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Does absolute convergence imply integrability?
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For a measurable function $f$ on $[1,+infty)$, which is bounded on bounded sets, define $a_n=intlimits_n^n+1f dm$ for each natural number $n$. Is is true that $ f $ is integrable over $[1,+infty)$ if and only if the series $sumlimits_n=1^inftya_n$ converges absolutely?
I have done like below:
We have $sumlimits_n=1^infty|a_n|=sumlimits_n=1^infty|intlimits_n^n+1f ,dm|leq sumlimits_n=1^infty intlimits_[n,n+1)|f|,dm=sumlimits_n=1^inftyintlimits_[1,+infty)|f|chi_[n,n+1),dm=intlimits_[1,+infty)sumlimits_n=1^infty|f|chi_[n,n+1)=intlimits_[1,+infty)|f|,dm$. If $f$ is integrable, then $sumlimits_n=1^infty|a_n|<+infty$ and so $sumlimits_n=1^inftya_n$ converges absolutely.
But is the converse true? I have no answer. Would anyone please help me?
measure-theory lebesgue-integral lebesgue-measure absolute-continuity
edited Aug 6 at 7:34
Bernard
110k635103
asked Aug 6 at 6:07
Anupam
2,1931822
add a comment |Â
up vote
3
down vote
favorite
For a measurable function $f$ on $[1,+infty)$, which is bounded on bounded sets, define $a_n=intlimits_n^n+1f dm$ for each natural number $n$. Is is true that $ f $ is integrable over $[1,+infty)$ if and only if the series $sumlimits_n=1^inftya_n$ converges absolutely?
I have done like below:
We have $sumlimits_n=1^infty|a_n|=sumlimits_n=1^infty|intlimits_n^n+1f ,dm|leq sumlimits_n=1^infty intlimits_[n,n+1)|f|,dm=sumlimits_n=1^inftyintlimits_[1,+infty)|f|chi_[n,n+1),dm=intlimits_[1,+infty)sumlimits_n=1^infty|f|chi_[n,n+1)=intlimits_[1,+infty)|f|,dm$. If $f$ is integrable, then $sumlimits_n=1^infty|a_n|<+infty$ and so $sumlimits_n=1^inftya_n$ converges absolutely.
But is the converse true? I have no answer. Would anyone please help me?
measure-theory lebesgue-integral lebesgue-measure absolute-continuity
edited Aug 6 at 7:34
Bernard
110k635103
asked Aug 6 at 6:07
Anupam
2,1931822
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
For a measurable function $f$ on $[1,+infty)$, which is bounded on bounded sets, define $a_n=intlimits_n^n+1f dm$ for each natural number $n$. Is is true that $ f $ is integrable over $[1,+infty)$ if and only if the series $sumlimits_n=1^inftya_n$ converges absolutely?
I have done like below:
We have $sumlimits_n=1^infty|a_n|=sumlimits_n=1^infty|intlimits_n^n+1f ,dm|leq sumlimits_n=1^infty intlimits_[n,n+1)|f|,dm=sumlimits_n=1^inftyintlimits_[1,+infty)|f|chi_[n,n+1),dm=intlimits_[1,+infty)sumlimits_n=1^infty|f|chi_[n,n+1)=intlimits_[1,+infty)|f|,dm$. If $f$ is integrable, then $sumlimits_n=1^infty|a_n|<+infty$ and so $sumlimits_n=1^inftya_n$ converges absolutely.
But is the converse true? I have no answer. Would anyone please help me?
measure-theory lebesgue-integral lebesgue-measure absolute-continuity
edited Aug 6 at 7:34
Bernard
110k635103
asked Aug 6 at 6:07
Anupam
2,1931822
For a measurable function $f$ on $[1,+infty)$, which is bounded on bounded sets, define $a_n=intlimits_n^n+1f dm$ for each natural number $n$. Is is true that $ f $ is integrable over $[1,+infty)$ if and only if the series $sumlimits_n=1^inftya_n$ converges absolutely?
I have done like below:
We have $sumlimits_n=1^infty|a_n|=sumlimits_n=1^infty|intlimits_n^n+1f ,dm|leq sumlimits_n=1^infty intlimits_[n,n+1)|f|,dm=sumlimits_n=1^inftyintlimits_[1,+infty)|f|chi_[n,n+1),dm=intlimits_[1,+infty)sumlimits_n=1^infty|f|chi_[n,n+1)=intlimits_[1,+infty)|f|,dm$. If $f$ is integrable, then $sumlimits_n=1^infty|a_n|<+infty$ and so $sumlimits_n=1^inftya_n$ converges absolutely.
But is the converse true? I have no answer. Would anyone please help me?
measure-theory lebesgue-integral lebesgue-measure absolute-continuity
edited Aug 6 at 7:34
Bernard
110k635103
asked Aug 6 at 6:07
Anupam
2,1931822
edited Aug 6 at 7:34
Bernard
110k635103
edited Aug 6 at 7:34
Bernard
110k635103
edited Aug 6 at 7:34
Bernard
110k635103
110k635103
asked Aug 6 at 6:07
Anupam
2,1931822
asked Aug 6 at 6:07
Anupam
2,1931822
asked Aug 6 at 6:07
Anupam
2,1931822
2,1931822
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
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accepted
Consider $f(x)=sin(2pi x)$. Is it integrable over $[1,infty)$? What are the $a_n$?
For the other direction, consider $g(x)=sin(pi x)/x$. It is integrable over $[1,infty)$? What is $sum|a_n|$?
answered Aug 6 at 6:21
Arthur
99.1k793175
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up vote
3
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There exists $f$ such that $a_n=0$ for all $n$ and $int_n ^n+1 |f(t)| , dt =1$ for all $n$. So convergence of $sum |a_n|$ does not impliy that $f $ is integrable. (You can take $f$ to be of the form $c(x-n-frac 1 2)$ on $(n,n+1)$.
answered Aug 6 at 6:20
Kavi Rama Murthy
21k2830
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Consider $f(x)=sin(2pi x)$. Is it integrable over $[1,infty)$? What are the $a_n$?
For the other direction, consider $g(x)=sin(pi x)/x$. It is integrable over $[1,infty)$? What is $sum|a_n|$?
answered Aug 6 at 6:21
Arthur
99.1k793175
add a comment |Â
up vote
3
down vote
accepted
Consider $f(x)=sin(2pi x)$. Is it integrable over $[1,infty)$? What are the $a_n$?
For the other direction, consider $g(x)=sin(pi x)/x$. It is integrable over $[1,infty)$? What is $sum|a_n|$?
answered Aug 6 at 6:21
Arthur
99.1k793175
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Consider $f(x)=sin(2pi x)$. Is it integrable over $[1,infty)$? What are the $a_n$?
For the other direction, consider $g(x)=sin(pi x)/x$. It is integrable over $[1,infty)$? What is $sum|a_n|$?
answered Aug 6 at 6:21
Arthur
99.1k793175
Consider $f(x)=sin(2pi x)$. Is it integrable over $[1,infty)$? What are the $a_n$?
For the other direction, consider $g(x)=sin(pi x)/x$. It is integrable over $[1,infty)$? What is $sum|a_n|$?
answered Aug 6 at 6:21
Arthur
99.1k793175
edited Aug 6 at 8:32
answered Aug 6 at 6:21
Arthur
99.1k793175
answered Aug 6 at 6:21
Arthur
99.1k793175
answered Aug 6 at 6:21
Arthur
99.1k793175
99.1k793175
add a comment |Â
add a comment |Â
up vote
3
down vote
There exists $f$ such that $a_n=0$ for all $n$ and $int_n ^n+1 |f(t)| , dt =1$ for all $n$. So convergence of $sum |a_n|$ does not impliy that $f $ is integrable. (You can take $f$ to be of the form $c(x-n-frac 1 2)$ on $(n,n+1)$.
answered Aug 6 at 6:20
Kavi Rama Murthy
21k2830
add a comment |Â
up vote
3
down vote
There exists $f$ such that $a_n=0$ for all $n$ and $int_n ^n+1 |f(t)| , dt =1$ for all $n$. So convergence of $sum |a_n|$ does not impliy that $f $ is integrable. (You can take $f$ to be of the form $c(x-n-frac 1 2)$ on $(n,n+1)$.
answered Aug 6 at 6:20
Kavi Rama Murthy
21k2830
add a comment |Â
up vote
3
down vote
up vote
3
down vote
There exists $f$ such that $a_n=0$ for all $n$ and $int_n ^n+1 |f(t)| , dt =1$ for all $n$. So convergence of $sum |a_n|$ does not impliy that $f $ is integrable. (You can take $f$ to be of the form $c(x-n-frac 1 2)$ on $(n,n+1)$.
answered Aug 6 at 6:20
Kavi Rama Murthy
21k2830
There exists $f$ such that $a_n=0$ for all $n$ and $int_n ^n+1 |f(t)| , dt =1$ for all $n$. So convergence of $sum |a_n|$ does not impliy that $f $ is integrable. (You can take $f$ to be of the form $c(x-n-frac 1 2)$ on $(n,n+1)$.
answered Aug 6 at 6:20
Kavi Rama Murthy
21k2830
answered Aug 6 at 6:20
Kavi Rama Murthy
21k2830
answered Aug 6 at 6:20
Kavi Rama Murthy
21k2830
answered Aug 6 at 6:20
Kavi Rama Murthy
21k2830
21k2830
add a comment |Â
add a comment |Â
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