convergence in $L^p$ implies convergence in measure












6














I am trying to show that if $f_n$ converges to $f$ in $L^p(X,mu)$ then $f_nto f$ in $L^p$ in measure, where $1le p le infty$.



Here is my attempt for $p>1$ - Let $varepsilon>0$ and define $A_{n,varepsilon}=lbrace x: vert f_n(x)-f(x) vert ge varepsilonrbrace$. I want to show $mu (A_{n,varepsilon})to 0$. $Vert f_n-f Vert_p=(int _X vert f_n-fvert ^p)^{1/p}ge int _{A_{n,varepsilon}}(vert f_n-fvert ^p)^{1/p}ge varepsilon mu (A_{n,varepsilon})^{1/p}$ so that $mu (A_{n,varepsilon})le (frac{Vert f_n-fVert }{varepsilon})^{1/p}$ and the RHS tends to $0$ as $f_nto f$ in the $L^p$ norm.



How can I deal with the case $p=infty$?






real-analysis analysis proof-verification lp-spaces







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  • 1




    Perhaps I'm missing something, but shouldn't the $L^{infty}$ case be the simplest? Since $||f_n-f||_{infty}<epsilon implies mu(|f_n-f|geq epsilon) = 0$.
    – Shalop
    May 14 '15 at 5:17


















6














I am trying to show that if $f_n$ converges to $f$ in $L^p(X,mu)$ then $f_nto f$ in $L^p$ in measure, where $1le p le infty$.



Here is my attempt for $p>1$ - Let $varepsilon>0$ and define $A_{n,varepsilon}=lbrace x: vert f_n(x)-f(x) vert ge varepsilonrbrace$. I want to show $mu (A_{n,varepsilon})to 0$. $Vert f_n-f Vert_p=(int _X vert f_n-fvert ^p)^{1/p}ge int _{A_{n,varepsilon}}(vert f_n-fvert ^p)^{1/p}ge varepsilon mu (A_{n,varepsilon})^{1/p}$ so that $mu (A_{n,varepsilon})le (frac{Vert f_n-fVert }{varepsilon})^{1/p}$ and the RHS tends to $0$ as $f_nto f$ in the $L^p$ norm.



How can I deal with the case $p=infty$?






real-analysis analysis proof-verification lp-spaces







share|cite|improve this question


















  • 1




    Perhaps I'm missing something, but shouldn't the $L^{infty}$ case be the simplest? Since $||f_n-f||_{infty}<epsilon implies mu(|f_n-f|geq epsilon) = 0$.
    – Shalop
    May 14 '15 at 5:17
















6












6








6


2





I am trying to show that if $f_n$ converges to $f$ in $L^p(X,mu)$ then $f_nto f$ in $L^p$ in measure, where $1le p le infty$.



Here is my attempt for $p>1$ - Let $varepsilon>0$ and define $A_{n,varepsilon}=lbrace x: vert f_n(x)-f(x) vert ge varepsilonrbrace$. I want to show $mu (A_{n,varepsilon})to 0$. $Vert f_n-f Vert_p=(int _X vert f_n-fvert ^p)^{1/p}ge int _{A_{n,varepsilon}}(vert f_n-fvert ^p)^{1/p}ge varepsilon mu (A_{n,varepsilon})^{1/p}$ so that $mu (A_{n,varepsilon})le (frac{Vert f_n-fVert }{varepsilon})^{1/p}$ and the RHS tends to $0$ as $f_nto f$ in the $L^p$ norm.



How can I deal with the case $p=infty$?






real-analysis analysis proof-verification lp-spaces







share|cite|improve this question













I am trying to show that if $f_n$ converges to $f$ in $L^p(X,mu)$ then $f_nto f$ in $L^p$ in measure, where $1le p le infty$.



Here is my attempt for $p>1$ - Let $varepsilon>0$ and define $A_{n,varepsilon}=lbrace x: vert f_n(x)-f(x) vert ge varepsilonrbrace$. I want to show $mu (A_{n,varepsilon})to 0$. $Vert f_n-f Vert_p=(int _X vert f_n-fvert ^p)^{1/p}ge int _{A_{n,varepsilon}}(vert f_n-fvert ^p)^{1/p}ge varepsilon mu (A_{n,varepsilon})^{1/p}$ so that $mu (A_{n,varepsilon})le (frac{Vert f_n-fVert }{varepsilon})^{1/p}$ and the RHS tends to $0$ as $f_nto f$ in the $L^p$ norm.



How can I deal with the case $p=infty$?






real-analysis analysis proof-verification lp-spaces




real-analysis analysis proof-verification lp-spaces






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked May 14 '15 at 5:10









capcap

7083926




7083926








  • 1




    Perhaps I'm missing something, but shouldn't the $L^{infty}$ case be the simplest? Since $||f_n-f||_{infty}<epsilon implies mu(|f_n-f|geq epsilon) = 0$.
    – Shalop
    May 14 '15 at 5:17
















  • 1




    Perhaps I'm missing something, but shouldn't the $L^{infty}$ case be the simplest? Since $||f_n-f||_{infty}<epsilon implies mu(|f_n-f|geq epsilon) = 0$.
    – Shalop
    May 14 '15 at 5:17










1




1




Perhaps I'm missing something, but shouldn't the $L^{infty}$ case be the simplest? Since $||f_n-f||_{infty}<epsilon implies mu(|f_n-f|geq epsilon) = 0$.
– Shalop
May 14 '15 at 5:17






Perhaps I'm missing something, but shouldn't the $L^{infty}$ case be the simplest? Since $||f_n-f||_{infty}<epsilon implies mu(|f_n-f|geq epsilon) = 0$.
– Shalop
May 14 '15 at 5:17












2 Answers
2






active

oldest

votes


















1














The case $p = infty$ is the simplest since $|f_n-f|_{infty} < epsilon$ means that $|f_n-f|$ is less than $epsilon$ almost everywhere, and therefore $mubig(|f_n-f|geepsilonbig) = 0$.



If $1le p < infty$, you can use Tchebychev's inequality:
$$
mubig(|f_n-f|geepsilonbig) = mubig(|f_n-f|^pgeepsilon^pbig) le frac{1}{epsilon^p}int|f_n-f|^p,dmu = frac{1}{epsilon^p}|f_n-f|_p^p,
$$
where the last expression can be made arbitrarily small by taking $n$ sufficiently large.






share|cite|improve this answer





























    -2














    If the underlying measure has finite variation then convergence in L-infinity implies convergence in Lp and the result follows. However, this is not true for non-finite measures. The constant functions x->1/n defined on the real line converge in L-infinity but not in measure.






    share|cite|improve this answer





















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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1














      The case $p = infty$ is the simplest since $|f_n-f|_{infty} < epsilon$ means that $|f_n-f|$ is less than $epsilon$ almost everywhere, and therefore $mubig(|f_n-f|geepsilonbig) = 0$.



      If $1le p < infty$, you can use Tchebychev's inequality:
      $$
      mubig(|f_n-f|geepsilonbig) = mubig(|f_n-f|^pgeepsilon^pbig) le frac{1}{epsilon^p}int|f_n-f|^p,dmu = frac{1}{epsilon^p}|f_n-f|_p^p,
      $$
      where the last expression can be made arbitrarily small by taking $n$ sufficiently large.






      share|cite|improve this answer


























        1














        The case $p = infty$ is the simplest since $|f_n-f|_{infty} < epsilon$ means that $|f_n-f|$ is less than $epsilon$ almost everywhere, and therefore $mubig(|f_n-f|geepsilonbig) = 0$.



        If $1le p < infty$, you can use Tchebychev's inequality:
        $$
        mubig(|f_n-f|geepsilonbig) = mubig(|f_n-f|^pgeepsilon^pbig) le frac{1}{epsilon^p}int|f_n-f|^p,dmu = frac{1}{epsilon^p}|f_n-f|_p^p,
        $$
        where the last expression can be made arbitrarily small by taking $n$ sufficiently large.






        share|cite|improve this answer
























          1












          1








          1






          The case $p = infty$ is the simplest since $|f_n-f|_{infty} < epsilon$ means that $|f_n-f|$ is less than $epsilon$ almost everywhere, and therefore $mubig(|f_n-f|geepsilonbig) = 0$.



          If $1le p < infty$, you can use Tchebychev's inequality:
          $$
          mubig(|f_n-f|geepsilonbig) = mubig(|f_n-f|^pgeepsilon^pbig) le frac{1}{epsilon^p}int|f_n-f|^p,dmu = frac{1}{epsilon^p}|f_n-f|_p^p,
          $$
          where the last expression can be made arbitrarily small by taking $n$ sufficiently large.






          share|cite|improve this answer












          The case $p = infty$ is the simplest since $|f_n-f|_{infty} < epsilon$ means that $|f_n-f|$ is less than $epsilon$ almost everywhere, and therefore $mubig(|f_n-f|geepsilonbig) = 0$.



          If $1le p < infty$, you can use Tchebychev's inequality:
          $$
          mubig(|f_n-f|geepsilonbig) = mubig(|f_n-f|^pgeepsilon^pbig) le frac{1}{epsilon^p}int|f_n-f|^p,dmu = frac{1}{epsilon^p}|f_n-f|_p^p,
          $$
          where the last expression can be made arbitrarily small by taking $n$ sufficiently large.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Oct 30 '18 at 19:23









          AOrtizAOrtiz

          10.5k21341




          10.5k21341























              -2














              If the underlying measure has finite variation then convergence in L-infinity implies convergence in Lp and the result follows. However, this is not true for non-finite measures. The constant functions x->1/n defined on the real line converge in L-infinity but not in measure.






              share|cite|improve this answer


























                -2














                If the underlying measure has finite variation then convergence in L-infinity implies convergence in Lp and the result follows. However, this is not true for non-finite measures. The constant functions x->1/n defined on the real line converge in L-infinity but not in measure.






                share|cite|improve this answer
























                  -2












                  -2








                  -2






                  If the underlying measure has finite variation then convergence in L-infinity implies convergence in Lp and the result follows. However, this is not true for non-finite measures. The constant functions x->1/n defined on the real line converge in L-infinity but not in measure.






                  share|cite|improve this answer












                  If the underlying measure has finite variation then convergence in L-infinity implies convergence in Lp and the result follows. However, this is not true for non-finite measures. The constant functions x->1/n defined on the real line converge in L-infinity but not in measure.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Oct 30 '18 at 18:41









                  user610395user610395

                  1




                  1






























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