Check if $supleft{int_J|f(x)|chi_B(x), dx:lambda(B)le1/kright}to 0$












2














Let $JsubsetBbb R$ be a perfect interval and let $finmathcal L_1(J,E)$. Set



$$R_k:=supleft{int_J|f(x)|chi_B(x), dx:lambda(B)le1/k, Bsubset Jright}tag1$$



I want to see if it is true that $(R_k)to 0$.



What I tried: let $(A_j)_j$ be any sequence of sets such that $lim_{jtoinfty}lambda(A_j)=0$, then
$$
lim_{ktoinfty}int_Jchi_{A_k}(x), dx=lim_{ktoinfty}lambda(A_k)=0tag2
$$
which implies, by a previous result, that there is a subsequence of $(chi_{A_k})_k$ such that $(chi_{A_{k_j}})_jto 0$ point-wise a.e.



Then for any sequence $(A_j)_j$ such that $lambda(A_j)le 1/j$ and $int_{A_j}|f(x)|dxgeint_{A_{j+1}}|f(x)|, dx$, we find that there is a subsequence $(A_{k_j})_j$ such that



$$lim_{jtoinfty}int_{A_{k_j}}|f(x)|=0tag3,$$



and because the sequence $Big(int_{A_k}|f(x)|, dxBig)_k$ is monotone then we also find that $lim_kint_{A_k}|f(x)|, dx=0$. Then note that $(R_k)_k$ is also monotone, and by the above we can conclude that $(R_k)to 0$ also.



Two questions here:




  1. It is the above correct or there is some flaw? In particular I find difficult to show formally the statement "and by the above we can conclude that $(R_k)to 0$ also", that is, if for each $xin A$ the statements above holds then it also holds for $sup A$. This, intuitively, seems not right.


  2. If the statement of the title is true, there is a simpler way to show the same result?







analysis measure-theory proof-verification lebesgue-integral alternative-proof







share|cite|improve this question





























    2














    Let $JsubsetBbb R$ be a perfect interval and let $finmathcal L_1(J,E)$. Set



    $$R_k:=supleft{int_J|f(x)|chi_B(x), dx:lambda(B)le1/k, Bsubset Jright}tag1$$



    I want to see if it is true that $(R_k)to 0$.



    What I tried: let $(A_j)_j$ be any sequence of sets such that $lim_{jtoinfty}lambda(A_j)=0$, then
    $$
    lim_{ktoinfty}int_Jchi_{A_k}(x), dx=lim_{ktoinfty}lambda(A_k)=0tag2
    $$
    which implies, by a previous result, that there is a subsequence of $(chi_{A_k})_k$ such that $(chi_{A_{k_j}})_jto 0$ point-wise a.e.



    Then for any sequence $(A_j)_j$ such that $lambda(A_j)le 1/j$ and $int_{A_j}|f(x)|dxgeint_{A_{j+1}}|f(x)|, dx$, we find that there is a subsequence $(A_{k_j})_j$ such that



    $$lim_{jtoinfty}int_{A_{k_j}}|f(x)|=0tag3,$$



    and because the sequence $Big(int_{A_k}|f(x)|, dxBig)_k$ is monotone then we also find that $lim_kint_{A_k}|f(x)|, dx=0$. Then note that $(R_k)_k$ is also monotone, and by the above we can conclude that $(R_k)to 0$ also.



    Two questions here:




    1. It is the above correct or there is some flaw? In particular I find difficult to show formally the statement "and by the above we can conclude that $(R_k)to 0$ also", that is, if for each $xin A$ the statements above holds then it also holds for $sup A$. This, intuitively, seems not right.


    2. If the statement of the title is true, there is a simpler way to show the same result?







    analysis measure-theory proof-verification lebesgue-integral alternative-proof







    share|cite|improve this question



























      2












      2








      2


      1





      Let $JsubsetBbb R$ be a perfect interval and let $finmathcal L_1(J,E)$. Set



      $$R_k:=supleft{int_J|f(x)|chi_B(x), dx:lambda(B)le1/k, Bsubset Jright}tag1$$



      I want to see if it is true that $(R_k)to 0$.



      What I tried: let $(A_j)_j$ be any sequence of sets such that $lim_{jtoinfty}lambda(A_j)=0$, then
      $$
      lim_{ktoinfty}int_Jchi_{A_k}(x), dx=lim_{ktoinfty}lambda(A_k)=0tag2
      $$
      which implies, by a previous result, that there is a subsequence of $(chi_{A_k})_k$ such that $(chi_{A_{k_j}})_jto 0$ point-wise a.e.



      Then for any sequence $(A_j)_j$ such that $lambda(A_j)le 1/j$ and $int_{A_j}|f(x)|dxgeint_{A_{j+1}}|f(x)|, dx$, we find that there is a subsequence $(A_{k_j})_j$ such that



      $$lim_{jtoinfty}int_{A_{k_j}}|f(x)|=0tag3,$$



      and because the sequence $Big(int_{A_k}|f(x)|, dxBig)_k$ is monotone then we also find that $lim_kint_{A_k}|f(x)|, dx=0$. Then note that $(R_k)_k$ is also monotone, and by the above we can conclude that $(R_k)to 0$ also.



      Two questions here:




      1. It is the above correct or there is some flaw? In particular I find difficult to show formally the statement "and by the above we can conclude that $(R_k)to 0$ also", that is, if for each $xin A$ the statements above holds then it also holds for $sup A$. This, intuitively, seems not right.


      2. If the statement of the title is true, there is a simpler way to show the same result?







      analysis measure-theory proof-verification lebesgue-integral alternative-proof







      share|cite|improve this question















      Let $JsubsetBbb R$ be a perfect interval and let $finmathcal L_1(J,E)$. Set



      $$R_k:=supleft{int_J|f(x)|chi_B(x), dx:lambda(B)le1/k, Bsubset Jright}tag1$$



      I want to see if it is true that $(R_k)to 0$.



      What I tried: let $(A_j)_j$ be any sequence of sets such that $lim_{jtoinfty}lambda(A_j)=0$, then
      $$
      lim_{ktoinfty}int_Jchi_{A_k}(x), dx=lim_{ktoinfty}lambda(A_k)=0tag2
      $$
      which implies, by a previous result, that there is a subsequence of $(chi_{A_k})_k$ such that $(chi_{A_{k_j}})_jto 0$ point-wise a.e.



      Then for any sequence $(A_j)_j$ such that $lambda(A_j)le 1/j$ and $int_{A_j}|f(x)|dxgeint_{A_{j+1}}|f(x)|, dx$, we find that there is a subsequence $(A_{k_j})_j$ such that



      $$lim_{jtoinfty}int_{A_{k_j}}|f(x)|=0tag3,$$



      and because the sequence $Big(int_{A_k}|f(x)|, dxBig)_k$ is monotone then we also find that $lim_kint_{A_k}|f(x)|, dx=0$. Then note that $(R_k)_k$ is also monotone, and by the above we can conclude that $(R_k)to 0$ also.



      Two questions here:




      1. It is the above correct or there is some flaw? In particular I find difficult to show formally the statement "and by the above we can conclude that $(R_k)to 0$ also", that is, if for each $xin A$ the statements above holds then it also holds for $sup A$. This, intuitively, seems not right.


      2. If the statement of the title is true, there is a simpler way to show the same result?







      analysis measure-theory proof-verification lebesgue-integral alternative-proof




      analysis measure-theory proof-verification lebesgue-integral alternative-proof






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 9 at 14:19









      Did

      246k23220454




      246k23220454










      asked Dec 9 at 7:52









      Masacroso

      12.9k41746




      12.9k41746






















          1 Answer
          1






          active

          oldest

          votes


















          2














          Firt of all, you cannot expect that $A_{j+1} subset A_j$. In the definition of $R_k$ you don't have any dependence on $R_{k-1}$. In fact, the "local peaks" of $f$ can be in two different points leading to the disjointness of $A_j$ and $A_{j+1}$.



          The statement is equivalent to the absolute continuity of the finite measure
          $$mu(A) := int_A |f(x)| , dx quad quad (A in mathcal{B}(I))$$
          with respect to the Lebesgue-measure. Since $f$ is integrable, there exists $K>0$ with
          $$int_{{|f|>K}} |f(x)| , dx < varepsilon/2.$$
          For any measurable set $A subset I$ with $lambda(A) < epsilon/(2K)$ we find that
          begin{align}
          int_A |f(x)| , dx &le int_{{|f|>K}} |f(x)| , dx + int_{A cap {|f| le K}} |f(x)| , dx \
          & le varepsilon/2 + K lambda(A) < varepsilon.
          end{align}
          In particular, we have for all $k in mathbb{N}$ with $1/k < varepsilon/(2K)$ that $R_k le varepsilon$. This proves $lim_{k rightarrow infty} R_k =0$.






          share|cite|improve this answer























          • there is a theorem that says that for any Cauchy sequence of simple functions then there is a subsequence that converges point-wise a.e. to some function. I used this theorem, not something related to monotone sequence of sets
            – Masacroso
            Dec 9 at 19:39












          • Ah, okay! However, my argumentation is much easier. :-)
            – p4sch
            Dec 9 at 20:02










          • Yes, of course! Typo corrected. :-)
            – p4sch
            Dec 9 at 20:28











          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%2f3032129%2fcheck-if-sup-left-int-jfx-chi-bx-dx-lambdab-le1-k-right-to-0%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2














          Firt of all, you cannot expect that $A_{j+1} subset A_j$. In the definition of $R_k$ you don't have any dependence on $R_{k-1}$. In fact, the "local peaks" of $f$ can be in two different points leading to the disjointness of $A_j$ and $A_{j+1}$.



          The statement is equivalent to the absolute continuity of the finite measure
          $$mu(A) := int_A |f(x)| , dx quad quad (A in mathcal{B}(I))$$
          with respect to the Lebesgue-measure. Since $f$ is integrable, there exists $K>0$ with
          $$int_{{|f|>K}} |f(x)| , dx < varepsilon/2.$$
          For any measurable set $A subset I$ with $lambda(A) < epsilon/(2K)$ we find that
          begin{align}
          int_A |f(x)| , dx &le int_{{|f|>K}} |f(x)| , dx + int_{A cap {|f| le K}} |f(x)| , dx \
          & le varepsilon/2 + K lambda(A) < varepsilon.
          end{align}
          In particular, we have for all $k in mathbb{N}$ with $1/k < varepsilon/(2K)$ that $R_k le varepsilon$. This proves $lim_{k rightarrow infty} R_k =0$.






          share|cite|improve this answer























          • there is a theorem that says that for any Cauchy sequence of simple functions then there is a subsequence that converges point-wise a.e. to some function. I used this theorem, not something related to monotone sequence of sets
            – Masacroso
            Dec 9 at 19:39












          • Ah, okay! However, my argumentation is much easier. :-)
            – p4sch
            Dec 9 at 20:02










          • Yes, of course! Typo corrected. :-)
            – p4sch
            Dec 9 at 20:28
















          2














          Firt of all, you cannot expect that $A_{j+1} subset A_j$. In the definition of $R_k$ you don't have any dependence on $R_{k-1}$. In fact, the "local peaks" of $f$ can be in two different points leading to the disjointness of $A_j$ and $A_{j+1}$.



          The statement is equivalent to the absolute continuity of the finite measure
          $$mu(A) := int_A |f(x)| , dx quad quad (A in mathcal{B}(I))$$
          with respect to the Lebesgue-measure. Since $f$ is integrable, there exists $K>0$ with
          $$int_{{|f|>K}} |f(x)| , dx < varepsilon/2.$$
          For any measurable set $A subset I$ with $lambda(A) < epsilon/(2K)$ we find that
          begin{align}
          int_A |f(x)| , dx &le int_{{|f|>K}} |f(x)| , dx + int_{A cap {|f| le K}} |f(x)| , dx \
          & le varepsilon/2 + K lambda(A) < varepsilon.
          end{align}
          In particular, we have for all $k in mathbb{N}$ with $1/k < varepsilon/(2K)$ that $R_k le varepsilon$. This proves $lim_{k rightarrow infty} R_k =0$.






          share|cite|improve this answer























          • there is a theorem that says that for any Cauchy sequence of simple functions then there is a subsequence that converges point-wise a.e. to some function. I used this theorem, not something related to monotone sequence of sets
            – Masacroso
            Dec 9 at 19:39












          • Ah, okay! However, my argumentation is much easier. :-)
            – p4sch
            Dec 9 at 20:02










          • Yes, of course! Typo corrected. :-)
            – p4sch
            Dec 9 at 20:28














          2












          2








          2






          Firt of all, you cannot expect that $A_{j+1} subset A_j$. In the definition of $R_k$ you don't have any dependence on $R_{k-1}$. In fact, the "local peaks" of $f$ can be in two different points leading to the disjointness of $A_j$ and $A_{j+1}$.



          The statement is equivalent to the absolute continuity of the finite measure
          $$mu(A) := int_A |f(x)| , dx quad quad (A in mathcal{B}(I))$$
          with respect to the Lebesgue-measure. Since $f$ is integrable, there exists $K>0$ with
          $$int_{{|f|>K}} |f(x)| , dx < varepsilon/2.$$
          For any measurable set $A subset I$ with $lambda(A) < epsilon/(2K)$ we find that
          begin{align}
          int_A |f(x)| , dx &le int_{{|f|>K}} |f(x)| , dx + int_{A cap {|f| le K}} |f(x)| , dx \
          & le varepsilon/2 + K lambda(A) < varepsilon.
          end{align}
          In particular, we have for all $k in mathbb{N}$ with $1/k < varepsilon/(2K)$ that $R_k le varepsilon$. This proves $lim_{k rightarrow infty} R_k =0$.






          share|cite|improve this answer














          Firt of all, you cannot expect that $A_{j+1} subset A_j$. In the definition of $R_k$ you don't have any dependence on $R_{k-1}$. In fact, the "local peaks" of $f$ can be in two different points leading to the disjointness of $A_j$ and $A_{j+1}$.



          The statement is equivalent to the absolute continuity of the finite measure
          $$mu(A) := int_A |f(x)| , dx quad quad (A in mathcal{B}(I))$$
          with respect to the Lebesgue-measure. Since $f$ is integrable, there exists $K>0$ with
          $$int_{{|f|>K}} |f(x)| , dx < varepsilon/2.$$
          For any measurable set $A subset I$ with $lambda(A) < epsilon/(2K)$ we find that
          begin{align}
          int_A |f(x)| , dx &le int_{{|f|>K}} |f(x)| , dx + int_{A cap {|f| le K}} |f(x)| , dx \
          & le varepsilon/2 + K lambda(A) < varepsilon.
          end{align}
          In particular, we have for all $k in mathbb{N}$ with $1/k < varepsilon/(2K)$ that $R_k le varepsilon$. This proves $lim_{k rightarrow infty} R_k =0$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 9 at 20:27

























          answered Dec 9 at 13:25









          p4sch

          4,760217




          4,760217












          • there is a theorem that says that for any Cauchy sequence of simple functions then there is a subsequence that converges point-wise a.e. to some function. I used this theorem, not something related to monotone sequence of sets
            – Masacroso
            Dec 9 at 19:39












          • Ah, okay! However, my argumentation is much easier. :-)
            – p4sch
            Dec 9 at 20:02










          • Yes, of course! Typo corrected. :-)
            – p4sch
            Dec 9 at 20:28


















          • there is a theorem that says that for any Cauchy sequence of simple functions then there is a subsequence that converges point-wise a.e. to some function. I used this theorem, not something related to monotone sequence of sets
            – Masacroso
            Dec 9 at 19:39












          • Ah, okay! However, my argumentation is much easier. :-)
            – p4sch
            Dec 9 at 20:02










          • Yes, of course! Typo corrected. :-)
            – p4sch
            Dec 9 at 20:28
















          there is a theorem that says that for any Cauchy sequence of simple functions then there is a subsequence that converges point-wise a.e. to some function. I used this theorem, not something related to monotone sequence of sets
          – Masacroso
          Dec 9 at 19:39






          there is a theorem that says that for any Cauchy sequence of simple functions then there is a subsequence that converges point-wise a.e. to some function. I used this theorem, not something related to monotone sequence of sets
          – Masacroso
          Dec 9 at 19:39














          Ah, okay! However, my argumentation is much easier. :-)
          – p4sch
          Dec 9 at 20:02




          Ah, okay! However, my argumentation is much easier. :-)
          – p4sch
          Dec 9 at 20:02












          Yes, of course! Typo corrected. :-)
          – p4sch
          Dec 9 at 20:28




          Yes, of course! Typo corrected. :-)
          – p4sch
          Dec 9 at 20:28


















          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%2f3032129%2fcheck-if-sup-left-int-jfx-chi-bx-dx-lambdab-le1-k-right-to-0%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

          Bressuire

          Image
          .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
          Read more

          Cabo Verde

          Image
            Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
          Read more

          Gyllenstierna

          Image
          .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
          Read more