finite in this proof.



          We first construct $f_0$ that holds for $g$ to be simple functions, i.e. we want $f_0$ satisfies
          $$int_Ef dmu=int_E f_0 dmu,mbox{ for any }Einmathcal{A}_0tag{1}$$



          Consider the $sigma$-finite measure space $(X,mathcal{A}_0,mu)$. Let $v$ be a measure defined on this space by $$v(E)=int_Ef dmumbox{ for my }Ein mathcal{A}_0$$



          It is clear that $v$ is a $sigma$-finite measure and $v<<mu$. Radon-Nikodym theorem allows us to
          construct an $mathcal{A}_0$-measurable function $f_0=dfrac{dv}{dmu}ge0$.



          In the case when $gge0$ is $mathcal{A}_0$-measurable, we can find an increasing sequence of $mathcal{A}_0$-simple functions $psi_n$ converges to $g$ pointwise. Then $fpsi _n$ and $f_0psi _n$ monotonically converges to $fg$ and $f_0g$ respectively. By previous case and the Monotone convergence theorem$$int_Xfg d mu=limint_Xfpsi _n dmu=limint_Xf_0psi _n dmu=int_Xf_0g dmu$$



          Therefore, $f_0$ satisfies all the hypotheses. By Radon-Nikodym construction, $f_0$ is unique up to $muupharpoonright mathcal{A}_0$






          share|cite|improve this answer









          $endgroup$



          Notice that the above problem is false if we do not require $(X,mathcal{A}_0,mu)$ to be $sigma$-finite. As a counter-example, take the usual Lebesgue measure on $mathbb{R}$, and $mathcal{A}_0={∅,mathbb{R}}$. Let $g = 1$ and $f =X_{[0,1]}$. Then LHS is $1$ while RHS is either $0$ and $infty ($contradiction $)$. Thus we will assume $(X, mathcal{A}_0, mu)$ is σ-finite in this proof.



          We first construct $f_0$ that holds for $g$ to be simple functions, i.e. we want $f_0$ satisfies
          $$int_Ef dmu=int_E f_0 dmu,mbox{ for any }Einmathcal{A}_0tag{1}$$



          Consider the $sigma$-finite measure space $(X,mathcal{A}_0,mu)$. Let $v$ be a measure defined on this space by $$v(E)=int_Ef dmumbox{ for my }Ein mathcal{A}_0$$



          It is clear that $v$ is a $sigma$-finite measure and $v<<mu$. Radon-Nikodym theorem allows us to
          construct an $mathcal{A}_0$-measurable function $f_0=dfrac{dv}{dmu}ge0$.



          In the case when $gge0$ is $mathcal{A}_0$-measurable, we can find an increasing sequence of $mathcal{A}_0$-simple functions $psi_n$ converges to $g$ pointwise. Then $fpsi _n$ and $f_0psi _n$ monotonically converges to $fg$ and $f_0g$ respectively. By previous case and the Monotone convergence theorem$$int_Xfg d mu=limint_Xfpsi _n dmu=limint_Xf_0psi _n dmu=int_Xf_0g dmu$$



          Therefore, $f_0$ satisfies all the hypotheses. By Radon-Nikodym construction, $f_0$ is unique up to $muupharpoonright mathcal{A}_0$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 23 at 4:40









          Key FlexKey Flex

          8,38261233




          8,38261233








          • 1




            $begingroup$
            I will add the fact that $mathcal{A}_0$ is $sigma$-finite. So the key was with the Radon-Nikodym Theorem. I'll have to do the reading again, I don't think I remember how to use the Theorem too well.
            $endgroup$
            – Darel
            Jan 23 at 4:45








          • 1




            $begingroup$
            @Darel Yes, the main point is to notice that $mathcal{A}_0$ is $sigma$-finite and to use Radon-Nikodym Theorem.
            $endgroup$
            – Key Flex
            Jan 23 at 4:48












          • $begingroup$
            Just to make this question a little more useful to anyone else reading, do you have any suggested readings?
            $endgroup$
            – Darel
            Jan 23 at 4:49
















          • 1




            $begingroup$
            I will add the fact that $mathcal{A}_0$ is $sigma$-finite. So the key was with the Radon-Nikodym Theorem. I'll have to do the reading again, I don't think I remember how to use the Theorem too well.
            $endgroup$
            – Darel
            Jan 23 at 4:45








          • 1




            $begingroup$
            @Darel Yes, the main point is to notice that $mathcal{A}_0$ is $sigma$-finite and to use Radon-Nikodym Theorem.
            $endgroup$
            – Key Flex
            Jan 23 at 4:48












          • $begingroup$
            Just to make this question a little more useful to anyone else reading, do you have any suggested readings?
            $endgroup$
            – Darel
            Jan 23 at 4:49










          1




          1




          $begingroup$
          I will add the fact that $mathcal{A}_0$ is $sigma$-finite. So the key was with the Radon-Nikodym Theorem. I'll have to do the reading again, I don't think I remember how to use the Theorem too well.
          $endgroup$
          – Darel
          Jan 23 at 4:45






          $begingroup$
          I will add the fact that $mathcal{A}_0$ is $sigma$-finite. So the key was with the Radon-Nikodym Theorem. I'll have to do the reading again, I don't think I remember how to use the Theorem too well.
          $endgroup$
          – Darel
          Jan 23 at 4:45






          1




          1




          $begingroup$
          @Darel Yes, the main point is to notice that $mathcal{A}_0$ is $sigma$-finite and to use Radon-Nikodym Theorem.
          $endgroup$
          – Key Flex
          Jan 23 at 4:48






          $begingroup$
          @Darel Yes, the main point is to notice that $mathcal{A}_0$ is $sigma$-finite and to use Radon-Nikodym Theorem.
          $endgroup$
          – Key Flex
          Jan 23 at 4:48














          $begingroup$
          Just to make this question a little more useful to anyone else reading, do you have any suggested readings?
          $endgroup$
          – Darel
          Jan 23 at 4:49






          $begingroup$
          Just to make this question a little more useful to anyone else reading, do you have any suggested readings?
          $endgroup$
          – Darel
          Jan 23 at 4:49




















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