Example $f_n = mathbf1_[n,infty)$ in Fatou's LemmaUnderstanding the assumptions in the Reverse Fatou's LemmaExtension of Fatou's lemmaCorollary of Fatou's LemmaFatou's Lemma in nonpositive functionFatou Lemma: Why is $liminf f_n = 0$ where $f_n = chi_[n,n+1]$$lim _ nrightarrow infty inf f_n leq t =lim _ nrightarrow infty sup f_n leq t $ Proof and IntuitionReal Analysis, 2.18 (Fatou's Lemma) Integration of Nonnegative functionsShowing that the Fatou's lemma inequality can be strict.$f_n geq f $ a.e implies $Longrightarrow int _X lim inf f_n d muleq lim inf int _X f_n dmu$Question about lim inf or lim sup
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Example $f_n = mathbf1_[n,infty)$ in Fatou's Lemma
Understanding the assumptions in the Reverse Fatou's LemmaExtension of Fatou's lemmaCorollary of Fatou's LemmaFatou's Lemma in nonpositive functionFatou Lemma: Why is $liminf f_n = 0$ where $f_n = chi_[n,n+1]$$lim _ nrightarrow infty inf f_n leq t =lim _ nrightarrow infty sup f_n leq t $ Proof and IntuitionReal Analysis, 2.18 (Fatou's Lemma) Integration of Nonnegative functionsShowing that the Fatou's lemma inequality can be strict.$f_n geq f $ a.e implies $Longrightarrow int _X lim inf f_n d muleq lim inf int _X f_n dmu$Question about lim inf or lim sup
$begingroup$
In the class, my professor gave the following example:
Let $f_n = mathbf1_[n,infty)$, then we have
$$int_X liminf f_n = 0, text since liminf f_n = 0,$$
and
$$liminf int_X f_n = infty, text since int f_n = infty, forall n.$$
I can understand "$int f_n = infty, forall n$". But even so, what is the key reason that $$int_X liminf f_n = 0 , text but , liminf int_X f_n = infty$$
To me for the second case, when $nrightarrow infty$, it should be $0$. I am confused on this point.
real-analysis limsup-and-liminf
asked Mar 29 at 16:52
sleeve chensleeve chen
3,18142255
$endgroup$
|
show 1 more comment
$begingroup$
In the class, my professor gave the following example:
Let $f_n = mathbf1_[n,infty)$, then we have
$$int_X liminf f_n = 0, text since liminf f_n = 0,$$
and
$$liminf int_X f_n = infty, text since int f_n = infty, forall n.$$
I can understand "$int f_n = infty, forall n$". But even so, what is the key reason that $$int_X liminf f_n = 0 , text but , liminf int_X f_n = infty$$
To me for the second case, when $nrightarrow infty$, it should be $0$. I am confused on this point.
real-analysis limsup-and-liminf
asked Mar 29 at 16:52
sleeve chensleeve chen
3,18142255
$endgroup$
3
$begingroup$
Fatou's lemma is an inequality.
$endgroup$
– Lord Shark the Unknown
Mar 29 at 16:54
3
$begingroup$
$liminf infty=infty$.
$endgroup$
– Eclipse Sun
Mar 29 at 16:55
$begingroup$
@LordSharktheUnknown Yes, so I think this example shows $<$ not just $leq$.
$endgroup$
– sleeve chen
Mar 29 at 16:56
2
$begingroup$
A less contentious example is $f_n = 1_[n,n+1)$. Then $liminf_n f_n = 0$ and $int f_n = 1$, so $0 = int liminf_n f_n le liminf_n int f_n =1 $.
$endgroup$
– copper.hat
Mar 29 at 17:01
$begingroup$
For all $n$, $int_X f_n = infty$.
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 29 at 17:38
|
show 1 more comment
$begingroup$
In the class, my professor gave the following example:
Let $f_n = mathbf1_[n,infty)$, then we have
$$int_X liminf f_n = 0, text since liminf f_n = 0,$$
and
$$liminf int_X f_n = infty, text since int f_n = infty, forall n.$$
I can understand "$int f_n = infty, forall n$". But even so, what is the key reason that $$int_X liminf f_n = 0 , text but , liminf int_X f_n = infty$$
To me for the second case, when $nrightarrow infty$, it should be $0$. I am confused on this point.
real-analysis limsup-and-liminf
asked Mar 29 at 16:52
sleeve chensleeve chen
3,18142255
$endgroup$
In the class, my professor gave the following example:
Let $f_n = mathbf1_[n,infty)$, then we have
$$int_X liminf f_n = 0, text since liminf f_n = 0,$$
and
$$liminf int_X f_n = infty, text since int f_n = infty, forall n.$$
I can understand "$int f_n = infty, forall n$". But even so, what is the key reason that $$int_X liminf f_n = 0 , text but , liminf int_X f_n = infty$$
To me for the second case, when $nrightarrow infty$, it should be $0$. I am confused on this point.
real-analysis limsup-and-liminf
real-analysis limsup-and-liminf
asked Mar 29 at 16:52
sleeve chensleeve chen
3,18142255
asked Mar 29 at 16:52
sleeve chensleeve chen
3,18142255
asked Mar 29 at 16:52
sleeve chensleeve chen
3,18142255
asked Mar 29 at 16:52
sleeve chensleeve chen
3,18142255
asked Mar 29 at 16:52
sleeve chensleeve chen
3,18142255
3,18142255
3
$begingroup$
Fatou's lemma is an inequality.
$endgroup$
– Lord Shark the Unknown
Mar 29 at 16:54
3
$begingroup$
$liminf infty=infty$.
$endgroup$
– Eclipse Sun
Mar 29 at 16:55
$begingroup$
@LordSharktheUnknown Yes, so I think this example shows $<$ not just $leq$.
$endgroup$
– sleeve chen
Mar 29 at 16:56
2
$begingroup$
A less contentious example is $f_n = 1_[n,n+1)$. Then $liminf_n f_n = 0$ and $int f_n = 1$, so $0 = int liminf_n f_n le liminf_n int f_n =1 $.
$endgroup$
– copper.hat
Mar 29 at 17:01
$begingroup$
For all $n$, $int_X f_n = infty$.
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 29 at 17:38
|
show 1 more comment
3
$begingroup$
Fatou's lemma is an inequality.
$endgroup$
– Lord Shark the Unknown
Mar 29 at 16:54
3
$begingroup$
$liminf infty=infty$.
$endgroup$
– Eclipse Sun
Mar 29 at 16:55
$begingroup$
@LordSharktheUnknown Yes, so I think this example shows $<$ not just $leq$.
$endgroup$
– sleeve chen
Mar 29 at 16:56
2
$begingroup$
A less contentious example is $f_n = 1_[n,n+1)$. Then $liminf_n f_n = 0$ and $int f_n = 1$, so $0 = int liminf_n f_n le liminf_n int f_n =1 $.
$endgroup$
– copper.hat
Mar 29 at 17:01
$begingroup$
For all $n$, $int_X f_n = infty$.
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 29 at 17:38
3
3
$begingroup$
Fatou's lemma is an inequality.
$endgroup$
– Lord Shark the Unknown
Mar 29 at 16:54
$begingroup$
Fatou's lemma is an inequality.
$endgroup$
– Lord Shark the Unknown
Mar 29 at 16:54
3
3
$begingroup$
$liminf infty=infty$.
$endgroup$
– Eclipse Sun
Mar 29 at 16:55
$begingroup$
$liminf infty=infty$.
$endgroup$
– Eclipse Sun
Mar 29 at 16:55
$begingroup$
@LordSharktheUnknown Yes, so I think this example shows $<$ not just $leq$.
$endgroup$
– sleeve chen
Mar 29 at 16:56
$begingroup$
@LordSharktheUnknown Yes, so I think this example shows $<$ not just $leq$.
$endgroup$
– sleeve chen
Mar 29 at 16:56
2
2
$begingroup$
A less contentious example is $f_n = 1_[n,n+1)$. Then $liminf_n f_n = 0$ and $int f_n = 1$, so $0 = int liminf_n f_n le liminf_n int f_n =1 $.
$endgroup$
– copper.hat
Mar 29 at 17:01
$begingroup$
A less contentious example is $f_n = 1_[n,n+1)$. Then $liminf_n f_n = 0$ and $int f_n = 1$, so $0 = int liminf_n f_n le liminf_n int f_n =1 $.
$endgroup$
– copper.hat
Mar 29 at 17:01
$begingroup$
For all $n$, $int_X f_n = infty$.
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 29 at 17:38
$begingroup$
For all $n$, $int_X f_n = infty$.
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 29 at 17:38
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
Well you are considering the sequence of functions $f_n$ where the $n^th$ function of the sequence is given by $$f_n(x):=begincases0 & textif xin (-infty,n) \ 1 & textotherwise .endcases$$
We have the pointwise limit $lim_n to infty f_n(x)=0$, and so $int lim f_n =0$. But for each $n in mathbbN$ we have $int f_n=infty$, and so $lim int f_n=infty$ (limit of a constant sequence).
The difference comes from what limit we are evaluating. On one hand we have the integral of a pointwise limit function, where the pointwise limit happens to be constantly $0$. And on the other hand we have the limit of a constant sequence of extended real numbers.
answered Mar 29 at 20:18
Matt A PeltoMatt A Pelto
2,657621
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Well you are considering the sequence of functions $f_n$ where the $n^th$ function of the sequence is given by $$f_n(x):=begincases0 & textif xin (-infty,n) \ 1 & textotherwise .endcases$$
We have the pointwise limit $lim_n to infty f_n(x)=0$, and so $int lim f_n =0$. But for each $n in mathbbN$ we have $int f_n=infty$, and so $lim int f_n=infty$ (limit of a constant sequence).
The difference comes from what limit we are evaluating. On one hand we have the integral of a pointwise limit function, where the pointwise limit happens to be constantly $0$. And on the other hand we have the limit of a constant sequence of extended real numbers.
answered Mar 29 at 20:18
Matt A PeltoMatt A Pelto
2,657621
$endgroup$
add a comment |
$begingroup$
Well you are considering the sequence of functions $f_n$ where the $n^th$ function of the sequence is given by $$f_n(x):=begincases0 & textif xin (-infty,n) \ 1 & textotherwise .endcases$$
We have the pointwise limit $lim_n to infty f_n(x)=0$, and so $int lim f_n =0$. But for each $n in mathbbN$ we have $int f_n=infty$, and so $lim int f_n=infty$ (limit of a constant sequence).
The difference comes from what limit we are evaluating. On one hand we have the integral of a pointwise limit function, where the pointwise limit happens to be constantly $0$. And on the other hand we have the limit of a constant sequence of extended real numbers.
answered Mar 29 at 20:18
Matt A PeltoMatt A Pelto
2,657621
$endgroup$
add a comment |
$begingroup$
Well you are considering the sequence of functions $f_n$ where the $n^th$ function of the sequence is given by $$f_n(x):=begincases0 & textif xin (-infty,n) \ 1 & textotherwise .endcases$$
We have the pointwise limit $lim_n to infty f_n(x)=0$, and so $int lim f_n =0$. But for each $n in mathbbN$ we have $int f_n=infty$, and so $lim int f_n=infty$ (limit of a constant sequence).
The difference comes from what limit we are evaluating. On one hand we have the integral of a pointwise limit function, where the pointwise limit happens to be constantly $0$. And on the other hand we have the limit of a constant sequence of extended real numbers.
answered Mar 29 at 20:18
Matt A PeltoMatt A Pelto
2,657621
$endgroup$
Well you are considering the sequence of functions $f_n$ where the $n^th$ function of the sequence is given by $$f_n(x):=begincases0 & textif xin (-infty,n) \ 1 & textotherwise .endcases$$
We have the pointwise limit $lim_n to infty f_n(x)=0$, and so $int lim f_n =0$. But for each $n in mathbbN$ we have $int f_n=infty$, and so $lim int f_n=infty$ (limit of a constant sequence).
The difference comes from what limit we are evaluating. On one hand we have the integral of a pointwise limit function, where the pointwise limit happens to be constantly $0$. And on the other hand we have the limit of a constant sequence of extended real numbers.
answered Mar 29 at 20:18
Matt A PeltoMatt A Pelto
2,657621
answered Mar 29 at 20:18
Matt A PeltoMatt A Pelto
2,657621
answered Mar 29 at 20:18
Matt A PeltoMatt A Pelto
2,657621
answered Mar 29 at 20:18
Matt A PeltoMatt A Pelto
2,657621
2,657621
add a comment |
add a comment |
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3
$begingroup$
Fatou's lemma is an inequality.
$endgroup$
– Lord Shark the Unknown
Mar 29 at 16:54
3
$begingroup$
$liminf infty=infty$.
$endgroup$
– Eclipse Sun
Mar 29 at 16:55
$begingroup$
@LordSharktheUnknown Yes, so I think this example shows $<$ not just $leq$.
$endgroup$
– sleeve chen
Mar 29 at 16:56
2
$begingroup$
A less contentious example is $f_n = 1_[n,n+1)$. Then $liminf_n f_n = 0$ and $int f_n = 1$, so $0 = int liminf_n f_n le liminf_n int f_n =1 $.
$endgroup$
– copper.hat
Mar 29 at 17:01
$begingroup$
For all $n$, $int_X f_n = infty$.
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 29 at 17:38