Convergence in $L^1$ implies there exists a subsequence that converges almost everywhere.











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In Folland, Theorem 2.32 states




If $f_n to f$ in $L^1$ then there is a subsequence ${f_{n_j}}$ such that $f_{n_j} to f$ almost everywhere.




The proof given just says to "combine propositions 2.29 and 2.30" which say




(2.29) If $f_n to f$ in $L^1$ then $f_n to f$ in measure.




and




(2.30) Suppose ${f_n}$ is Cauchy in measure then there is a measureable $f$ such that $f_n to f$ in measure and there is a subsequence $f_{n_j}$ that converges to $f$ almost everywhere. Moreover if $f_n to g$ in measure then $f =g$ almost everywhere.




I'm unsure how how this is implied? I think part of this is because I don't fully understand 2.30. I know that given $f_n to f$ in L^1 then by 2.29 we know that $f_n to f$ in measure. 2.30, however, deals with a sequence that is Cauchy in measure. However, in general convergence in measure does not imply Cauchy in measure, correct?






real-analysis analysis measure-theory convergence measurable-functions







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    down vote

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    In Folland, Theorem 2.32 states




    If $f_n to f$ in $L^1$ then there is a subsequence ${f_{n_j}}$ such that $f_{n_j} to f$ almost everywhere.




    The proof given just says to "combine propositions 2.29 and 2.30" which say




    (2.29) If $f_n to f$ in $L^1$ then $f_n to f$ in measure.




    and




    (2.30) Suppose ${f_n}$ is Cauchy in measure then there is a measureable $f$ such that $f_n to f$ in measure and there is a subsequence $f_{n_j}$ that converges to $f$ almost everywhere. Moreover if $f_n to g$ in measure then $f =g$ almost everywhere.




    I'm unsure how how this is implied? I think part of this is because I don't fully understand 2.30. I know that given $f_n to f$ in L^1 then by 2.29 we know that $f_n to f$ in measure. 2.30, however, deals with a sequence that is Cauchy in measure. However, in general convergence in measure does not imply Cauchy in measure, correct?






    real-analysis analysis measure-theory convergence measurable-functions







    share|cite|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      In Folland, Theorem 2.32 states




      If $f_n to f$ in $L^1$ then there is a subsequence ${f_{n_j}}$ such that $f_{n_j} to f$ almost everywhere.




      The proof given just says to "combine propositions 2.29 and 2.30" which say




      (2.29) If $f_n to f$ in $L^1$ then $f_n to f$ in measure.




      and




      (2.30) Suppose ${f_n}$ is Cauchy in measure then there is a measureable $f$ such that $f_n to f$ in measure and there is a subsequence $f_{n_j}$ that converges to $f$ almost everywhere. Moreover if $f_n to g$ in measure then $f =g$ almost everywhere.




      I'm unsure how how this is implied? I think part of this is because I don't fully understand 2.30. I know that given $f_n to f$ in L^1 then by 2.29 we know that $f_n to f$ in measure. 2.30, however, deals with a sequence that is Cauchy in measure. However, in general convergence in measure does not imply Cauchy in measure, correct?






      real-analysis analysis measure-theory convergence measurable-functions







      share|cite|improve this question













      In Folland, Theorem 2.32 states




      If $f_n to f$ in $L^1$ then there is a subsequence ${f_{n_j}}$ such that $f_{n_j} to f$ almost everywhere.




      The proof given just says to "combine propositions 2.29 and 2.30" which say




      (2.29) If $f_n to f$ in $L^1$ then $f_n to f$ in measure.




      and




      (2.30) Suppose ${f_n}$ is Cauchy in measure then there is a measureable $f$ such that $f_n to f$ in measure and there is a subsequence $f_{n_j}$ that converges to $f$ almost everywhere. Moreover if $f_n to g$ in measure then $f =g$ almost everywhere.




      I'm unsure how how this is implied? I think part of this is because I don't fully understand 2.30. I know that given $f_n to f$ in L^1 then by 2.29 we know that $f_n to f$ in measure. 2.30, however, deals with a sequence that is Cauchy in measure. However, in general convergence in measure does not imply Cauchy in measure, correct?






      real-analysis analysis measure-theory convergence measurable-functions




      real-analysis analysis measure-theory convergence measurable-functions






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      asked Nov 29 at 5:03









      carsandpulsars

      34117




      34117






















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          Convergence in measure implies Cauchy in measure.



          Let our measure be $mu$. if $f_n to f$ in measure, then given $epsilon > 0$, we have $mu({x : |f_n(x) - f(x)| > epsilon/2})to 0$ as $n to infty$. But as $|f_n(x) - f_m(x)| leq |f_n(x) - f(x)| + |f(x) - f_m(x)|$, we have
          $$ {x: |f_n(x) - f_m(x)| > epsilon} subseteq {x: |f_n(x) - f(x)| > epsilon/2} cup {x : |f(x) - f_m(x)| > epsilon/2}.$$
          Monotonicity of measure then gives that $mu({x: |f_n(x) - f_m(x)| > epsilon})$ is arbitrarily small as $n,m to infty$. Concretely: given $delta > 0$, there exists $N$ such that if $n geq N$, then $mu({x : |f_n(x) - f(x)|> epsilon/2}) < delta/2$. For this $N$, if $n,m geq N$, then also $mu({x: |f_n(x) - f_m(x)| > epsilon}) < delta$.






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          • I feel like this should have been stated as a theorem/corollary on its own! Thank you very much. This answers my question perfectly.
            – carsandpulsars
            Nov 29 at 5:37











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          Convergence in measure implies Cauchy in measure.



          Let our measure be $mu$. if $f_n to f$ in measure, then given $epsilon > 0$, we have $mu({x : |f_n(x) - f(x)| > epsilon/2})to 0$ as $n to infty$. But as $|f_n(x) - f_m(x)| leq |f_n(x) - f(x)| + |f(x) - f_m(x)|$, we have
          $$ {x: |f_n(x) - f_m(x)| > epsilon} subseteq {x: |f_n(x) - f(x)| > epsilon/2} cup {x : |f(x) - f_m(x)| > epsilon/2}.$$
          Monotonicity of measure then gives that $mu({x: |f_n(x) - f_m(x)| > epsilon})$ is arbitrarily small as $n,m to infty$. Concretely: given $delta > 0$, there exists $N$ such that if $n geq N$, then $mu({x : |f_n(x) - f(x)|> epsilon/2}) < delta/2$. For this $N$, if $n,m geq N$, then also $mu({x: |f_n(x) - f_m(x)| > epsilon}) < delta$.






          share|cite|improve this answer





















          • I feel like this should have been stated as a theorem/corollary on its own! Thank you very much. This answers my question perfectly.
            – carsandpulsars
            Nov 29 at 5:37















          up vote
          1
          down vote



          accepted










          Convergence in measure implies Cauchy in measure.



          Let our measure be $mu$. if $f_n to f$ in measure, then given $epsilon > 0$, we have $mu({x : |f_n(x) - f(x)| > epsilon/2})to 0$ as $n to infty$. But as $|f_n(x) - f_m(x)| leq |f_n(x) - f(x)| + |f(x) - f_m(x)|$, we have
          $$ {x: |f_n(x) - f_m(x)| > epsilon} subseteq {x: |f_n(x) - f(x)| > epsilon/2} cup {x : |f(x) - f_m(x)| > epsilon/2}.$$
          Monotonicity of measure then gives that $mu({x: |f_n(x) - f_m(x)| > epsilon})$ is arbitrarily small as $n,m to infty$. Concretely: given $delta > 0$, there exists $N$ such that if $n geq N$, then $mu({x : |f_n(x) - f(x)|> epsilon/2}) < delta/2$. For this $N$, if $n,m geq N$, then also $mu({x: |f_n(x) - f_m(x)| > epsilon}) < delta$.






          share|cite|improve this answer





















          • I feel like this should have been stated as a theorem/corollary on its own! Thank you very much. This answers my question perfectly.
            – carsandpulsars
            Nov 29 at 5:37













          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          Convergence in measure implies Cauchy in measure.



          Let our measure be $mu$. if $f_n to f$ in measure, then given $epsilon > 0$, we have $mu({x : |f_n(x) - f(x)| > epsilon/2})to 0$ as $n to infty$. But as $|f_n(x) - f_m(x)| leq |f_n(x) - f(x)| + |f(x) - f_m(x)|$, we have
          $$ {x: |f_n(x) - f_m(x)| > epsilon} subseteq {x: |f_n(x) - f(x)| > epsilon/2} cup {x : |f(x) - f_m(x)| > epsilon/2}.$$
          Monotonicity of measure then gives that $mu({x: |f_n(x) - f_m(x)| > epsilon})$ is arbitrarily small as $n,m to infty$. Concretely: given $delta > 0$, there exists $N$ such that if $n geq N$, then $mu({x : |f_n(x) - f(x)|> epsilon/2}) < delta/2$. For this $N$, if $n,m geq N$, then also $mu({x: |f_n(x) - f_m(x)| > epsilon}) < delta$.






          share|cite|improve this answer












          Convergence in measure implies Cauchy in measure.



          Let our measure be $mu$. if $f_n to f$ in measure, then given $epsilon > 0$, we have $mu({x : |f_n(x) - f(x)| > epsilon/2})to 0$ as $n to infty$. But as $|f_n(x) - f_m(x)| leq |f_n(x) - f(x)| + |f(x) - f_m(x)|$, we have
          $$ {x: |f_n(x) - f_m(x)| > epsilon} subseteq {x: |f_n(x) - f(x)| > epsilon/2} cup {x : |f(x) - f_m(x)| > epsilon/2}.$$
          Monotonicity of measure then gives that $mu({x: |f_n(x) - f_m(x)| > epsilon})$ is arbitrarily small as $n,m to infty$. Concretely: given $delta > 0$, there exists $N$ such that if $n geq N$, then $mu({x : |f_n(x) - f(x)|> epsilon/2}) < delta/2$. For this $N$, if $n,m geq N$, then also $mu({x: |f_n(x) - f_m(x)| > epsilon}) < delta$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 29 at 5:14









          Joshua Mundinger

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          2,4801026












          • I feel like this should have been stated as a theorem/corollary on its own! Thank you very much. This answers my question perfectly.
            – carsandpulsars
            Nov 29 at 5:37


















          • I feel like this should have been stated as a theorem/corollary on its own! Thank you very much. This answers my question perfectly.
            – carsandpulsars
            Nov 29 at 5:37
















          I feel like this should have been stated as a theorem/corollary on its own! Thank you very much. This answers my question perfectly.
          – carsandpulsars
          Nov 29 at 5:37




          I feel like this should have been stated as a theorem/corollary on its own! Thank you very much. This answers my question perfectly.
          – carsandpulsars
          Nov 29 at 5:37


















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