Sigma-finite measure (countable sequence)












0














Let $(X,mathcal{A},mu)$ be a measure space, $g:X to mathbb{R}$ a $mu$-integrable map and $A=X$ $g^{-1}(0)$.



How to show that it exists a countable sequence $(A_i)_{i in mathbb{N}}$ of $mathcal{A}$ such that



$A=bigcuplimits_{i=1}^{infty} A_{i}$ and $mu(A_i)<infty, forall n in mathbb{N}$?



I got:



Since $Ainmathcal{A}Rightarrow (X$ $g^{-1}(0))^c in mathcal{A}$



So $X=A dotcup (X$ $g^{-1}(0))^c$



By finite additivity of measures: $mu(X)=mu(A)+mu(X$ $g^{-1}(0))^c$



By non-negativity: $mu(A)<infty$



I'm not sure if this works for $A_i$ and how to show that $(A_i)_{i in mathbb{N}}$ exists with $A=bigcuplimits_{i=1}^{infty} A_{i}$.






measure-theory







share|cite|improve this question



























    0














    Let $(X,mathcal{A},mu)$ be a measure space, $g:X to mathbb{R}$ a $mu$-integrable map and $A=X$ $g^{-1}(0)$.



    How to show that it exists a countable sequence $(A_i)_{i in mathbb{N}}$ of $mathcal{A}$ such that



    $A=bigcuplimits_{i=1}^{infty} A_{i}$ and $mu(A_i)<infty, forall n in mathbb{N}$?



    I got:



    Since $Ainmathcal{A}Rightarrow (X$ $g^{-1}(0))^c in mathcal{A}$



    So $X=A dotcup (X$ $g^{-1}(0))^c$



    By finite additivity of measures: $mu(X)=mu(A)+mu(X$ $g^{-1}(0))^c$



    By non-negativity: $mu(A)<infty$



    I'm not sure if this works for $A_i$ and how to show that $(A_i)_{i in mathbb{N}}$ exists with $A=bigcuplimits_{i=1}^{infty} A_{i}$.






    measure-theory







    share|cite|improve this question

























      0












      0








      0







      Let $(X,mathcal{A},mu)$ be a measure space, $g:X to mathbb{R}$ a $mu$-integrable map and $A=X$ $g^{-1}(0)$.



      How to show that it exists a countable sequence $(A_i)_{i in mathbb{N}}$ of $mathcal{A}$ such that



      $A=bigcuplimits_{i=1}^{infty} A_{i}$ and $mu(A_i)<infty, forall n in mathbb{N}$?



      I got:



      Since $Ainmathcal{A}Rightarrow (X$ $g^{-1}(0))^c in mathcal{A}$



      So $X=A dotcup (X$ $g^{-1}(0))^c$



      By finite additivity of measures: $mu(X)=mu(A)+mu(X$ $g^{-1}(0))^c$



      By non-negativity: $mu(A)<infty$



      I'm not sure if this works for $A_i$ and how to show that $(A_i)_{i in mathbb{N}}$ exists with $A=bigcuplimits_{i=1}^{infty} A_{i}$.






      measure-theory







      share|cite|improve this question













      Let $(X,mathcal{A},mu)$ be a measure space, $g:X to mathbb{R}$ a $mu$-integrable map and $A=X$ $g^{-1}(0)$.



      How to show that it exists a countable sequence $(A_i)_{i in mathbb{N}}$ of $mathcal{A}$ such that



      $A=bigcuplimits_{i=1}^{infty} A_{i}$ and $mu(A_i)<infty, forall n in mathbb{N}$?



      I got:



      Since $Ainmathcal{A}Rightarrow (X$ $g^{-1}(0))^c in mathcal{A}$



      So $X=A dotcup (X$ $g^{-1}(0))^c$



      By finite additivity of measures: $mu(X)=mu(A)+mu(X$ $g^{-1}(0))^c$



      By non-negativity: $mu(A)<infty$



      I'm not sure if this works for $A_i$ and how to show that $(A_i)_{i in mathbb{N}}$ exists with $A=bigcuplimits_{i=1}^{infty} A_{i}$.






      measure-theory




      measure-theory






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 24 at 11:53









      Tartulop

      656




      656






















          2 Answers
          2






          active

          oldest

          votes


















          1














          You can take $A_i={xin Xmid |g(x)|geqfrac1i}$.



          Observe that $mu(A_i)=intmathbf1_{A_i};dmuleqint imathbf|g|;dmuleq iintmathbf|g|;dmu<infty$.



          Also the $A_i$ are preimages of Borel sets wrt measurable $|g|$ so are elements of $mathcal A$.



          It is evident that $bigcup_{i=1}^{infty}A_i={xin Xmid g(x)neq0}=Xsetminus g^{-1}({0})=A$.






          share|cite|improve this answer























          • Does ${A_i}$ cover $X$? I think $cup_{I=1}^{infty} A_i={x in X mid |g(x)|>0} neq X$.
            – sharpe
            Nov 24 at 12:11












          • It is not demanded that they cover $X$, but that their union is $A$.
            – drhab
            Nov 24 at 12:13










          • Sorry. I understood.
            – sharpe
            Nov 24 at 12:14



















          0














          The function $g$ here doesn't care so much: the space is $sigma$-finite, so there is a countable cover of measurable sets $(X_j)$ of $X$ such that $mu(X_j)<infty$ for each $jinBbb N$.



          Then the sequence defined by $X_jcap A$ cover $A$ and necessarily $mu(X_jcap A)lemu(X_j)$ because any measure is increasing, that is, if $Asubset B$ then $mu(A)lemu(B)$.






          share|cite|improve this answer























          • No. $X$ is just a measure space.
            – sharpe
            Nov 24 at 12:08










          • @sharpe it is the same.
            – Masacroso
            Nov 24 at 12:10










          • Probably, your argument is mistaken.
            – sharpe
            Nov 24 at 12:22













          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',
          autoActivateHeartbeat: false,
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          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%2f3011466%2fsigma-finite-measure-countable-sequence%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1














          You can take $A_i={xin Xmid |g(x)|geqfrac1i}$.



          Observe that $mu(A_i)=intmathbf1_{A_i};dmuleqint imathbf|g|;dmuleq iintmathbf|g|;dmu<infty$.



          Also the $A_i$ are preimages of Borel sets wrt measurable $|g|$ so are elements of $mathcal A$.



          It is evident that $bigcup_{i=1}^{infty}A_i={xin Xmid g(x)neq0}=Xsetminus g^{-1}({0})=A$.






          share|cite|improve this answer























          • Does ${A_i}$ cover $X$? I think $cup_{I=1}^{infty} A_i={x in X mid |g(x)|>0} neq X$.
            – sharpe
            Nov 24 at 12:11












          • It is not demanded that they cover $X$, but that their union is $A$.
            – drhab
            Nov 24 at 12:13










          • Sorry. I understood.
            – sharpe
            Nov 24 at 12:14
















          1














          You can take $A_i={xin Xmid |g(x)|geqfrac1i}$.



          Observe that $mu(A_i)=intmathbf1_{A_i};dmuleqint imathbf|g|;dmuleq iintmathbf|g|;dmu<infty$.



          Also the $A_i$ are preimages of Borel sets wrt measurable $|g|$ so are elements of $mathcal A$.



          It is evident that $bigcup_{i=1}^{infty}A_i={xin Xmid g(x)neq0}=Xsetminus g^{-1}({0})=A$.






          share|cite|improve this answer























          • Does ${A_i}$ cover $X$? I think $cup_{I=1}^{infty} A_i={x in X mid |g(x)|>0} neq X$.
            – sharpe
            Nov 24 at 12:11












          • It is not demanded that they cover $X$, but that their union is $A$.
            – drhab
            Nov 24 at 12:13










          • Sorry. I understood.
            – sharpe
            Nov 24 at 12:14














          1












          1








          1






          You can take $A_i={xin Xmid |g(x)|geqfrac1i}$.



          Observe that $mu(A_i)=intmathbf1_{A_i};dmuleqint imathbf|g|;dmuleq iintmathbf|g|;dmu<infty$.



          Also the $A_i$ are preimages of Borel sets wrt measurable $|g|$ so are elements of $mathcal A$.



          It is evident that $bigcup_{i=1}^{infty}A_i={xin Xmid g(x)neq0}=Xsetminus g^{-1}({0})=A$.






          share|cite|improve this answer














          You can take $A_i={xin Xmid |g(x)|geqfrac1i}$.



          Observe that $mu(A_i)=intmathbf1_{A_i};dmuleqint imathbf|g|;dmuleq iintmathbf|g|;dmu<infty$.



          Also the $A_i$ are preimages of Borel sets wrt measurable $|g|$ so are elements of $mathcal A$.



          It is evident that $bigcup_{i=1}^{infty}A_i={xin Xmid g(x)neq0}=Xsetminus g^{-1}({0})=A$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 24 at 12:39

























          answered Nov 24 at 12:08









          drhab

          96.5k544127




          96.5k544127












          • Does ${A_i}$ cover $X$? I think $cup_{I=1}^{infty} A_i={x in X mid |g(x)|>0} neq X$.
            – sharpe
            Nov 24 at 12:11












          • It is not demanded that they cover $X$, but that their union is $A$.
            – drhab
            Nov 24 at 12:13










          • Sorry. I understood.
            – sharpe
            Nov 24 at 12:14


















          • Does ${A_i}$ cover $X$? I think $cup_{I=1}^{infty} A_i={x in X mid |g(x)|>0} neq X$.
            – sharpe
            Nov 24 at 12:11












          • It is not demanded that they cover $X$, but that their union is $A$.
            – drhab
            Nov 24 at 12:13










          • Sorry. I understood.
            – sharpe
            Nov 24 at 12:14
















          Does ${A_i}$ cover $X$? I think $cup_{I=1}^{infty} A_i={x in X mid |g(x)|>0} neq X$.
          – sharpe
          Nov 24 at 12:11






          Does ${A_i}$ cover $X$? I think $cup_{I=1}^{infty} A_i={x in X mid |g(x)|>0} neq X$.
          – sharpe
          Nov 24 at 12:11














          It is not demanded that they cover $X$, but that their union is $A$.
          – drhab
          Nov 24 at 12:13




          It is not demanded that they cover $X$, but that their union is $A$.
          – drhab
          Nov 24 at 12:13












          Sorry. I understood.
          – sharpe
          Nov 24 at 12:14




          Sorry. I understood.
          – sharpe
          Nov 24 at 12:14











          0














          The function $g$ here doesn't care so much: the space is $sigma$-finite, so there is a countable cover of measurable sets $(X_j)$ of $X$ such that $mu(X_j)<infty$ for each $jinBbb N$.



          Then the sequence defined by $X_jcap A$ cover $A$ and necessarily $mu(X_jcap A)lemu(X_j)$ because any measure is increasing, that is, if $Asubset B$ then $mu(A)lemu(B)$.






          share|cite|improve this answer























          • No. $X$ is just a measure space.
            – sharpe
            Nov 24 at 12:08










          • @sharpe it is the same.
            – Masacroso
            Nov 24 at 12:10










          • Probably, your argument is mistaken.
            – sharpe
            Nov 24 at 12:22


















          0














          The function $g$ here doesn't care so much: the space is $sigma$-finite, so there is a countable cover of measurable sets $(X_j)$ of $X$ such that $mu(X_j)<infty$ for each $jinBbb N$.



          Then the sequence defined by $X_jcap A$ cover $A$ and necessarily $mu(X_jcap A)lemu(X_j)$ because any measure is increasing, that is, if $Asubset B$ then $mu(A)lemu(B)$.






          share|cite|improve this answer























          • No. $X$ is just a measure space.
            – sharpe
            Nov 24 at 12:08










          • @sharpe it is the same.
            – Masacroso
            Nov 24 at 12:10










          • Probably, your argument is mistaken.
            – sharpe
            Nov 24 at 12:22
















          0












          0








          0






          The function $g$ here doesn't care so much: the space is $sigma$-finite, so there is a countable cover of measurable sets $(X_j)$ of $X$ such that $mu(X_j)<infty$ for each $jinBbb N$.



          Then the sequence defined by $X_jcap A$ cover $A$ and necessarily $mu(X_jcap A)lemu(X_j)$ because any measure is increasing, that is, if $Asubset B$ then $mu(A)lemu(B)$.






          share|cite|improve this answer














          The function $g$ here doesn't care so much: the space is $sigma$-finite, so there is a countable cover of measurable sets $(X_j)$ of $X$ such that $mu(X_j)<infty$ for each $jinBbb N$.



          Then the sequence defined by $X_jcap A$ cover $A$ and necessarily $mu(X_jcap A)lemu(X_j)$ because any measure is increasing, that is, if $Asubset B$ then $mu(A)lemu(B)$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 24 at 12:08

























          answered Nov 24 at 12:07









          Masacroso

          12.8k41746




          12.8k41746












          • No. $X$ is just a measure space.
            – sharpe
            Nov 24 at 12:08










          • @sharpe it is the same.
            – Masacroso
            Nov 24 at 12:10










          • Probably, your argument is mistaken.
            – sharpe
            Nov 24 at 12:22




















          • No. $X$ is just a measure space.
            – sharpe
            Nov 24 at 12:08










          • @sharpe it is the same.
            – Masacroso
            Nov 24 at 12:10










          • Probably, your argument is mistaken.
            – sharpe
            Nov 24 at 12:22


















          No. $X$ is just a measure space.
          – sharpe
          Nov 24 at 12:08




          No. $X$ is just a measure space.
          – sharpe
          Nov 24 at 12:08












          @sharpe it is the same.
          – Masacroso
          Nov 24 at 12:10




          @sharpe it is the same.
          – Masacroso
          Nov 24 at 12:10












          Probably, your argument is mistaken.
          – sharpe
          Nov 24 at 12:22






          Probably, your argument is mistaken.
          – sharpe
          Nov 24 at 12:22




















          draft saved

          draft discarded




















































          Thanks for contributing an answer to Mathematics Stack Exchange!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid …



          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.


          Use MathJax to format equations. MathJax reference.


          To learn more, see our tips on writing great answers.





          Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


          Please pay close attention to the following guidance:


          • Please be sure to answer the question. Provide details and share your research!

          But avoid …



          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3011466%2fsigma-finite-measure-countable-sequence%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          -

          Popular posts from this blog

          Ellipse (mathématiques)

          Image
          Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
          Read more

          Quarter-circle Tiles

          Image
          up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
          Read more

          Mont Emei

          Image
          Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
          Read more