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





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          1 Answer
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          active

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          active

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          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


















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