Completion of measure spaces - uniqueness

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
4
down vote

favorite













If $(X,mathcalA,mu)$ is a measure space,
$(X,mathcalB,overlinemu)$ is complete,
$mathcalBsupsetmathcalA$, and $overlinemu(A)=mu(A)$ for
every $AinmathcalA$, is $(X,mathcalB,overlinemu)$
necessarily the completion of $(X,mathcalA,mu)$?




My definition of completion is:



The completion of $mathcalA$ is the smallest $sigma$-algebra $mathcalB$ containing $mathcalA$ such that $(X,mathcalB,mu)$ is complete.



It seems like the answer to my question is no, because it isn't clear that $mathcalB$ is necessarily the smallest $sigma$-algebra satisfying the required properties. But I have been looking at the Completion Theorem where they seem to assume that the answer to my question is yes. How can I see that the smallest $sigma$-algebra satisfying the required properties is the only $sigma$-algebra satisfying the required properties?










share|cite|improve this question

























    up vote
    4
    down vote

    favorite













    If $(X,mathcalA,mu)$ is a measure space,
    $(X,mathcalB,overlinemu)$ is complete,
    $mathcalBsupsetmathcalA$, and $overlinemu(A)=mu(A)$ for
    every $AinmathcalA$, is $(X,mathcalB,overlinemu)$
    necessarily the completion of $(X,mathcalA,mu)$?




    My definition of completion is:



    The completion of $mathcalA$ is the smallest $sigma$-algebra $mathcalB$ containing $mathcalA$ such that $(X,mathcalB,mu)$ is complete.



    It seems like the answer to my question is no, because it isn't clear that $mathcalB$ is necessarily the smallest $sigma$-algebra satisfying the required properties. But I have been looking at the Completion Theorem where they seem to assume that the answer to my question is yes. How can I see that the smallest $sigma$-algebra satisfying the required properties is the only $sigma$-algebra satisfying the required properties?










    share|cite|improve this question























      up vote
      4
      down vote

      favorite









      up vote
      4
      down vote

      favorite












      If $(X,mathcalA,mu)$ is a measure space,
      $(X,mathcalB,overlinemu)$ is complete,
      $mathcalBsupsetmathcalA$, and $overlinemu(A)=mu(A)$ for
      every $AinmathcalA$, is $(X,mathcalB,overlinemu)$
      necessarily the completion of $(X,mathcalA,mu)$?




      My definition of completion is:



      The completion of $mathcalA$ is the smallest $sigma$-algebra $mathcalB$ containing $mathcalA$ such that $(X,mathcalB,mu)$ is complete.



      It seems like the answer to my question is no, because it isn't clear that $mathcalB$ is necessarily the smallest $sigma$-algebra satisfying the required properties. But I have been looking at the Completion Theorem where they seem to assume that the answer to my question is yes. How can I see that the smallest $sigma$-algebra satisfying the required properties is the only $sigma$-algebra satisfying the required properties?










      share|cite|improve this question














      If $(X,mathcalA,mu)$ is a measure space,
      $(X,mathcalB,overlinemu)$ is complete,
      $mathcalBsupsetmathcalA$, and $overlinemu(A)=mu(A)$ for
      every $AinmathcalA$, is $(X,mathcalB,overlinemu)$
      necessarily the completion of $(X,mathcalA,mu)$?




      My definition of completion is:



      The completion of $mathcalA$ is the smallest $sigma$-algebra $mathcalB$ containing $mathcalA$ such that $(X,mathcalB,mu)$ is complete.



      It seems like the answer to my question is no, because it isn't clear that $mathcalB$ is necessarily the smallest $sigma$-algebra satisfying the required properties. But I have been looking at the Completion Theorem where they seem to assume that the answer to my question is yes. How can I see that the smallest $sigma$-algebra satisfying the required properties is the only $sigma$-algebra satisfying the required properties?







      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 Aug 17 at 8:43









      rbird

      1,12313




      1,12313




















          3 Answers
          3






          active

          oldest

          votes

















          up vote
          3
          down vote



          accepted










          The completion theorem is first an existence theorem. The proof gives an explicit construction of the completion.



          If you look into the proof, then you see that the $sigma$-algebra $Sigma^*$ constructed there is the smallest possible.
          It contains the original $mathcal A$ and all subsets of sets of zero measure.



          Completion of measure space is usually defined without the requirement of smallest possible $sigma$-algebra. The completion theorem gives you this smallest completion for free.






          share|cite|improve this answer



























            up vote
            4
            down vote













            As stated the result is surely false. If $(X,mathcal A, mu)$ is already complete you are asserting that we cannot extend the measure to a larger sigma algebra. Take $mathcal A$ to be trivial sigma algebra (consisting of just the empty set and the whole space) to get obvious counter-examples.






            share|cite|improve this answer



























              up vote
              2
              down vote













              In general, the extension you are looking for is not unique.
              Let $X=[0,1]$, for example, and consider the trivial $sigma$-algebra $mathcal A=emptyset, X$. Let $mu(emptyset)=0$ and $mu(X)=1$.



              This is complete, but there are a lot of different ways to extend it to a complete measure on a finer $sigma$-algebra.
              For example, $mathcal B$ can be all subsets of $X$, and the measure can be any Dirac measure.



              Or the Lebesgue measure is also good.






              share|cite|improve this answer




















                Your Answer




                StackExchange.ifUsing("editor", function ()
                return StackExchange.using("mathjaxEditing", function ()
                StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
                StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
                );
                );
                , "mathjax-editing");

                StackExchange.ready(function()
                var channelOptions =
                tags: "".split(" "),
                id: "69"
                ;
                initTagRenderer("".split(" "), "".split(" "), channelOptions);

                StackExchange.using("externalEditor", function()
                // Have to fire editor after snippets, if snippets enabled
                if (StackExchange.settings.snippets.snippetsEnabled)
                StackExchange.using("snippets", function()
                createEditor();
                );

                else
                createEditor();

                );

                function createEditor()
                StackExchange.prepareEditor(
                heartbeatType: 'answer',
                convertImagesToLinks: true,
                noModals: false,
                showLowRepImageUploadWarning: true,
                reputationToPostImages: 10,
                bindNavPrevention: true,
                postfix: "",
                noCode: true, onDemand: true,
                discardSelector: ".discard-answer"
                ,immediatelyShowMarkdownHelp:true
                );



                );













                 

                draft saved


                draft discarded


















                StackExchange.ready(
                function ()
                StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2885548%2fcompletion-of-measure-spaces-uniqueness%23new-answer', 'question_page');

                );

                Post as a guest






























                3 Answers
                3






                active

                oldest

                votes








                3 Answers
                3






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes








                up vote
                3
                down vote



                accepted










                The completion theorem is first an existence theorem. The proof gives an explicit construction of the completion.



                If you look into the proof, then you see that the $sigma$-algebra $Sigma^*$ constructed there is the smallest possible.
                It contains the original $mathcal A$ and all subsets of sets of zero measure.



                Completion of measure space is usually defined without the requirement of smallest possible $sigma$-algebra. The completion theorem gives you this smallest completion for free.






                share|cite|improve this answer
























                  up vote
                  3
                  down vote



                  accepted










                  The completion theorem is first an existence theorem. The proof gives an explicit construction of the completion.



                  If you look into the proof, then you see that the $sigma$-algebra $Sigma^*$ constructed there is the smallest possible.
                  It contains the original $mathcal A$ and all subsets of sets of zero measure.



                  Completion of measure space is usually defined without the requirement of smallest possible $sigma$-algebra. The completion theorem gives you this smallest completion for free.






                  share|cite|improve this answer






















                    up vote
                    3
                    down vote



                    accepted







                    up vote
                    3
                    down vote



                    accepted






                    The completion theorem is first an existence theorem. The proof gives an explicit construction of the completion.



                    If you look into the proof, then you see that the $sigma$-algebra $Sigma^*$ constructed there is the smallest possible.
                    It contains the original $mathcal A$ and all subsets of sets of zero measure.



                    Completion of measure space is usually defined without the requirement of smallest possible $sigma$-algebra. The completion theorem gives you this smallest completion for free.






                    share|cite|improve this answer












                    The completion theorem is first an existence theorem. The proof gives an explicit construction of the completion.



                    If you look into the proof, then you see that the $sigma$-algebra $Sigma^*$ constructed there is the smallest possible.
                    It contains the original $mathcal A$ and all subsets of sets of zero measure.



                    Completion of measure space is usually defined without the requirement of smallest possible $sigma$-algebra. The completion theorem gives you this smallest completion for free.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Aug 17 at 8:53









                    daw

                    22.2k1542




                    22.2k1542




















                        up vote
                        4
                        down vote













                        As stated the result is surely false. If $(X,mathcal A, mu)$ is already complete you are asserting that we cannot extend the measure to a larger sigma algebra. Take $mathcal A$ to be trivial sigma algebra (consisting of just the empty set and the whole space) to get obvious counter-examples.






                        share|cite|improve this answer
























                          up vote
                          4
                          down vote













                          As stated the result is surely false. If $(X,mathcal A, mu)$ is already complete you are asserting that we cannot extend the measure to a larger sigma algebra. Take $mathcal A$ to be trivial sigma algebra (consisting of just the empty set and the whole space) to get obvious counter-examples.






                          share|cite|improve this answer






















                            up vote
                            4
                            down vote










                            up vote
                            4
                            down vote









                            As stated the result is surely false. If $(X,mathcal A, mu)$ is already complete you are asserting that we cannot extend the measure to a larger sigma algebra. Take $mathcal A$ to be trivial sigma algebra (consisting of just the empty set and the whole space) to get obvious counter-examples.






                            share|cite|improve this answer












                            As stated the result is surely false. If $(X,mathcal A, mu)$ is already complete you are asserting that we cannot extend the measure to a larger sigma algebra. Take $mathcal A$ to be trivial sigma algebra (consisting of just the empty set and the whole space) to get obvious counter-examples.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Aug 17 at 9:00









                            Kavi Rama Murthy

                            26.5k31438




                            26.5k31438




















                                up vote
                                2
                                down vote













                                In general, the extension you are looking for is not unique.
                                Let $X=[0,1]$, for example, and consider the trivial $sigma$-algebra $mathcal A=emptyset, X$. Let $mu(emptyset)=0$ and $mu(X)=1$.



                                This is complete, but there are a lot of different ways to extend it to a complete measure on a finer $sigma$-algebra.
                                For example, $mathcal B$ can be all subsets of $X$, and the measure can be any Dirac measure.



                                Or the Lebesgue measure is also good.






                                share|cite|improve this answer
























                                  up vote
                                  2
                                  down vote













                                  In general, the extension you are looking for is not unique.
                                  Let $X=[0,1]$, for example, and consider the trivial $sigma$-algebra $mathcal A=emptyset, X$. Let $mu(emptyset)=0$ and $mu(X)=1$.



                                  This is complete, but there are a lot of different ways to extend it to a complete measure on a finer $sigma$-algebra.
                                  For example, $mathcal B$ can be all subsets of $X$, and the measure can be any Dirac measure.



                                  Or the Lebesgue measure is also good.






                                  share|cite|improve this answer






















                                    up vote
                                    2
                                    down vote










                                    up vote
                                    2
                                    down vote









                                    In general, the extension you are looking for is not unique.
                                    Let $X=[0,1]$, for example, and consider the trivial $sigma$-algebra $mathcal A=emptyset, X$. Let $mu(emptyset)=0$ and $mu(X)=1$.



                                    This is complete, but there are a lot of different ways to extend it to a complete measure on a finer $sigma$-algebra.
                                    For example, $mathcal B$ can be all subsets of $X$, and the measure can be any Dirac measure.



                                    Or the Lebesgue measure is also good.






                                    share|cite|improve this answer












                                    In general, the extension you are looking for is not unique.
                                    Let $X=[0,1]$, for example, and consider the trivial $sigma$-algebra $mathcal A=emptyset, X$. Let $mu(emptyset)=0$ and $mu(X)=1$.



                                    This is complete, but there are a lot of different ways to extend it to a complete measure on a finer $sigma$-algebra.
                                    For example, $mathcal B$ can be all subsets of $X$, and the measure can be any Dirac measure.



                                    Or the Lebesgue measure is also good.







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Aug 17 at 9:00









                                    A. Pongrácz

                                    4,450725




                                    4,450725



























                                         

                                        draft saved


                                        draft discarded















































                                         


                                        draft saved


                                        draft discarded














                                        StackExchange.ready(
                                        function ()
                                        StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2885548%2fcompletion-of-measure-spaces-uniqueness%23new-answer', 'question_page');

                                        );

                                        Post as a guest













































































                                        Popular posts from this blog

                                        How to check contact read email or not when send email to Individual?

                                        Displaying single band from multi-band raster using QGIS

                                        How many registers does an x86_64 CPU actually have?