Why is it ok to swap modulus and sum in this case?












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$begingroup$


I am trying to solve Exercise 3.21 in Folland's real analysis. I am watching the solution to this question in the following video.



https://www.youtube.com/watch?v=E8FwgGn1e_o



At about 2:31 in the video he says something along the lines of, "we can bring the sum inside the modulus since the sets are disjoint"



That is to say that if $E_1, E_2,...$ are disjoint measurable sets such that $E=cup_{i=1}^{infty}E_i$ then



$sum_{i=1}^{infty}lvert nu (E_i) rvert =sum_{i=1}^{infty}lvert int_{E}chi_{E_i}dnu rvert=lvert int_{E}(sum_{i=1}^{infty}chi_{E_1})dnurvert$. The first equality is fine but the second equality does not seem true to me. At the end of the day those integrals are complex numbers, so it does not seem reasonable that this should be true. Note that $nu$ is a complex measure.






real-analysis measure-theory







share|cite|improve this question









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    1












    $begingroup$


    I am trying to solve Exercise 3.21 in Folland's real analysis. I am watching the solution to this question in the following video.



    https://www.youtube.com/watch?v=E8FwgGn1e_o



    At about 2:31 in the video he says something along the lines of, "we can bring the sum inside the modulus since the sets are disjoint"



    That is to say that if $E_1, E_2,...$ are disjoint measurable sets such that $E=cup_{i=1}^{infty}E_i$ then



    $sum_{i=1}^{infty}lvert nu (E_i) rvert =sum_{i=1}^{infty}lvert int_{E}chi_{E_i}dnu rvert=lvert int_{E}(sum_{i=1}^{infty}chi_{E_1})dnurvert$. The first equality is fine but the second equality does not seem true to me. At the end of the day those integrals are complex numbers, so it does not seem reasonable that this should be true. Note that $nu$ is a complex measure.






    real-analysis measure-theory







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      I am trying to solve Exercise 3.21 in Folland's real analysis. I am watching the solution to this question in the following video.



      https://www.youtube.com/watch?v=E8FwgGn1e_o



      At about 2:31 in the video he says something along the lines of, "we can bring the sum inside the modulus since the sets are disjoint"



      That is to say that if $E_1, E_2,...$ are disjoint measurable sets such that $E=cup_{i=1}^{infty}E_i$ then



      $sum_{i=1}^{infty}lvert nu (E_i) rvert =sum_{i=1}^{infty}lvert int_{E}chi_{E_i}dnu rvert=lvert int_{E}(sum_{i=1}^{infty}chi_{E_1})dnurvert$. The first equality is fine but the second equality does not seem true to me. At the end of the day those integrals are complex numbers, so it does not seem reasonable that this should be true. Note that $nu$ is a complex measure.






      real-analysis measure-theory







      share|cite|improve this question









      $endgroup$




      I am trying to solve Exercise 3.21 in Folland's real analysis. I am watching the solution to this question in the following video.



      https://www.youtube.com/watch?v=E8FwgGn1e_o



      At about 2:31 in the video he says something along the lines of, "we can bring the sum inside the modulus since the sets are disjoint"



      That is to say that if $E_1, E_2,...$ are disjoint measurable sets such that $E=cup_{i=1}^{infty}E_i$ then



      $sum_{i=1}^{infty}lvert nu (E_i) rvert =sum_{i=1}^{infty}lvert int_{E}chi_{E_i}dnu rvert=lvert int_{E}(sum_{i=1}^{infty}chi_{E_1})dnurvert$. The first equality is fine but the second equality does not seem true to me. At the end of the day those integrals are complex numbers, so it does not seem reasonable that this should be true. Note that $nu$ is a complex measure.






      real-analysis measure-theory




      real-analysis measure-theory






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 6 '18 at 22:14









      DamoDamo

      491210




      491210






















          1 Answer
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          $begingroup$

          The reasoning is definitely not correct, or at least not worded properly. But we can amend this as follows:



          Given $E_1,E_2,ldots$ disjoint with $E=cup_jE_j$, define $g:Xtomathbb C$ by
          $$g=sum_{j=1}^inftyoverline{operatorname{sgn}(nu(E_j))}chi_{E_j}.$$
          Then $|g|leq1$, and
          $$sum_{j=1}^infty|nu(E_j)|=left|int_Eg dnuright|leqsupleft{left|int f dnuright|:|f|leq 1right}.$$
          Taking the supremum over all such partitions ${E_j}$ of $E$, we obtain $mu_2(E)leqmu_3(E)$.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            I think when you define $g$ is should involve the conjugate of $operatorname{sgn(nu(E_j))}$
            $endgroup$
            – Damo
            Dec 7 '18 at 1:18






          • 1




            $begingroup$
            Also, for anyone reading this, one will need to invoke dominated convergence theorem to swap the integral and the sum. It will also involve splitting the complex measure into the positive and negative components of the real and imaginary parts.
            $endgroup$
            – Damo
            Dec 7 '18 at 1:24










          • $begingroup$
            You're completely correct, thanks for pointing that out.
            $endgroup$
            – Aweygan
            Dec 7 '18 at 1:42











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          1 Answer
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          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          The reasoning is definitely not correct, or at least not worded properly. But we can amend this as follows:



          Given $E_1,E_2,ldots$ disjoint with $E=cup_jE_j$, define $g:Xtomathbb C$ by
          $$g=sum_{j=1}^inftyoverline{operatorname{sgn}(nu(E_j))}chi_{E_j}.$$
          Then $|g|leq1$, and
          $$sum_{j=1}^infty|nu(E_j)|=left|int_Eg dnuright|leqsupleft{left|int f dnuright|:|f|leq 1right}.$$
          Taking the supremum over all such partitions ${E_j}$ of $E$, we obtain $mu_2(E)leqmu_3(E)$.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            I think when you define $g$ is should involve the conjugate of $operatorname{sgn(nu(E_j))}$
            $endgroup$
            – Damo
            Dec 7 '18 at 1:18






          • 1




            $begingroup$
            Also, for anyone reading this, one will need to invoke dominated convergence theorem to swap the integral and the sum. It will also involve splitting the complex measure into the positive and negative components of the real and imaginary parts.
            $endgroup$
            – Damo
            Dec 7 '18 at 1:24










          • $begingroup$
            You're completely correct, thanks for pointing that out.
            $endgroup$
            – Aweygan
            Dec 7 '18 at 1:42
















          1












          $begingroup$

          The reasoning is definitely not correct, or at least not worded properly. But we can amend this as follows:



          Given $E_1,E_2,ldots$ disjoint with $E=cup_jE_j$, define $g:Xtomathbb C$ by
          $$g=sum_{j=1}^inftyoverline{operatorname{sgn}(nu(E_j))}chi_{E_j}.$$
          Then $|g|leq1$, and
          $$sum_{j=1}^infty|nu(E_j)|=left|int_Eg dnuright|leqsupleft{left|int f dnuright|:|f|leq 1right}.$$
          Taking the supremum over all such partitions ${E_j}$ of $E$, we obtain $mu_2(E)leqmu_3(E)$.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            I think when you define $g$ is should involve the conjugate of $operatorname{sgn(nu(E_j))}$
            $endgroup$
            – Damo
            Dec 7 '18 at 1:18






          • 1




            $begingroup$
            Also, for anyone reading this, one will need to invoke dominated convergence theorem to swap the integral and the sum. It will also involve splitting the complex measure into the positive and negative components of the real and imaginary parts.
            $endgroup$
            – Damo
            Dec 7 '18 at 1:24










          • $begingroup$
            You're completely correct, thanks for pointing that out.
            $endgroup$
            – Aweygan
            Dec 7 '18 at 1:42














          1












          1








          1





          $begingroup$

          The reasoning is definitely not correct, or at least not worded properly. But we can amend this as follows:



          Given $E_1,E_2,ldots$ disjoint with $E=cup_jE_j$, define $g:Xtomathbb C$ by
          $$g=sum_{j=1}^inftyoverline{operatorname{sgn}(nu(E_j))}chi_{E_j}.$$
          Then $|g|leq1$, and
          $$sum_{j=1}^infty|nu(E_j)|=left|int_Eg dnuright|leqsupleft{left|int f dnuright|:|f|leq 1right}.$$
          Taking the supremum over all such partitions ${E_j}$ of $E$, we obtain $mu_2(E)leqmu_3(E)$.






          share|cite|improve this answer











          $endgroup$



          The reasoning is definitely not correct, or at least not worded properly. But we can amend this as follows:



          Given $E_1,E_2,ldots$ disjoint with $E=cup_jE_j$, define $g:Xtomathbb C$ by
          $$g=sum_{j=1}^inftyoverline{operatorname{sgn}(nu(E_j))}chi_{E_j}.$$
          Then $|g|leq1$, and
          $$sum_{j=1}^infty|nu(E_j)|=left|int_Eg dnuright|leqsupleft{left|int f dnuright|:|f|leq 1right}.$$
          Taking the supremum over all such partitions ${E_j}$ of $E$, we obtain $mu_2(E)leqmu_3(E)$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 7 '18 at 1:41

























          answered Dec 6 '18 at 22:31









          AweyganAweygan

          14.5k21442




          14.5k21442








          • 1




            $begingroup$
            I think when you define $g$ is should involve the conjugate of $operatorname{sgn(nu(E_j))}$
            $endgroup$
            – Damo
            Dec 7 '18 at 1:18






          • 1




            $begingroup$
            Also, for anyone reading this, one will need to invoke dominated convergence theorem to swap the integral and the sum. It will also involve splitting the complex measure into the positive and negative components of the real and imaginary parts.
            $endgroup$
            – Damo
            Dec 7 '18 at 1:24










          • $begingroup$
            You're completely correct, thanks for pointing that out.
            $endgroup$
            – Aweygan
            Dec 7 '18 at 1:42














          • 1




            $begingroup$
            I think when you define $g$ is should involve the conjugate of $operatorname{sgn(nu(E_j))}$
            $endgroup$
            – Damo
            Dec 7 '18 at 1:18






          • 1




            $begingroup$
            Also, for anyone reading this, one will need to invoke dominated convergence theorem to swap the integral and the sum. It will also involve splitting the complex measure into the positive and negative components of the real and imaginary parts.
            $endgroup$
            – Damo
            Dec 7 '18 at 1:24










          • $begingroup$
            You're completely correct, thanks for pointing that out.
            $endgroup$
            – Aweygan
            Dec 7 '18 at 1:42








          1




          1




          $begingroup$
          I think when you define $g$ is should involve the conjugate of $operatorname{sgn(nu(E_j))}$
          $endgroup$
          – Damo
          Dec 7 '18 at 1:18




          $begingroup$
          I think when you define $g$ is should involve the conjugate of $operatorname{sgn(nu(E_j))}$
          $endgroup$
          – Damo
          Dec 7 '18 at 1:18




          1




          1




          $begingroup$
          Also, for anyone reading this, one will need to invoke dominated convergence theorem to swap the integral and the sum. It will also involve splitting the complex measure into the positive and negative components of the real and imaginary parts.
          $endgroup$
          – Damo
          Dec 7 '18 at 1:24




          $begingroup$
          Also, for anyone reading this, one will need to invoke dominated convergence theorem to swap the integral and the sum. It will also involve splitting the complex measure into the positive and negative components of the real and imaginary parts.
          $endgroup$
          – Damo
          Dec 7 '18 at 1:24












          $begingroup$
          You're completely correct, thanks for pointing that out.
          $endgroup$
          – Aweygan
          Dec 7 '18 at 1:42




          $begingroup$
          You're completely correct, thanks for pointing that out.
          $endgroup$
          – Aweygan
          Dec 7 '18 at 1:42


















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