What's wrong with the proof using Fatou's Lemma?












1












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Fatou's Lemma states that given any measure space $(Omega,Sigma,mu)$ and $XinSigma,$ let $(f_n)_{n=1}^infty$ be a sequence of real-valued nonnegative measurable functions $f_n:Xto [0,+infty]$ converges to $f$ point wise.
Then
$$int_X f dmuleq liminf int_X f_n,dmu.$$




If we assume that $g_n$ is integrable for each $n,$ then its pointwise limit $g$ may not be integrable.



However, what goes wrong with the 'proof' below?




Since each $g_n$ is integrable, so
$$int_X |g_n| ,dmu<infty. $$
Since $(|g_n|)$ is a sequence of nonnegative measurable functions, by Fatou's Lemma, we have
$$int_X |g|,dmuleq liminf int_X |g_n|,dmu<infty.$$
Therefore, $g$ is integrable.







real-analysis measure-theory proof-explanation lebesgue-integral







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$endgroup$








  • 5




    $begingroup$
    Why is it true that $liminf int |g_n|$ is finite?
    $endgroup$
    – D. Brogan
    Jan 9 at 4:35










  • $begingroup$
    $x_n = n < infty$ and $liminf x_n = infty$.
    $endgroup$
    – RRL
    Jan 9 at 4:38
















1












$begingroup$


Fatou's Lemma states that given any measure space $(Omega,Sigma,mu)$ and $XinSigma,$ let $(f_n)_{n=1}^infty$ be a sequence of real-valued nonnegative measurable functions $f_n:Xto [0,+infty]$ converges to $f$ point wise.
Then
$$int_X f dmuleq liminf int_X f_n,dmu.$$




If we assume that $g_n$ is integrable for each $n,$ then its pointwise limit $g$ may not be integrable.



However, what goes wrong with the 'proof' below?




Since each $g_n$ is integrable, so
$$int_X |g_n| ,dmu<infty. $$
Since $(|g_n|)$ is a sequence of nonnegative measurable functions, by Fatou's Lemma, we have
$$int_X |g|,dmuleq liminf int_X |g_n|,dmu<infty.$$
Therefore, $g$ is integrable.







real-analysis measure-theory proof-explanation lebesgue-integral







share|cite|improve this question









$endgroup$








  • 5




    $begingroup$
    Why is it true that $liminf int |g_n|$ is finite?
    $endgroup$
    – D. Brogan
    Jan 9 at 4:35










  • $begingroup$
    $x_n = n < infty$ and $liminf x_n = infty$.
    $endgroup$
    – RRL
    Jan 9 at 4:38














1












1








1





$begingroup$


Fatou's Lemma states that given any measure space $(Omega,Sigma,mu)$ and $XinSigma,$ let $(f_n)_{n=1}^infty$ be a sequence of real-valued nonnegative measurable functions $f_n:Xto [0,+infty]$ converges to $f$ point wise.
Then
$$int_X f dmuleq liminf int_X f_n,dmu.$$




If we assume that $g_n$ is integrable for each $n,$ then its pointwise limit $g$ may not be integrable.



However, what goes wrong with the 'proof' below?




Since each $g_n$ is integrable, so
$$int_X |g_n| ,dmu<infty. $$
Since $(|g_n|)$ is a sequence of nonnegative measurable functions, by Fatou's Lemma, we have
$$int_X |g|,dmuleq liminf int_X |g_n|,dmu<infty.$$
Therefore, $g$ is integrable.







real-analysis measure-theory proof-explanation lebesgue-integral







share|cite|improve this question









$endgroup$




Fatou's Lemma states that given any measure space $(Omega,Sigma,mu)$ and $XinSigma,$ let $(f_n)_{n=1}^infty$ be a sequence of real-valued nonnegative measurable functions $f_n:Xto [0,+infty]$ converges to $f$ point wise.
Then
$$int_X f dmuleq liminf int_X f_n,dmu.$$




If we assume that $g_n$ is integrable for each $n,$ then its pointwise limit $g$ may not be integrable.



However, what goes wrong with the 'proof' below?




Since each $g_n$ is integrable, so
$$int_X |g_n| ,dmu<infty. $$
Since $(|g_n|)$ is a sequence of nonnegative measurable functions, by Fatou's Lemma, we have
$$int_X |g|,dmuleq liminf int_X |g_n|,dmu<infty.$$
Therefore, $g$ is integrable.







real-analysis measure-theory proof-explanation lebesgue-integral




real-analysis measure-theory proof-explanation lebesgue-integral






share|cite|improve this question













share|cite|improve this question











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asked Jan 9 at 4:32









IdonknowIdonknow

2,620950119




2,620950119








  • 5




    $begingroup$
    Why is it true that $liminf int |g_n|$ is finite?
    $endgroup$
    – D. Brogan
    Jan 9 at 4:35










  • $begingroup$
    $x_n = n < infty$ and $liminf x_n = infty$.
    $endgroup$
    – RRL
    Jan 9 at 4:38














  • 5




    $begingroup$
    Why is it true that $liminf int |g_n|$ is finite?
    $endgroup$
    – D. Brogan
    Jan 9 at 4:35










  • $begingroup$
    $x_n = n < infty$ and $liminf x_n = infty$.
    $endgroup$
    – RRL
    Jan 9 at 4:38








5




5




$begingroup$
Why is it true that $liminf int |g_n|$ is finite?
$endgroup$
– D. Brogan
Jan 9 at 4:35




$begingroup$
Why is it true that $liminf int |g_n|$ is finite?
$endgroup$
– D. Brogan
Jan 9 at 4:35












$begingroup$
$x_n = n < infty$ and $liminf x_n = infty$.
$endgroup$
– RRL
Jan 9 at 4:38




$begingroup$
$x_n = n < infty$ and $liminf x_n = infty$.
$endgroup$
– RRL
Jan 9 at 4:38










1 Answer
1






active

oldest

votes


















2












$begingroup$

Take $X=mathbb{R}$ equipped with the Lebesgue Measure. Set $g_n(x)=chi_{[-n,n]}(x)$ i.e.
$$ g_n(x)=
begin{cases}
1&xin [-n,n]\
0&xnotin[-n,n].
end{cases}$$
$g_nin L^1(mathbb{R})$ for all $n$. Moreover,
$$ int_mathbb{R} g_n(x)dx=2n.$$
The pointwise limit $g(x)$ exists, and is the constant $1$ function. However, if we check our bounds, we see that
$$ liminf_{ntoinfty} int_{mathbb{R}} lvert g_n(x)rvert dx=liminf_{ntoinfty} :2n=infty$$
and as such provides no "bound" on $int g(x)dx$. So, your argument was flawed because you assumed that $liminf int lvert g_nrvert<infty$.






share|cite|improve this answer









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    active

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    1 Answer
    1






    active

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    2












    $begingroup$

    Take $X=mathbb{R}$ equipped with the Lebesgue Measure. Set $g_n(x)=chi_{[-n,n]}(x)$ i.e.
    $$ g_n(x)=
    begin{cases}
    1&xin [-n,n]\
    0&xnotin[-n,n].
    end{cases}$$
    $g_nin L^1(mathbb{R})$ for all $n$. Moreover,
    $$ int_mathbb{R} g_n(x)dx=2n.$$
    The pointwise limit $g(x)$ exists, and is the constant $1$ function. However, if we check our bounds, we see that
    $$ liminf_{ntoinfty} int_{mathbb{R}} lvert g_n(x)rvert dx=liminf_{ntoinfty} :2n=infty$$
    and as such provides no "bound" on $int g(x)dx$. So, your argument was flawed because you assumed that $liminf int lvert g_nrvert<infty$.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Take $X=mathbb{R}$ equipped with the Lebesgue Measure. Set $g_n(x)=chi_{[-n,n]}(x)$ i.e.
      $$ g_n(x)=
      begin{cases}
      1&xin [-n,n]\
      0&xnotin[-n,n].
      end{cases}$$
      $g_nin L^1(mathbb{R})$ for all $n$. Moreover,
      $$ int_mathbb{R} g_n(x)dx=2n.$$
      The pointwise limit $g(x)$ exists, and is the constant $1$ function. However, if we check our bounds, we see that
      $$ liminf_{ntoinfty} int_{mathbb{R}} lvert g_n(x)rvert dx=liminf_{ntoinfty} :2n=infty$$
      and as such provides no "bound" on $int g(x)dx$. So, your argument was flawed because you assumed that $liminf int lvert g_nrvert<infty$.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Take $X=mathbb{R}$ equipped with the Lebesgue Measure. Set $g_n(x)=chi_{[-n,n]}(x)$ i.e.
        $$ g_n(x)=
        begin{cases}
        1&xin [-n,n]\
        0&xnotin[-n,n].
        end{cases}$$
        $g_nin L^1(mathbb{R})$ for all $n$. Moreover,
        $$ int_mathbb{R} g_n(x)dx=2n.$$
        The pointwise limit $g(x)$ exists, and is the constant $1$ function. However, if we check our bounds, we see that
        $$ liminf_{ntoinfty} int_{mathbb{R}} lvert g_n(x)rvert dx=liminf_{ntoinfty} :2n=infty$$
        and as such provides no "bound" on $int g(x)dx$. So, your argument was flawed because you assumed that $liminf int lvert g_nrvert<infty$.






        share|cite|improve this answer









        $endgroup$



        Take $X=mathbb{R}$ equipped with the Lebesgue Measure. Set $g_n(x)=chi_{[-n,n]}(x)$ i.e.
        $$ g_n(x)=
        begin{cases}
        1&xin [-n,n]\
        0&xnotin[-n,n].
        end{cases}$$
        $g_nin L^1(mathbb{R})$ for all $n$. Moreover,
        $$ int_mathbb{R} g_n(x)dx=2n.$$
        The pointwise limit $g(x)$ exists, and is the constant $1$ function. However, if we check our bounds, we see that
        $$ liminf_{ntoinfty} int_{mathbb{R}} lvert g_n(x)rvert dx=liminf_{ntoinfty} :2n=infty$$
        and as such provides no "bound" on $int g(x)dx$. So, your argument was flawed because you assumed that $liminf int lvert g_nrvert<infty$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 9 at 6:27









        Antonios-Alexandros RobotisAntonios-Alexandros Robotis

        10.6k41741




        10.6k41741






























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