Show that $mu(bigcaplimits_n=1^inftyF_n)=limmu(F_n)$ can fail if $mu(F_1) = +infty$ [closed]

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Show that lemma 3.4 (b) in The Elements of Integration by Bartle may fail if the finiteness condition $mu(F_1)<+infty$ is dropped.




Lemma 3.4(b): Let $mu$ be a measure defined on a $sigma$-algebra $X$. If $(F_n)$ is decreasing sequence in $X$ and if $mu(F_1)<+infty$, then
$$mu Bigg(bigcaplimits_n=1^inftyF_nBigg)=limmu(F_n)$$











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closed as off-topic by Theoretical Economist, amWhy, Xander Henderson, Bungo, Jose Arnaldo Bebita Dris Sep 6 at 5:58


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Theoretical Economist, amWhy, Xander Henderson
If this question can be reworded to fit the rules in the help center, please edit the question.












  • I think there should be an intersection there...
    – Eduardo Longa
    Sep 6 at 1:26






  • 1




    Hello Max, here in math exchange you should explain more what thoughts did you have until here and try to show what you have done yet.
    – Robson
    Sep 6 at 3:57






  • 1




    Possible duplicate of Example of decreasing sequence of sets with first set having infinite measure
    – Bungo
    Sep 6 at 5:55














up vote
1
down vote

favorite












Show that lemma 3.4 (b) in The Elements of Integration by Bartle may fail if the finiteness condition $mu(F_1)<+infty$ is dropped.




Lemma 3.4(b): Let $mu$ be a measure defined on a $sigma$-algebra $X$. If $(F_n)$ is decreasing sequence in $X$ and if $mu(F_1)<+infty$, then
$$mu Bigg(bigcaplimits_n=1^inftyF_nBigg)=limmu(F_n)$$











share|cite|improve this question















closed as off-topic by Theoretical Economist, amWhy, Xander Henderson, Bungo, Jose Arnaldo Bebita Dris Sep 6 at 5:58


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Theoretical Economist, amWhy, Xander Henderson
If this question can be reworded to fit the rules in the help center, please edit the question.












  • I think there should be an intersection there...
    – Eduardo Longa
    Sep 6 at 1:26






  • 1




    Hello Max, here in math exchange you should explain more what thoughts did you have until here and try to show what you have done yet.
    – Robson
    Sep 6 at 3:57






  • 1




    Possible duplicate of Example of decreasing sequence of sets with first set having infinite measure
    – Bungo
    Sep 6 at 5:55












up vote
1
down vote

favorite









up vote
1
down vote

favorite











Show that lemma 3.4 (b) in The Elements of Integration by Bartle may fail if the finiteness condition $mu(F_1)<+infty$ is dropped.




Lemma 3.4(b): Let $mu$ be a measure defined on a $sigma$-algebra $X$. If $(F_n)$ is decreasing sequence in $X$ and if $mu(F_1)<+infty$, then
$$mu Bigg(bigcaplimits_n=1^inftyF_nBigg)=limmu(F_n)$$











share|cite|improve this question















Show that lemma 3.4 (b) in The Elements of Integration by Bartle may fail if the finiteness condition $mu(F_1)<+infty$ is dropped.




Lemma 3.4(b): Let $mu$ be a measure defined on a $sigma$-algebra $X$. If $(F_n)$ is decreasing sequence in $X$ and if $mu(F_1)<+infty$, then
$$mu Bigg(bigcaplimits_n=1^inftyF_nBigg)=limmu(F_n)$$








measure-theory lebesgue-measure






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edited Sep 6 at 5:50









Xander Henderson

13.3k83250




13.3k83250










asked Sep 5 at 23:01









Max_Quecano

214




214




closed as off-topic by Theoretical Economist, amWhy, Xander Henderson, Bungo, Jose Arnaldo Bebita Dris Sep 6 at 5:58


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Theoretical Economist, amWhy, Xander Henderson
If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by Theoretical Economist, amWhy, Xander Henderson, Bungo, Jose Arnaldo Bebita Dris Sep 6 at 5:58


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Theoretical Economist, amWhy, Xander Henderson
If this question can be reworded to fit the rules in the help center, please edit the question.











  • I think there should be an intersection there...
    – Eduardo Longa
    Sep 6 at 1:26






  • 1




    Hello Max, here in math exchange you should explain more what thoughts did you have until here and try to show what you have done yet.
    – Robson
    Sep 6 at 3:57






  • 1




    Possible duplicate of Example of decreasing sequence of sets with first set having infinite measure
    – Bungo
    Sep 6 at 5:55
















  • I think there should be an intersection there...
    – Eduardo Longa
    Sep 6 at 1:26






  • 1




    Hello Max, here in math exchange you should explain more what thoughts did you have until here and try to show what you have done yet.
    – Robson
    Sep 6 at 3:57






  • 1




    Possible duplicate of Example of decreasing sequence of sets with first set having infinite measure
    – Bungo
    Sep 6 at 5:55















I think there should be an intersection there...
– Eduardo Longa
Sep 6 at 1:26




I think there should be an intersection there...
– Eduardo Longa
Sep 6 at 1:26




1




1




Hello Max, here in math exchange you should explain more what thoughts did you have until here and try to show what you have done yet.
– Robson
Sep 6 at 3:57




Hello Max, here in math exchange you should explain more what thoughts did you have until here and try to show what you have done yet.
– Robson
Sep 6 at 3:57




1




1




Possible duplicate of Example of decreasing sequence of sets with first set having infinite measure
– Bungo
Sep 6 at 5:55




Possible duplicate of Example of decreasing sequence of sets with first set having infinite measure
– Bungo
Sep 6 at 5:55










2 Answers
2






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1
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You can see that in the proof of Lema 3.4(b) when he defines the sequence of sets $E_n=F_1 - F_n $ , you can't say that $mu (E_n) = mu (F_1) - mu (F_n)$ if $mu(F_1)$ is not finite.



This could get you in trouble because expression like $infty - infty$ aren't well defined in this context.



For a counter-example, you could use $X=mathbbR$ as the set; the power set $mathcalP(mathbbR)$ as the $sigma$-álgebra and the following measure:



$$mu (E)=begincases vert E vert mbox, if $E$ is finite\
+inftymbox, otherwise
endcases$$



Then use the sequence $F_n=(-frac1n,frac1n)$.



$mu(bigcaplimits _n=1^inftyF_n)=mu (0)=1 $, but $limmu(F_n)=+infty$.






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  • 1




    Thank you sou much
    – Max_Quecano
    Sep 6 at 5:57










  • Not begging for upvote but if you are already satisfied with the answers here you can mark the answer as correct and upvote it if you want to...
    – Robson
    Sep 6 at 5:59










  • You're welcome! Hope you enjoy math stackexchange!
    – Robson
    Sep 6 at 5:59


















up vote
0
down vote













As Robson said, please include your thoughts on the question. You'll get better responses if we know what you are thinking already.



However, I thought the question was interesting, so:



Hint: Can any of the sets have finite measure if the lemma is to fail?






share|cite|improve this answer



























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    1
    down vote



    accepted










    You can see that in the proof of Lema 3.4(b) when he defines the sequence of sets $E_n=F_1 - F_n $ , you can't say that $mu (E_n) = mu (F_1) - mu (F_n)$ if $mu(F_1)$ is not finite.



    This could get you in trouble because expression like $infty - infty$ aren't well defined in this context.



    For a counter-example, you could use $X=mathbbR$ as the set; the power set $mathcalP(mathbbR)$ as the $sigma$-álgebra and the following measure:



    $$mu (E)=begincases vert E vert mbox, if $E$ is finite\
    +inftymbox, otherwise
    endcases$$



    Then use the sequence $F_n=(-frac1n,frac1n)$.



    $mu(bigcaplimits _n=1^inftyF_n)=mu (0)=1 $, but $limmu(F_n)=+infty$.






    share|cite|improve this answer


















    • 1




      Thank you sou much
      – Max_Quecano
      Sep 6 at 5:57










    • Not begging for upvote but if you are already satisfied with the answers here you can mark the answer as correct and upvote it if you want to...
      – Robson
      Sep 6 at 5:59










    • You're welcome! Hope you enjoy math stackexchange!
      – Robson
      Sep 6 at 5:59















    up vote
    1
    down vote



    accepted










    You can see that in the proof of Lema 3.4(b) when he defines the sequence of sets $E_n=F_1 - F_n $ , you can't say that $mu (E_n) = mu (F_1) - mu (F_n)$ if $mu(F_1)$ is not finite.



    This could get you in trouble because expression like $infty - infty$ aren't well defined in this context.



    For a counter-example, you could use $X=mathbbR$ as the set; the power set $mathcalP(mathbbR)$ as the $sigma$-álgebra and the following measure:



    $$mu (E)=begincases vert E vert mbox, if $E$ is finite\
    +inftymbox, otherwise
    endcases$$



    Then use the sequence $F_n=(-frac1n,frac1n)$.



    $mu(bigcaplimits _n=1^inftyF_n)=mu (0)=1 $, but $limmu(F_n)=+infty$.






    share|cite|improve this answer


















    • 1




      Thank you sou much
      – Max_Quecano
      Sep 6 at 5:57










    • Not begging for upvote but if you are already satisfied with the answers here you can mark the answer as correct and upvote it if you want to...
      – Robson
      Sep 6 at 5:59










    • You're welcome! Hope you enjoy math stackexchange!
      – Robson
      Sep 6 at 5:59













    up vote
    1
    down vote



    accepted







    up vote
    1
    down vote



    accepted






    You can see that in the proof of Lema 3.4(b) when he defines the sequence of sets $E_n=F_1 - F_n $ , you can't say that $mu (E_n) = mu (F_1) - mu (F_n)$ if $mu(F_1)$ is not finite.



    This could get you in trouble because expression like $infty - infty$ aren't well defined in this context.



    For a counter-example, you could use $X=mathbbR$ as the set; the power set $mathcalP(mathbbR)$ as the $sigma$-álgebra and the following measure:



    $$mu (E)=begincases vert E vert mbox, if $E$ is finite\
    +inftymbox, otherwise
    endcases$$



    Then use the sequence $F_n=(-frac1n,frac1n)$.



    $mu(bigcaplimits _n=1^inftyF_n)=mu (0)=1 $, but $limmu(F_n)=+infty$.






    share|cite|improve this answer














    You can see that in the proof of Lema 3.4(b) when he defines the sequence of sets $E_n=F_1 - F_n $ , you can't say that $mu (E_n) = mu (F_1) - mu (F_n)$ if $mu(F_1)$ is not finite.



    This could get you in trouble because expression like $infty - infty$ aren't well defined in this context.



    For a counter-example, you could use $X=mathbbR$ as the set; the power set $mathcalP(mathbbR)$ as the $sigma$-álgebra and the following measure:



    $$mu (E)=begincases vert E vert mbox, if $E$ is finite\
    +inftymbox, otherwise
    endcases$$



    Then use the sequence $F_n=(-frac1n,frac1n)$.



    $mu(bigcaplimits _n=1^inftyF_n)=mu (0)=1 $, but $limmu(F_n)=+infty$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Sep 6 at 4:45

























    answered Sep 6 at 4:30









    Robson

    47920




    47920







    • 1




      Thank you sou much
      – Max_Quecano
      Sep 6 at 5:57










    • Not begging for upvote but if you are already satisfied with the answers here you can mark the answer as correct and upvote it if you want to...
      – Robson
      Sep 6 at 5:59










    • You're welcome! Hope you enjoy math stackexchange!
      – Robson
      Sep 6 at 5:59













    • 1




      Thank you sou much
      – Max_Quecano
      Sep 6 at 5:57










    • Not begging for upvote but if you are already satisfied with the answers here you can mark the answer as correct and upvote it if you want to...
      – Robson
      Sep 6 at 5:59










    • You're welcome! Hope you enjoy math stackexchange!
      – Robson
      Sep 6 at 5:59








    1




    1




    Thank you sou much
    – Max_Quecano
    Sep 6 at 5:57




    Thank you sou much
    – Max_Quecano
    Sep 6 at 5:57












    Not begging for upvote but if you are already satisfied with the answers here you can mark the answer as correct and upvote it if you want to...
    – Robson
    Sep 6 at 5:59




    Not begging for upvote but if you are already satisfied with the answers here you can mark the answer as correct and upvote it if you want to...
    – Robson
    Sep 6 at 5:59












    You're welcome! Hope you enjoy math stackexchange!
    – Robson
    Sep 6 at 5:59





    You're welcome! Hope you enjoy math stackexchange!
    – Robson
    Sep 6 at 5:59











    up vote
    0
    down vote













    As Robson said, please include your thoughts on the question. You'll get better responses if we know what you are thinking already.



    However, I thought the question was interesting, so:



    Hint: Can any of the sets have finite measure if the lemma is to fail?






    share|cite|improve this answer
























      up vote
      0
      down vote













      As Robson said, please include your thoughts on the question. You'll get better responses if we know what you are thinking already.



      However, I thought the question was interesting, so:



      Hint: Can any of the sets have finite measure if the lemma is to fail?






      share|cite|improve this answer






















        up vote
        0
        down vote










        up vote
        0
        down vote









        As Robson said, please include your thoughts on the question. You'll get better responses if we know what you are thinking already.



        However, I thought the question was interesting, so:



        Hint: Can any of the sets have finite measure if the lemma is to fail?






        share|cite|improve this answer












        As Robson said, please include your thoughts on the question. You'll get better responses if we know what you are thinking already.



        However, I thought the question was interesting, so:



        Hint: Can any of the sets have finite measure if the lemma is to fail?







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Sep 6 at 4:02









        jgon

        8,57011536




        8,57011536












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