Prove Lebesgue integrability












1












$begingroup$


Let $f_n in L_0(X), |f_n|leqslantphi in L_1(X), n in N$ and $f_n rightarrow f$ in measure. Prove that $f in L_1(X)$ and
$$lim_{ntoinfty}int_{X}{f_n}dmu = int_{X} fdmu$$



Here $L_0(X)$ stands for Lebesgue measurability, and $L_1(X)$ — for Lebesgue integrability on $X$.



As far as I understand, we should use the fact that $f_n$ is bounded by a Lebesgue integrable function being itself measurable, so $f$ is also Lebesgue integrable. Is that correct? I remember similar reasoning being used in the classroom.
If the first part is correct, how do I prove the equality?






measure-theory lebesgue-integral lebesgue-measure







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    Let $f_n in L_0(X), |f_n|leqslantphi in L_1(X), n in N$ and $f_n rightarrow f$ in measure. Prove that $f in L_1(X)$ and
    $$lim_{ntoinfty}int_{X}{f_n}dmu = int_{X} fdmu$$



    Here $L_0(X)$ stands for Lebesgue measurability, and $L_1(X)$ — for Lebesgue integrability on $X$.



    As far as I understand, we should use the fact that $f_n$ is bounded by a Lebesgue integrable function being itself measurable, so $f$ is also Lebesgue integrable. Is that correct? I remember similar reasoning being used in the classroom.
    If the first part is correct, how do I prove the equality?






    measure-theory lebesgue-integral lebesgue-measure







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Let $f_n in L_0(X), |f_n|leqslantphi in L_1(X), n in N$ and $f_n rightarrow f$ in measure. Prove that $f in L_1(X)$ and
      $$lim_{ntoinfty}int_{X}{f_n}dmu = int_{X} fdmu$$



      Here $L_0(X)$ stands for Lebesgue measurability, and $L_1(X)$ — for Lebesgue integrability on $X$.



      As far as I understand, we should use the fact that $f_n$ is bounded by a Lebesgue integrable function being itself measurable, so $f$ is also Lebesgue integrable. Is that correct? I remember similar reasoning being used in the classroom.
      If the first part is correct, how do I prove the equality?






      measure-theory lebesgue-integral lebesgue-measure







      share|cite|improve this question









      $endgroup$




      Let $f_n in L_0(X), |f_n|leqslantphi in L_1(X), n in N$ and $f_n rightarrow f$ in measure. Prove that $f in L_1(X)$ and
      $$lim_{ntoinfty}int_{X}{f_n}dmu = int_{X} fdmu$$



      Here $L_0(X)$ stands for Lebesgue measurability, and $L_1(X)$ — for Lebesgue integrability on $X$.



      As far as I understand, we should use the fact that $f_n$ is bounded by a Lebesgue integrable function being itself measurable, so $f$ is also Lebesgue integrable. Is that correct? I remember similar reasoning being used in the classroom.
      If the first part is correct, how do I prove the equality?






      measure-theory lebesgue-integral lebesgue-measure




      measure-theory lebesgue-integral lebesgue-measure






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 25 '18 at 16:17









      Don DraperDon Draper

      87110




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          1 Answer
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          2












          $begingroup$

          Let ${f_{n_k}}_{k=1}^infty$ be an arbitrary subsequence of ${f_n}_{n=1}^infty$. Then $f_{n_k} to f$ in measure, so one may extract a further subsequence ${f_{n_{k_j}}}_{j=1}^infty$ of ${f_{n_k}}_{k=1}^infty$ such that $f_{n_{k_j}}to f$ pointwise a.e. in $X$. By Fatou's lemma $fin L_1(X)$; by the dominated convergence theorem $int_X f, dmu = limlimits_{jto infty} int_X f_{n_{k_j}}, dmu$. Since ${f_{n_k}}_{k=1}^infty$ was arbitrary, the result follows.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Why does the result follow from $f_{n_k}$ being arbitrary?
            $endgroup$
            – Guacho Perez
            Dec 25 '18 at 19:29










          • $begingroup$
            @Gaucho If every subsequence of a sequence of real numbers converges to some number $A$, then the sequence converges to $A$. In this case, the sequence under consideration is $int_X f_n, dmu$.
            $endgroup$
            – kobe
            Dec 25 '18 at 19:30












          • $begingroup$
            But it is only proved that a subsequence of every subsequence converges to $int f$.
            $endgroup$
            – Guacho Perez
            Dec 25 '18 at 19:45










          • $begingroup$
            @GuachoPerez no, it has been shown that every subsequence of $int_X f_n, dmu$ has a further subsequence which converges to $int_X f, dmu$, so $int_X f_n, dmu to int_X f, dmu$. Indeed, since $int_X f_{n_j}, dmu$ has a subsequence which converges to $int_X f, dmu$, then $int_X f_{n_j}, dmu$ to converges to the same number. But since $int_X f_{n_j}, dmu$ is an arbitrary subsequence of $int_X f_n, dmu$, $lim_n int_X f_n, dmu = int_X f, dmu$.
            $endgroup$
            – kobe
            Dec 25 '18 at 19:53












          • $begingroup$
            @Gaucho take a look here: math.stackexchange.com/questions/397978/….
            $endgroup$
            – kobe
            Dec 25 '18 at 19:55












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          1 Answer
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          1 Answer
          1






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          active

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          2












          $begingroup$

          Let ${f_{n_k}}_{k=1}^infty$ be an arbitrary subsequence of ${f_n}_{n=1}^infty$. Then $f_{n_k} to f$ in measure, so one may extract a further subsequence ${f_{n_{k_j}}}_{j=1}^infty$ of ${f_{n_k}}_{k=1}^infty$ such that $f_{n_{k_j}}to f$ pointwise a.e. in $X$. By Fatou's lemma $fin L_1(X)$; by the dominated convergence theorem $int_X f, dmu = limlimits_{jto infty} int_X f_{n_{k_j}}, dmu$. Since ${f_{n_k}}_{k=1}^infty$ was arbitrary, the result follows.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Why does the result follow from $f_{n_k}$ being arbitrary?
            $endgroup$
            – Guacho Perez
            Dec 25 '18 at 19:29










          • $begingroup$
            @Gaucho If every subsequence of a sequence of real numbers converges to some number $A$, then the sequence converges to $A$. In this case, the sequence under consideration is $int_X f_n, dmu$.
            $endgroup$
            – kobe
            Dec 25 '18 at 19:30












          • $begingroup$
            But it is only proved that a subsequence of every subsequence converges to $int f$.
            $endgroup$
            – Guacho Perez
            Dec 25 '18 at 19:45










          • $begingroup$
            @GuachoPerez no, it has been shown that every subsequence of $int_X f_n, dmu$ has a further subsequence which converges to $int_X f, dmu$, so $int_X f_n, dmu to int_X f, dmu$. Indeed, since $int_X f_{n_j}, dmu$ has a subsequence which converges to $int_X f, dmu$, then $int_X f_{n_j}, dmu$ to converges to the same number. But since $int_X f_{n_j}, dmu$ is an arbitrary subsequence of $int_X f_n, dmu$, $lim_n int_X f_n, dmu = int_X f, dmu$.
            $endgroup$
            – kobe
            Dec 25 '18 at 19:53












          • $begingroup$
            @Gaucho take a look here: math.stackexchange.com/questions/397978/….
            $endgroup$
            – kobe
            Dec 25 '18 at 19:55
















          2












          $begingroup$

          Let ${f_{n_k}}_{k=1}^infty$ be an arbitrary subsequence of ${f_n}_{n=1}^infty$. Then $f_{n_k} to f$ in measure, so one may extract a further subsequence ${f_{n_{k_j}}}_{j=1}^infty$ of ${f_{n_k}}_{k=1}^infty$ such that $f_{n_{k_j}}to f$ pointwise a.e. in $X$. By Fatou's lemma $fin L_1(X)$; by the dominated convergence theorem $int_X f, dmu = limlimits_{jto infty} int_X f_{n_{k_j}}, dmu$. Since ${f_{n_k}}_{k=1}^infty$ was arbitrary, the result follows.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Why does the result follow from $f_{n_k}$ being arbitrary?
            $endgroup$
            – Guacho Perez
            Dec 25 '18 at 19:29










          • $begingroup$
            @Gaucho If every subsequence of a sequence of real numbers converges to some number $A$, then the sequence converges to $A$. In this case, the sequence under consideration is $int_X f_n, dmu$.
            $endgroup$
            – kobe
            Dec 25 '18 at 19:30












          • $begingroup$
            But it is only proved that a subsequence of every subsequence converges to $int f$.
            $endgroup$
            – Guacho Perez
            Dec 25 '18 at 19:45










          • $begingroup$
            @GuachoPerez no, it has been shown that every subsequence of $int_X f_n, dmu$ has a further subsequence which converges to $int_X f, dmu$, so $int_X f_n, dmu to int_X f, dmu$. Indeed, since $int_X f_{n_j}, dmu$ has a subsequence which converges to $int_X f, dmu$, then $int_X f_{n_j}, dmu$ to converges to the same number. But since $int_X f_{n_j}, dmu$ is an arbitrary subsequence of $int_X f_n, dmu$, $lim_n int_X f_n, dmu = int_X f, dmu$.
            $endgroup$
            – kobe
            Dec 25 '18 at 19:53












          • $begingroup$
            @Gaucho take a look here: math.stackexchange.com/questions/397978/….
            $endgroup$
            – kobe
            Dec 25 '18 at 19:55














          2












          2








          2





          $begingroup$

          Let ${f_{n_k}}_{k=1}^infty$ be an arbitrary subsequence of ${f_n}_{n=1}^infty$. Then $f_{n_k} to f$ in measure, so one may extract a further subsequence ${f_{n_{k_j}}}_{j=1}^infty$ of ${f_{n_k}}_{k=1}^infty$ such that $f_{n_{k_j}}to f$ pointwise a.e. in $X$. By Fatou's lemma $fin L_1(X)$; by the dominated convergence theorem $int_X f, dmu = limlimits_{jto infty} int_X f_{n_{k_j}}, dmu$. Since ${f_{n_k}}_{k=1}^infty$ was arbitrary, the result follows.






          share|cite|improve this answer









          $endgroup$



          Let ${f_{n_k}}_{k=1}^infty$ be an arbitrary subsequence of ${f_n}_{n=1}^infty$. Then $f_{n_k} to f$ in measure, so one may extract a further subsequence ${f_{n_{k_j}}}_{j=1}^infty$ of ${f_{n_k}}_{k=1}^infty$ such that $f_{n_{k_j}}to f$ pointwise a.e. in $X$. By Fatou's lemma $fin L_1(X)$; by the dominated convergence theorem $int_X f, dmu = limlimits_{jto infty} int_X f_{n_{k_j}}, dmu$. Since ${f_{n_k}}_{k=1}^infty$ was arbitrary, the result follows.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 25 '18 at 17:21









          kobekobe

          35.1k22248




          35.1k22248












          • $begingroup$
            Why does the result follow from $f_{n_k}$ being arbitrary?
            $endgroup$
            – Guacho Perez
            Dec 25 '18 at 19:29










          • $begingroup$
            @Gaucho If every subsequence of a sequence of real numbers converges to some number $A$, then the sequence converges to $A$. In this case, the sequence under consideration is $int_X f_n, dmu$.
            $endgroup$
            – kobe
            Dec 25 '18 at 19:30












          • $begingroup$
            But it is only proved that a subsequence of every subsequence converges to $int f$.
            $endgroup$
            – Guacho Perez
            Dec 25 '18 at 19:45










          • $begingroup$
            @GuachoPerez no, it has been shown that every subsequence of $int_X f_n, dmu$ has a further subsequence which converges to $int_X f, dmu$, so $int_X f_n, dmu to int_X f, dmu$. Indeed, since $int_X f_{n_j}, dmu$ has a subsequence which converges to $int_X f, dmu$, then $int_X f_{n_j}, dmu$ to converges to the same number. But since $int_X f_{n_j}, dmu$ is an arbitrary subsequence of $int_X f_n, dmu$, $lim_n int_X f_n, dmu = int_X f, dmu$.
            $endgroup$
            – kobe
            Dec 25 '18 at 19:53












          • $begingroup$
            @Gaucho take a look here: math.stackexchange.com/questions/397978/….
            $endgroup$
            – kobe
            Dec 25 '18 at 19:55


















          • $begingroup$
            Why does the result follow from $f_{n_k}$ being arbitrary?
            $endgroup$
            – Guacho Perez
            Dec 25 '18 at 19:29










          • $begingroup$
            @Gaucho If every subsequence of a sequence of real numbers converges to some number $A$, then the sequence converges to $A$. In this case, the sequence under consideration is $int_X f_n, dmu$.
            $endgroup$
            – kobe
            Dec 25 '18 at 19:30












          • $begingroup$
            But it is only proved that a subsequence of every subsequence converges to $int f$.
            $endgroup$
            – Guacho Perez
            Dec 25 '18 at 19:45










          • $begingroup$
            @GuachoPerez no, it has been shown that every subsequence of $int_X f_n, dmu$ has a further subsequence which converges to $int_X f, dmu$, so $int_X f_n, dmu to int_X f, dmu$. Indeed, since $int_X f_{n_j}, dmu$ has a subsequence which converges to $int_X f, dmu$, then $int_X f_{n_j}, dmu$ to converges to the same number. But since $int_X f_{n_j}, dmu$ is an arbitrary subsequence of $int_X f_n, dmu$, $lim_n int_X f_n, dmu = int_X f, dmu$.
            $endgroup$
            – kobe
            Dec 25 '18 at 19:53












          • $begingroup$
            @Gaucho take a look here: math.stackexchange.com/questions/397978/….
            $endgroup$
            – kobe
            Dec 25 '18 at 19:55
















          $begingroup$
          Why does the result follow from $f_{n_k}$ being arbitrary?
          $endgroup$
          – Guacho Perez
          Dec 25 '18 at 19:29




          $begingroup$
          Why does the result follow from $f_{n_k}$ being arbitrary?
          $endgroup$
          – Guacho Perez
          Dec 25 '18 at 19:29












          $begingroup$
          @Gaucho If every subsequence of a sequence of real numbers converges to some number $A$, then the sequence converges to $A$. In this case, the sequence under consideration is $int_X f_n, dmu$.
          $endgroup$
          – kobe
          Dec 25 '18 at 19:30






          $begingroup$
          @Gaucho If every subsequence of a sequence of real numbers converges to some number $A$, then the sequence converges to $A$. In this case, the sequence under consideration is $int_X f_n, dmu$.
          $endgroup$
          – kobe
          Dec 25 '18 at 19:30














          $begingroup$
          But it is only proved that a subsequence of every subsequence converges to $int f$.
          $endgroup$
          – Guacho Perez
          Dec 25 '18 at 19:45




          $begingroup$
          But it is only proved that a subsequence of every subsequence converges to $int f$.
          $endgroup$
          – Guacho Perez
          Dec 25 '18 at 19:45












          $begingroup$
          @GuachoPerez no, it has been shown that every subsequence of $int_X f_n, dmu$ has a further subsequence which converges to $int_X f, dmu$, so $int_X f_n, dmu to int_X f, dmu$. Indeed, since $int_X f_{n_j}, dmu$ has a subsequence which converges to $int_X f, dmu$, then $int_X f_{n_j}, dmu$ to converges to the same number. But since $int_X f_{n_j}, dmu$ is an arbitrary subsequence of $int_X f_n, dmu$, $lim_n int_X f_n, dmu = int_X f, dmu$.
          $endgroup$
          – kobe
          Dec 25 '18 at 19:53






          $begingroup$
          @GuachoPerez no, it has been shown that every subsequence of $int_X f_n, dmu$ has a further subsequence which converges to $int_X f, dmu$, so $int_X f_n, dmu to int_X f, dmu$. Indeed, since $int_X f_{n_j}, dmu$ has a subsequence which converges to $int_X f, dmu$, then $int_X f_{n_j}, dmu$ to converges to the same number. But since $int_X f_{n_j}, dmu$ is an arbitrary subsequence of $int_X f_n, dmu$, $lim_n int_X f_n, dmu = int_X f, dmu$.
          $endgroup$
          – kobe
          Dec 25 '18 at 19:53














          $begingroup$
          @Gaucho take a look here: math.stackexchange.com/questions/397978/….
          $endgroup$
          – kobe
          Dec 25 '18 at 19:55




          $begingroup$
          @Gaucho take a look here: math.stackexchange.com/questions/397978/….
          $endgroup$
          – kobe
          Dec 25 '18 at 19:55


















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