Showing a measure is countably additive if it's finitely additive












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$begingroup$


Let $mu$ be a finitely additive nonnegative set function on some measurable space $(Omega, mathcal{F})$ with the continuity property $$B_nin mathcal{F},, B_ndownarrowemptyset, , mu(B_n)<inftyRightarrow mu(B_n)to 0 tag{$*$} $$



is countably additive when $mu(Omega)<infty$.



I am having difficulty using the property to show $mu$ is countably additive. I would like to prove for any disjoint collection of sets $(A_i)$, $lim mu (bigcup _{i=1} ^n A_n)=mu(bigcup _{i=1} ^infty A_n)$. WLOG, (by rearranging indices) suppose $(A_n)$ is decreasing. Then $A_ndownarrow emptyset$. By $(*)$ $mu(A_n)to 0$. I don't see how this helps or how this implies $sum_i mu (A_i)<infty$.




  1. How can I show $mu$ is countably additive?

  2. Doesn't the condition that $B_ndownarrow emptyset$ automatically imply $mu(B_n)to 0$?






measure-theory continuity







share|cite|improve this question











$endgroup$

















    3












    $begingroup$


    Let $mu$ be a finitely additive nonnegative set function on some measurable space $(Omega, mathcal{F})$ with the continuity property $$B_nin mathcal{F},, B_ndownarrowemptyset, , mu(B_n)<inftyRightarrow mu(B_n)to 0 tag{$*$} $$



    is countably additive when $mu(Omega)<infty$.



    I am having difficulty using the property to show $mu$ is countably additive. I would like to prove for any disjoint collection of sets $(A_i)$, $lim mu (bigcup _{i=1} ^n A_n)=mu(bigcup _{i=1} ^infty A_n)$. WLOG, (by rearranging indices) suppose $(A_n)$ is decreasing. Then $A_ndownarrow emptyset$. By $(*)$ $mu(A_n)to 0$. I don't see how this helps or how this implies $sum_i mu (A_i)<infty$.




    1. How can I show $mu$ is countably additive?

    2. Doesn't the condition that $B_ndownarrow emptyset$ automatically imply $mu(B_n)to 0$?






    measure-theory continuity







    share|cite|improve this question











    $endgroup$















      3












      3








      3


      0



      $begingroup$


      Let $mu$ be a finitely additive nonnegative set function on some measurable space $(Omega, mathcal{F})$ with the continuity property $$B_nin mathcal{F},, B_ndownarrowemptyset, , mu(B_n)<inftyRightarrow mu(B_n)to 0 tag{$*$} $$



      is countably additive when $mu(Omega)<infty$.



      I am having difficulty using the property to show $mu$ is countably additive. I would like to prove for any disjoint collection of sets $(A_i)$, $lim mu (bigcup _{i=1} ^n A_n)=mu(bigcup _{i=1} ^infty A_n)$. WLOG, (by rearranging indices) suppose $(A_n)$ is decreasing. Then $A_ndownarrow emptyset$. By $(*)$ $mu(A_n)to 0$. I don't see how this helps or how this implies $sum_i mu (A_i)<infty$.




      1. How can I show $mu$ is countably additive?

      2. Doesn't the condition that $B_ndownarrow emptyset$ automatically imply $mu(B_n)to 0$?






      measure-theory continuity







      share|cite|improve this question











      $endgroup$




      Let $mu$ be a finitely additive nonnegative set function on some measurable space $(Omega, mathcal{F})$ with the continuity property $$B_nin mathcal{F},, B_ndownarrowemptyset, , mu(B_n)<inftyRightarrow mu(B_n)to 0 tag{$*$} $$



      is countably additive when $mu(Omega)<infty$.



      I am having difficulty using the property to show $mu$ is countably additive. I would like to prove for any disjoint collection of sets $(A_i)$, $lim mu (bigcup _{i=1} ^n A_n)=mu(bigcup _{i=1} ^infty A_n)$. WLOG, (by rearranging indices) suppose $(A_n)$ is decreasing. Then $A_ndownarrow emptyset$. By $(*)$ $mu(A_n)to 0$. I don't see how this helps or how this implies $sum_i mu (A_i)<infty$.




      1. How can I show $mu$ is countably additive?

      2. Doesn't the condition that $B_ndownarrow emptyset$ automatically imply $mu(B_n)to 0$?






      measure-theory continuity




      measure-theory continuity






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Sep 27 '15 at 8:21







      cap

















      asked Sep 27 '15 at 8:07









      capcap

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      7183926






















          2 Answers
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          5












          $begingroup$

          Hint: note that
          $$cup_0^infty A_n = cup_0^N A_n cup left( cup_{N+1}^infty A_nright) $$
          is a finite union and that you can apply the assumption to the sequence $(cup_{N+1}^infty A_n)_N$






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$


            1. Take any ${E_i}$ a sequence of disjoint sets in $mathcal{F}$, and $Etriangleqbigcuplimits_{i=1}^{infty}E_i$. We have to show that $mu(E) = sumlimits_{i=1}^{infty}mu(E_i)$.
              Consider the sets $F_n = bigcuplimits_{i=1}^n E_i$. You can observe that the sequence $(Esetminus F_n)downarrowemptyset$. Note that the finiteness of $(Esetminus F_n)$ for every $n$ follows from the fact that $mu(Omega)<infty$.
              Now since $mu$ is finitely additive $mu(F_n)=sumlimits_{i=1}^n mu(E_i)$ .
              $$ mu(E)=mu(Esetminus F_n)+sumlimits_{i=1}^n mu(E_i)$$
              Take limits on both the sides;
              $$mu(E)=lim_{ntoinfty}mu(Esetminus F_n) + sumlimits_{i=1}^infty mu(E_i)$$
              By the property of $mu$ mentioned in the problem, $limlimits_{ntoinfty}mu(Esetminus F_n)=0$. And hence $mu(E)=sumlimits_{i=1}^infty mu(E_i)$.


            2. For your second doubt assume the measure (you can verify it is indeed a measure) $mu:mathcal{P}(mathbb{R})mapstomathbb{R}^*$ defined as
              $$mu(E)=begin{cases}
              infty & text{if $E$ is infinite}\n &text{if $E$ is contains n elements }end{cases}$$
              Now consider the sets $F_n triangleqleft(0,frac{1}{n}right]$. Note that though $F_ndownarrowemptyset$, $limlimits_{ntoinfty}
              mu(F_n)=infty$              .







            share|cite|improve this answer











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              2 Answers
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              2 Answers
              2






              active

              oldest

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              active

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              votes






              active

              oldest

              votes









              5












              $begingroup$

              Hint: note that
              $$cup_0^infty A_n = cup_0^N A_n cup left( cup_{N+1}^infty A_nright) $$
              is a finite union and that you can apply the assumption to the sequence $(cup_{N+1}^infty A_n)_N$






              share|cite|improve this answer









              $endgroup$


















                5












                $begingroup$

                Hint: note that
                $$cup_0^infty A_n = cup_0^N A_n cup left( cup_{N+1}^infty A_nright) $$
                is a finite union and that you can apply the assumption to the sequence $(cup_{N+1}^infty A_n)_N$






                share|cite|improve this answer









                $endgroup$
















                  5












                  5








                  5





                  $begingroup$

                  Hint: note that
                  $$cup_0^infty A_n = cup_0^N A_n cup left( cup_{N+1}^infty A_nright) $$
                  is a finite union and that you can apply the assumption to the sequence $(cup_{N+1}^infty A_n)_N$






                  share|cite|improve this answer









                  $endgroup$



                  Hint: note that
                  $$cup_0^infty A_n = cup_0^N A_n cup left( cup_{N+1}^infty A_nright) $$
                  is a finite union and that you can apply the assumption to the sequence $(cup_{N+1}^infty A_n)_N$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Sep 27 '15 at 8:24









                  ThomasThomas

                  16.9k21731




                  16.9k21731























                      1












                      $begingroup$


                      1. Take any ${E_i}$ a sequence of disjoint sets in $mathcal{F}$, and $Etriangleqbigcuplimits_{i=1}^{infty}E_i$. We have to show that $mu(E) = sumlimits_{i=1}^{infty}mu(E_i)$.
                        Consider the sets $F_n = bigcuplimits_{i=1}^n E_i$. You can observe that the sequence $(Esetminus F_n)downarrowemptyset$. Note that the finiteness of $(Esetminus F_n)$ for every $n$ follows from the fact that $mu(Omega)<infty$.
                        Now since $mu$ is finitely additive $mu(F_n)=sumlimits_{i=1}^n mu(E_i)$ .
                        $$ mu(E)=mu(Esetminus F_n)+sumlimits_{i=1}^n mu(E_i)$$
                        Take limits on both the sides;
                        $$mu(E)=lim_{ntoinfty}mu(Esetminus F_n) + sumlimits_{i=1}^infty mu(E_i)$$
                        By the property of $mu$ mentioned in the problem, $limlimits_{ntoinfty}mu(Esetminus F_n)=0$. And hence $mu(E)=sumlimits_{i=1}^infty mu(E_i)$.


                      2. For your second doubt assume the measure (you can verify it is indeed a measure) $mu:mathcal{P}(mathbb{R})mapstomathbb{R}^*$ defined as
                        $$mu(E)=begin{cases}
                        infty & text{if $E$ is infinite}\n &text{if $E$ is contains n elements }end{cases}$$
                        Now consider the sets $F_n triangleqleft(0,frac{1}{n}right]$. Note that though $F_ndownarrowemptyset$, $limlimits_{ntoinfty}
                        mu(F_n)=infty$                        .







                      share|cite|improve this answer











                      $endgroup$


















                        1












                        $begingroup$


                        1. Take any ${E_i}$ a sequence of disjoint sets in $mathcal{F}$, and $Etriangleqbigcuplimits_{i=1}^{infty}E_i$. We have to show that $mu(E) = sumlimits_{i=1}^{infty}mu(E_i)$.
                          Consider the sets $F_n = bigcuplimits_{i=1}^n E_i$. You can observe that the sequence $(Esetminus F_n)downarrowemptyset$. Note that the finiteness of $(Esetminus F_n)$ for every $n$ follows from the fact that $mu(Omega)<infty$.
                          Now since $mu$ is finitely additive $mu(F_n)=sumlimits_{i=1}^n mu(E_i)$ .
                          $$ mu(E)=mu(Esetminus F_n)+sumlimits_{i=1}^n mu(E_i)$$
                          Take limits on both the sides;
                          $$mu(E)=lim_{ntoinfty}mu(Esetminus F_n) + sumlimits_{i=1}^infty mu(E_i)$$
                          By the property of $mu$ mentioned in the problem, $limlimits_{ntoinfty}mu(Esetminus F_n)=0$. And hence $mu(E)=sumlimits_{i=1}^infty mu(E_i)$.


                        2. For your second doubt assume the measure (you can verify it is indeed a measure) $mu:mathcal{P}(mathbb{R})mapstomathbb{R}^*$ defined as
                          $$mu(E)=begin{cases}
                          infty & text{if $E$ is infinite}\n &text{if $E$ is contains n elements }end{cases}$$
                          Now consider the sets $F_n triangleqleft(0,frac{1}{n}right]$. Note that though $F_ndownarrowemptyset$, $limlimits_{ntoinfty}
                          mu(F_n)=infty$                          .







                        share|cite|improve this answer











                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$


                          1. Take any ${E_i}$ a sequence of disjoint sets in $mathcal{F}$, and $Etriangleqbigcuplimits_{i=1}^{infty}E_i$. We have to show that $mu(E) = sumlimits_{i=1}^{infty}mu(E_i)$.
                            Consider the sets $F_n = bigcuplimits_{i=1}^n E_i$. You can observe that the sequence $(Esetminus F_n)downarrowemptyset$. Note that the finiteness of $(Esetminus F_n)$ for every $n$ follows from the fact that $mu(Omega)<infty$.
                            Now since $mu$ is finitely additive $mu(F_n)=sumlimits_{i=1}^n mu(E_i)$ .
                            $$ mu(E)=mu(Esetminus F_n)+sumlimits_{i=1}^n mu(E_i)$$
                            Take limits on both the sides;
                            $$mu(E)=lim_{ntoinfty}mu(Esetminus F_n) + sumlimits_{i=1}^infty mu(E_i)$$
                            By the property of $mu$ mentioned in the problem, $limlimits_{ntoinfty}mu(Esetminus F_n)=0$. And hence $mu(E)=sumlimits_{i=1}^infty mu(E_i)$.


                          2. For your second doubt assume the measure (you can verify it is indeed a measure) $mu:mathcal{P}(mathbb{R})mapstomathbb{R}^*$ defined as
                            $$mu(E)=begin{cases}
                            infty & text{if $E$ is infinite}\n &text{if $E$ is contains n elements }end{cases}$$
                            Now consider the sets $F_n triangleqleft(0,frac{1}{n}right]$. Note that though $F_ndownarrowemptyset$, $limlimits_{ntoinfty}
                            mu(F_n)=infty$                            .







                          share|cite|improve this answer











                          $endgroup$




                          1. Take any ${E_i}$ a sequence of disjoint sets in $mathcal{F}$, and $Etriangleqbigcuplimits_{i=1}^{infty}E_i$. We have to show that $mu(E) = sumlimits_{i=1}^{infty}mu(E_i)$.
                            Consider the sets $F_n = bigcuplimits_{i=1}^n E_i$. You can observe that the sequence $(Esetminus F_n)downarrowemptyset$. Note that the finiteness of $(Esetminus F_n)$ for every $n$ follows from the fact that $mu(Omega)<infty$.
                            Now since $mu$ is finitely additive $mu(F_n)=sumlimits_{i=1}^n mu(E_i)$ .
                            $$ mu(E)=mu(Esetminus F_n)+sumlimits_{i=1}^n mu(E_i)$$
                            Take limits on both the sides;
                            $$mu(E)=lim_{ntoinfty}mu(Esetminus F_n) + sumlimits_{i=1}^infty mu(E_i)$$
                            By the property of $mu$ mentioned in the problem, $limlimits_{ntoinfty}mu(Esetminus F_n)=0$. And hence $mu(E)=sumlimits_{i=1}^infty mu(E_i)$.


                          2. For your second doubt assume the measure (you can verify it is indeed a measure) $mu:mathcal{P}(mathbb{R})mapstomathbb{R}^*$ defined as
                            $$mu(E)=begin{cases}
                            infty & text{if $E$ is infinite}\n &text{if $E$ is contains n elements }end{cases}$$
                            Now consider the sets $F_n triangleqleft(0,frac{1}{n}right]$. Note that though $F_ndownarrowemptyset$, $limlimits_{ntoinfty}
                            mu(F_n)=infty$                            .








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                          edited Jan 17 at 7:36









                          user128949

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                          answered Mar 31 '18 at 13:22









                          Soham ChatterjeeSoham Chatterjee

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                          134






























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