Beppo Levi's theorem, is this assertion correct?
$begingroup$
My notes report the following assertion for the theorem:
Beppo Levi's Theorem: Let $E$ be a measurable set and ${ f_n(x)}$ a sequence of integrable functions on E, such that $limlimits_{ntoinfty} f_n(x) = f(x)$ (pointwise convergence) almost everywhere on E, and $f_n(x)leq f(x)$. Then $f(x)$ is integrable on E and $limlimits_{ntoinfty} intlimits_E f_n(x) = intlimits_E f(x)$
Is this correct? Cause my book reports multiple versions of the theorem, but not this one.
lebesgue-integral
asked Jan 16 at 15:14
Baffo rastaBaffo rasta
14811
$endgroup$
|
show 1 more comment
$begingroup$
My notes report the following assertion for the theorem:
Beppo Levi's Theorem: Let $E$ be a measurable set and ${ f_n(x)}$ a sequence of integrable functions on E, such that $limlimits_{ntoinfty} f_n(x) = f(x)$ (pointwise convergence) almost everywhere on E, and $f_n(x)leq f(x)$. Then $f(x)$ is integrable on E and $limlimits_{ntoinfty} intlimits_E f_n(x) = intlimits_E f(x)$
Is this correct? Cause my book reports multiple versions of the theorem, but not this one.
lebesgue-integral
asked Jan 16 at 15:14
Baffo rastaBaffo rasta
14811
$endgroup$
$begingroup$
What does punctual convergence mean?
$endgroup$
– BigbearZzz
Jan 16 at 15:17
$begingroup$
Pointwise, my bad, translation error. Let me correct it
$endgroup$
– Baffo rasta
Jan 16 at 15:18
$begingroup$
It should be only $limlimits_{ntoinfty} f_n(x) = f(x)$ without "for every $n$" because $n$ is already in the limit sign.
$endgroup$
– BigbearZzz
Jan 16 at 15:20
$begingroup$
I don't know why, but my notes report both. I'll take your tip anyway!
$endgroup$
– Baffo rasta
Jan 16 at 15:21
$begingroup$
As is, the statement is wrong, because $E=mathbb{R}^+$, $f_n=1_{[-n,n]}$ seems to give your theorem a problem.
$endgroup$
– Mindlack
Jan 16 at 15:30
|
show 1 more comment
$begingroup$
My notes report the following assertion for the theorem:
Beppo Levi's Theorem: Let $E$ be a measurable set and ${ f_n(x)}$ a sequence of integrable functions on E, such that $limlimits_{ntoinfty} f_n(x) = f(x)$ (pointwise convergence) almost everywhere on E, and $f_n(x)leq f(x)$. Then $f(x)$ is integrable on E and $limlimits_{ntoinfty} intlimits_E f_n(x) = intlimits_E f(x)$
Is this correct? Cause my book reports multiple versions of the theorem, but not this one.
lebesgue-integral
asked Jan 16 at 15:14
Baffo rastaBaffo rasta
14811
$endgroup$
My notes report the following assertion for the theorem:
Beppo Levi's Theorem: Let $E$ be a measurable set and ${ f_n(x)}$ a sequence of integrable functions on E, such that $limlimits_{ntoinfty} f_n(x) = f(x)$ (pointwise convergence) almost everywhere on E, and $f_n(x)leq f(x)$. Then $f(x)$ is integrable on E and $limlimits_{ntoinfty} intlimits_E f_n(x) = intlimits_E f(x)$
Is this correct? Cause my book reports multiple versions of the theorem, but not this one.
lebesgue-integral
lebesgue-integral
asked Jan 16 at 15:14
Baffo rastaBaffo rasta
14811
asked Jan 16 at 15:14
Baffo rastaBaffo rasta
14811
edited Jan 16 at 15:21
Baffo rasta
asked Jan 16 at 15:14
Baffo rastaBaffo rasta
14811
asked Jan 16 at 15:14
Baffo rastaBaffo rasta
14811
asked Jan 16 at 15:14
Baffo rastaBaffo rasta
14811
14811
$begingroup$
What does punctual convergence mean?
$endgroup$
– BigbearZzz
Jan 16 at 15:17
$begingroup$
Pointwise, my bad, translation error. Let me correct it
$endgroup$
– Baffo rasta
Jan 16 at 15:18
$begingroup$
It should be only $limlimits_{ntoinfty} f_n(x) = f(x)$ without "for every $n$" because $n$ is already in the limit sign.
$endgroup$
– BigbearZzz
Jan 16 at 15:20
$begingroup$
I don't know why, but my notes report both. I'll take your tip anyway!
$endgroup$
– Baffo rasta
Jan 16 at 15:21
$begingroup$
As is, the statement is wrong, because $E=mathbb{R}^+$, $f_n=1_{[-n,n]}$ seems to give your theorem a problem.
$endgroup$
– Mindlack
Jan 16 at 15:30
|
show 1 more comment
$begingroup$
What does punctual convergence mean?
$endgroup$
– BigbearZzz
Jan 16 at 15:17
$begingroup$
Pointwise, my bad, translation error. Let me correct it
$endgroup$
– Baffo rasta
Jan 16 at 15:18
$begingroup$
It should be only $limlimits_{ntoinfty} f_n(x) = f(x)$ without "for every $n$" because $n$ is already in the limit sign.
$endgroup$
– BigbearZzz
Jan 16 at 15:20
$begingroup$
I don't know why, but my notes report both. I'll take your tip anyway!
$endgroup$
– Baffo rasta
Jan 16 at 15:21
$begingroup$
As is, the statement is wrong, because $E=mathbb{R}^+$, $f_n=1_{[-n,n]}$ seems to give your theorem a problem.
$endgroup$
– Mindlack
Jan 16 at 15:30
$begingroup$
What does punctual convergence mean?
$endgroup$
– BigbearZzz
Jan 16 at 15:17
$begingroup$
What does punctual convergence mean?
$endgroup$
– BigbearZzz
Jan 16 at 15:17
$begingroup$
Pointwise, my bad, translation error. Let me correct it
$endgroup$
– Baffo rasta
Jan 16 at 15:18
$begingroup$
Pointwise, my bad, translation error. Let me correct it
$endgroup$
– Baffo rasta
Jan 16 at 15:18
$begingroup$
It should be only $limlimits_{ntoinfty} f_n(x) = f(x)$ without "for every $n$" because $n$ is already in the limit sign.
$endgroup$
– BigbearZzz
Jan 16 at 15:20
$begingroup$
It should be only $limlimits_{ntoinfty} f_n(x) = f(x)$ without "for every $n$" because $n$ is already in the limit sign.
$endgroup$
– BigbearZzz
Jan 16 at 15:20
$begingroup$
I don't know why, but my notes report both. I'll take your tip anyway!
$endgroup$
– Baffo rasta
Jan 16 at 15:21
$begingroup$
I don't know why, but my notes report both. I'll take your tip anyway!
$endgroup$
– Baffo rasta
Jan 16 at 15:21
$begingroup$
As is, the statement is wrong, because $E=mathbb{R}^+$, $f_n=1_{[-n,n]}$ seems to give your theorem a problem.
$endgroup$
– Mindlack
Jan 16 at 15:30
$begingroup$
As is, the statement is wrong, because $E=mathbb{R}^+$, $f_n=1_{[-n,n]}$ seems to give your theorem a problem.
$endgroup$
– Mindlack
Jan 16 at 15:30
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
First this theorem is wrong
Consider $E = mathbb R$, $f=0$ and $f_n = -chi_{[n,n+1]}$. Where $chi_{[n,n+1]}$ is the indicator function of the interval $[n,n+1]$.
They satisfy the hypothesis, but the conclusion doesn't hold as
$$ -1 = int_{mathbb R} f_n neq int_{mathbb R} f = 0$$
while $(f_n)$ converges pointwise to the always vanishing function $f$.
Second this would more related to dominated convergence theorem
See Dominated convergence theorem if the hypothesis would be correctly set.
answered Jan 16 at 15:28
mathcounterexamples.netmathcounterexamples.net
26.9k22158
$endgroup$
add a comment |
$begingroup$
The theorem is usually known as the monotone convergence theorem and comes with an extra requirement that
$$
f_1(x)le f_2(x)ledots le f(x)
$$
for almost every $x$. Other than this the other parts of your statement is correct.
answered Jan 16 at 15:24
BigbearZzzBigbearZzz
8,96021652
$endgroup$
$begingroup$
My notes report that the classical statement requires that $f_1(x) < f_2(x) < dots < f_n(x) < dots$ but it can be proved not to be necessary... Isn't that right?
$endgroup$
– Baffo rasta
Jan 16 at 15:26
$begingroup$
That’s not the monotone convergence theorem. There is no assumption on the sequence $f_n$.
$endgroup$
– Mindlack
Jan 16 at 15:27
$begingroup$
@Bafforasta Then you'd need extra assumptions. As it is now the statement is not true.
$endgroup$
– BigbearZzz
Jan 16 at 15:30
$begingroup$
@Mindlack I know the statement in the OP is not the MCT that's why I said that it missed one assumption, i.e. monotonicity. The reason being the name Beppo Levi is associated with the MCT so I made the link.
$endgroup$
– BigbearZzz
Jan 16 at 15:35
$begingroup$
@BigbearZzz I just found this post: math.stackexchange.com/questions/1177788/beppo-levis-theorem this looks pretty similar
$endgroup$
– Baffo rasta
Jan 16 at 15:46
|
show 3 more comments
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
First this theorem is wrong
Consider $E = mathbb R$, $f=0$ and $f_n = -chi_{[n,n+1]}$. Where $chi_{[n,n+1]}$ is the indicator function of the interval $[n,n+1]$.
They satisfy the hypothesis, but the conclusion doesn't hold as
$$ -1 = int_{mathbb R} f_n neq int_{mathbb R} f = 0$$
while $(f_n)$ converges pointwise to the always vanishing function $f$.
Second this would more related to dominated convergence theorem
See Dominated convergence theorem if the hypothesis would be correctly set.
answered Jan 16 at 15:28
mathcounterexamples.netmathcounterexamples.net
26.9k22158
$endgroup$
add a comment |
$begingroup$
First this theorem is wrong
Consider $E = mathbb R$, $f=0$ and $f_n = -chi_{[n,n+1]}$. Where $chi_{[n,n+1]}$ is the indicator function of the interval $[n,n+1]$.
They satisfy the hypothesis, but the conclusion doesn't hold as
$$ -1 = int_{mathbb R} f_n neq int_{mathbb R} f = 0$$
while $(f_n)$ converges pointwise to the always vanishing function $f$.
Second this would more related to dominated convergence theorem
See Dominated convergence theorem if the hypothesis would be correctly set.
answered Jan 16 at 15:28
mathcounterexamples.netmathcounterexamples.net
26.9k22158
$endgroup$
add a comment |
$begingroup$
First this theorem is wrong
Consider $E = mathbb R$, $f=0$ and $f_n = -chi_{[n,n+1]}$. Where $chi_{[n,n+1]}$ is the indicator function of the interval $[n,n+1]$.
They satisfy the hypothesis, but the conclusion doesn't hold as
$$ -1 = int_{mathbb R} f_n neq int_{mathbb R} f = 0$$
while $(f_n)$ converges pointwise to the always vanishing function $f$.
Second this would more related to dominated convergence theorem
See Dominated convergence theorem if the hypothesis would be correctly set.
answered Jan 16 at 15:28
mathcounterexamples.netmathcounterexamples.net
26.9k22158
$endgroup$
First this theorem is wrong
Consider $E = mathbb R$, $f=0$ and $f_n = -chi_{[n,n+1]}$. Where $chi_{[n,n+1]}$ is the indicator function of the interval $[n,n+1]$.
They satisfy the hypothesis, but the conclusion doesn't hold as
$$ -1 = int_{mathbb R} f_n neq int_{mathbb R} f = 0$$
while $(f_n)$ converges pointwise to the always vanishing function $f$.
Second this would more related to dominated convergence theorem
See Dominated convergence theorem if the hypothesis would be correctly set.
answered Jan 16 at 15:28
mathcounterexamples.netmathcounterexamples.net
26.9k22158
answered Jan 16 at 15:28
mathcounterexamples.netmathcounterexamples.net
26.9k22158
answered Jan 16 at 15:28
mathcounterexamples.netmathcounterexamples.net
26.9k22158
answered Jan 16 at 15:28
mathcounterexamples.netmathcounterexamples.net
26.9k22158
26.9k22158
add a comment |
add a comment |
$begingroup$
The theorem is usually known as the monotone convergence theorem and comes with an extra requirement that
$$
f_1(x)le f_2(x)ledots le f(x)
$$
for almost every $x$. Other than this the other parts of your statement is correct.
answered Jan 16 at 15:24
BigbearZzzBigbearZzz
8,96021652
$endgroup$
$begingroup$
My notes report that the classical statement requires that $f_1(x) < f_2(x) < dots < f_n(x) < dots$ but it can be proved not to be necessary... Isn't that right?
$endgroup$
– Baffo rasta
Jan 16 at 15:26
$begingroup$
That’s not the monotone convergence theorem. There is no assumption on the sequence $f_n$.
$endgroup$
– Mindlack
Jan 16 at 15:27
$begingroup$
@Bafforasta Then you'd need extra assumptions. As it is now the statement is not true.
$endgroup$
– BigbearZzz
Jan 16 at 15:30
$begingroup$
@Mindlack I know the statement in the OP is not the MCT that's why I said that it missed one assumption, i.e. monotonicity. The reason being the name Beppo Levi is associated with the MCT so I made the link.
$endgroup$
– BigbearZzz
Jan 16 at 15:35
$begingroup$
@BigbearZzz I just found this post: math.stackexchange.com/questions/1177788/beppo-levis-theorem this looks pretty similar
$endgroup$
– Baffo rasta
Jan 16 at 15:46
|
show 3 more comments
$begingroup$
The theorem is usually known as the monotone convergence theorem and comes with an extra requirement that
$$
f_1(x)le f_2(x)ledots le f(x)
$$
for almost every $x$. Other than this the other parts of your statement is correct.
answered Jan 16 at 15:24
BigbearZzzBigbearZzz
8,96021652
$endgroup$
$begingroup$
My notes report that the classical statement requires that $f_1(x) < f_2(x) < dots < f_n(x) < dots$ but it can be proved not to be necessary... Isn't that right?
$endgroup$
– Baffo rasta
Jan 16 at 15:26
$begingroup$
That’s not the monotone convergence theorem. There is no assumption on the sequence $f_n$.
$endgroup$
– Mindlack
Jan 16 at 15:27
$begingroup$
@Bafforasta Then you'd need extra assumptions. As it is now the statement is not true.
$endgroup$
– BigbearZzz
Jan 16 at 15:30
$begingroup$
@Mindlack I know the statement in the OP is not the MCT that's why I said that it missed one assumption, i.e. monotonicity. The reason being the name Beppo Levi is associated with the MCT so I made the link.
$endgroup$
– BigbearZzz
Jan 16 at 15:35
$begingroup$
@BigbearZzz I just found this post: math.stackexchange.com/questions/1177788/beppo-levis-theorem this looks pretty similar
$endgroup$
– Baffo rasta
Jan 16 at 15:46
|
show 3 more comments
$begingroup$
The theorem is usually known as the monotone convergence theorem and comes with an extra requirement that
$$
f_1(x)le f_2(x)ledots le f(x)
$$
for almost every $x$. Other than this the other parts of your statement is correct.
answered Jan 16 at 15:24
BigbearZzzBigbearZzz
8,96021652
$endgroup$
The theorem is usually known as the monotone convergence theorem and comes with an extra requirement that
$$
f_1(x)le f_2(x)ledots le f(x)
$$
for almost every $x$. Other than this the other parts of your statement is correct.
answered Jan 16 at 15:24
BigbearZzzBigbearZzz
8,96021652
answered Jan 16 at 15:24
BigbearZzzBigbearZzz
8,96021652
answered Jan 16 at 15:24
BigbearZzzBigbearZzz
8,96021652
answered Jan 16 at 15:24
BigbearZzzBigbearZzz
8,96021652
8,96021652
$begingroup$
My notes report that the classical statement requires that $f_1(x) < f_2(x) < dots < f_n(x) < dots$ but it can be proved not to be necessary... Isn't that right?
$endgroup$
– Baffo rasta
Jan 16 at 15:26
$begingroup$
That’s not the monotone convergence theorem. There is no assumption on the sequence $f_n$.
$endgroup$
– Mindlack
Jan 16 at 15:27
$begingroup$
@Bafforasta Then you'd need extra assumptions. As it is now the statement is not true.
$endgroup$
– BigbearZzz
Jan 16 at 15:30
$begingroup$
@Mindlack I know the statement in the OP is not the MCT that's why I said that it missed one assumption, i.e. monotonicity. The reason being the name Beppo Levi is associated with the MCT so I made the link.
$endgroup$
– BigbearZzz
Jan 16 at 15:35
$begingroup$
@BigbearZzz I just found this post: math.stackexchange.com/questions/1177788/beppo-levis-theorem this looks pretty similar
$endgroup$
– Baffo rasta
Jan 16 at 15:46
|
show 3 more comments
$begingroup$
My notes report that the classical statement requires that $f_1(x) < f_2(x) < dots < f_n(x) < dots$ but it can be proved not to be necessary... Isn't that right?
$endgroup$
– Baffo rasta
Jan 16 at 15:26
$begingroup$
That’s not the monotone convergence theorem. There is no assumption on the sequence $f_n$.
$endgroup$
– Mindlack
Jan 16 at 15:27
$begingroup$
@Bafforasta Then you'd need extra assumptions. As it is now the statement is not true.
$endgroup$
– BigbearZzz
Jan 16 at 15:30
$begingroup$
@Mindlack I know the statement in the OP is not the MCT that's why I said that it missed one assumption, i.e. monotonicity. The reason being the name Beppo Levi is associated with the MCT so I made the link.
$endgroup$
– BigbearZzz
Jan 16 at 15:35
$begingroup$
@BigbearZzz I just found this post: math.stackexchange.com/questions/1177788/beppo-levis-theorem this looks pretty similar
$endgroup$
– Baffo rasta
Jan 16 at 15:46
$begingroup$
My notes report that the classical statement requires that $f_1(x) < f_2(x) < dots < f_n(x) < dots$ but it can be proved not to be necessary... Isn't that right?
$endgroup$
– Baffo rasta
Jan 16 at 15:26
$begingroup$
My notes report that the classical statement requires that $f_1(x) < f_2(x) < dots < f_n(x) < dots$ but it can be proved not to be necessary... Isn't that right?
$endgroup$
– Baffo rasta
Jan 16 at 15:26
$begingroup$
That’s not the monotone convergence theorem. There is no assumption on the sequence $f_n$.
$endgroup$
– Mindlack
Jan 16 at 15:27
$begingroup$
That’s not the monotone convergence theorem. There is no assumption on the sequence $f_n$.
$endgroup$
– Mindlack
Jan 16 at 15:27
$begingroup$
@Bafforasta Then you'd need extra assumptions. As it is now the statement is not true.
$endgroup$
– BigbearZzz
Jan 16 at 15:30
$begingroup$
@Bafforasta Then you'd need extra assumptions. As it is now the statement is not true.
$endgroup$
– BigbearZzz
Jan 16 at 15:30
$begingroup$
@Mindlack I know the statement in the OP is not the MCT that's why I said that it missed one assumption, i.e. monotonicity. The reason being the name Beppo Levi is associated with the MCT so I made the link.
$endgroup$
– BigbearZzz
Jan 16 at 15:35
$begingroup$
@Mindlack I know the statement in the OP is not the MCT that's why I said that it missed one assumption, i.e. monotonicity. The reason being the name Beppo Levi is associated with the MCT so I made the link.
$endgroup$
– BigbearZzz
Jan 16 at 15:35
$begingroup$
@BigbearZzz I just found this post: math.stackexchange.com/questions/1177788/beppo-levis-theorem this looks pretty similar
$endgroup$
– Baffo rasta
Jan 16 at 15:46
$begingroup$
@BigbearZzz I just found this post: math.stackexchange.com/questions/1177788/beppo-levis-theorem this looks pretty similar
$endgroup$
– Baffo rasta
Jan 16 at 15:46
|
show 3 more comments
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$begingroup$
What does punctual convergence mean?
$endgroup$
– BigbearZzz
Jan 16 at 15:17
$begingroup$
Pointwise, my bad, translation error. Let me correct it
$endgroup$
– Baffo rasta
Jan 16 at 15:18
$begingroup$
It should be only $limlimits_{ntoinfty} f_n(x) = f(x)$ without "for every $n$" because $n$ is already in the limit sign.
$endgroup$
– BigbearZzz
Jan 16 at 15:20
$begingroup$
I don't know why, but my notes report both. I'll take your tip anyway!
$endgroup$
– Baffo rasta
Jan 16 at 15:21
$begingroup$
As is, the statement is wrong, because $E=mathbb{R}^+$, $f_n=1_{[-n,n]}$ seems to give your theorem a problem.
$endgroup$
– Mindlack
Jan 16 at 15:30