What's wrong with the proof using Fatou's Lemma?
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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
asked Jan 9 at 4:32
IdonknowIdonknow
2,620950119
$endgroup$
add a comment |
$begingroup$
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
asked Jan 9 at 4:32
IdonknowIdonknow
2,620950119
$endgroup$
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
add a comment |
$begingroup$
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
asked Jan 9 at 4:32
IdonknowIdonknow
2,620950119
$endgroup$
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
asked Jan 9 at 4:32
IdonknowIdonknow
2,620950119
asked Jan 9 at 4:32
IdonknowIdonknow
2,620950119
asked Jan 9 at 4:32
IdonknowIdonknow
2,620950119
asked Jan 9 at 4:32
IdonknowIdonknow
2,620950119
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 $liminf int lvert g_nrvert<infty$.
answered Jan 9 at 6:27
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.6k41741
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
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 $liminf int lvert g_nrvert<infty$.
answered Jan 9 at 6:27
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.6k41741
$endgroup$
add a comment |
$begingroup$
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 $liminf int lvert g_nrvert<infty$.
answered Jan 9 at 6:27
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.6k41741
$endgroup$
add a comment |
$begingroup$
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 $liminf int lvert g_nrvert<infty$.
answered Jan 9 at 6:27
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.6k41741
$endgroup$
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 $liminf int lvert g_nrvert<infty$.
answered Jan 9 at 6:27
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.6k41741
answered Jan 9 at 6:27
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.6k41741
answered Jan 9 at 6:27
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.6k41741
answered Jan 9 at 6:27
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.6k41741
10.6k41741
add a comment |
add a comment |
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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