Intuition behind Riemann integrabillity and sequences of partitions [closed]
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I started to self-study measure theory and as a review the book Im using(Rana) begin explaining some theorems on Riemann Integrals and im trying to understand to the fullest the intuition and proofs of the next theorem:
Let $f$ be a Riemann integrble function on $[a,b]$, then there exists a sequence of partitions $(pi_n)_n$ such that $pi_nsubset pi_{n+1}$ for all $ngeq 1$.
Im not even sure how to start, besides noticing that since f is Riemann integrable $int_{a}^{b}f(x)dx=sup{L(pi,f)}=inf{U(pi,f)}$ where $pi$ is a partition of $[a,b]$. Then we can choose $(pi_{n}^{1})_{n},(pi_{n}^{2})_{n}$ such that $lim$ $U(pi_{n}^{1},f)$=$lim$ $L(pi_{n}^{2},f)=int_{a}^{b}f(x)dx$ as n tends to $infty$ and such that $lim$ $U(pi_{n},f)=$$lim$ $L(pi_{n},f)=int_{a}^{b}f(x)dx$.
After that Im stuck. Any help would be really appreciated.
real-analysis measure-theory riemann-integration
asked Dec 8 '18 at 3:17
DadadaveDadadave
9118
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closed as unclear what you're asking by Andrés E. Caicedo, Brahadeesh, Paramanand Singh, A. Pongrácz, RRL Dec 8 '18 at 23:48
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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show 2 more comments
$begingroup$
I started to self-study measure theory and as a review the book Im using(Rana) begin explaining some theorems on Riemann Integrals and im trying to understand to the fullest the intuition and proofs of the next theorem:
Let $f$ be a Riemann integrble function on $[a,b]$, then there exists a sequence of partitions $(pi_n)_n$ such that $pi_nsubset pi_{n+1}$ for all $ngeq 1$.
Im not even sure how to start, besides noticing that since f is Riemann integrable $int_{a}^{b}f(x)dx=sup{L(pi,f)}=inf{U(pi,f)}$ where $pi$ is a partition of $[a,b]$. Then we can choose $(pi_{n}^{1})_{n},(pi_{n}^{2})_{n}$ such that $lim$ $U(pi_{n}^{1},f)$=$lim$ $L(pi_{n}^{2},f)=int_{a}^{b}f(x)dx$ as n tends to $infty$ and such that $lim$ $U(pi_{n},f)=$$lim$ $L(pi_{n},f)=int_{a}^{b}f(x)dx$.
After that Im stuck. Any help would be really appreciated.
real-analysis measure-theory riemann-integration
asked Dec 8 '18 at 3:17
DadadaveDadadave
9118
$endgroup$
closed as unclear what you're asking by Andrés E. Caicedo, Brahadeesh, Paramanand Singh, A. Pongrácz, RRL Dec 8 '18 at 23:48
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
1
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Think about what happens to $L(pi, f)$ when you take a finer or a coarser partition and similarly for $U(pi, f)$.
$endgroup$
– twnly
Dec 8 '18 at 4:21
4
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What exactly is the full statement of the theorem? As written, there is no connection between $f$ and the sequence of partitions, and RRL's answer below seems to be the best interpretation of your question. I went through the first chapter of the book but I can't find any related theorem or exercise, so a more precise citation would also help.
$endgroup$
– Brahadeesh
Dec 8 '18 at 11:42
1
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Your question is incomplete. Do you mean to say that there is a sequence $pi_n$ such that $pi_nsubset pi_{n+1}$ and $lim_{ntoinfty} L(pi_n, f) =lim_{ntoinfty} U(pi_n, f) =int_{a} ^{b} f(x) , dx$?
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– Paramanand Singh
Dec 8 '18 at 15:22
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@ParamanandSingh Yes, thats the full question but I thought that there were separated statements, maybe is because of my english but I thought that the first part of the problem was to show that if f is Riemann integrable then I could find a sequence of partitions s.t $ pi_{n} subset pi_{n+1}$.
$endgroup$
– Dadadave
Dec 8 '18 at 16:32
1
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Ok then please edit the question accordingly. I have given a close vote with the reason "unclear what you are asking". Perhaps the close vote can be retracted after your edit.
$endgroup$
– Paramanand Singh
Dec 8 '18 at 16:40
|
show 2 more comments
$begingroup$
I started to self-study measure theory and as a review the book Im using(Rana) begin explaining some theorems on Riemann Integrals and im trying to understand to the fullest the intuition and proofs of the next theorem:
Let $f$ be a Riemann integrble function on $[a,b]$, then there exists a sequence of partitions $(pi_n)_n$ such that $pi_nsubset pi_{n+1}$ for all $ngeq 1$.
Im not even sure how to start, besides noticing that since f is Riemann integrable $int_{a}^{b}f(x)dx=sup{L(pi,f)}=inf{U(pi,f)}$ where $pi$ is a partition of $[a,b]$. Then we can choose $(pi_{n}^{1})_{n},(pi_{n}^{2})_{n}$ such that $lim$ $U(pi_{n}^{1},f)$=$lim$ $L(pi_{n}^{2},f)=int_{a}^{b}f(x)dx$ as n tends to $infty$ and such that $lim$ $U(pi_{n},f)=$$lim$ $L(pi_{n},f)=int_{a}^{b}f(x)dx$.
After that Im stuck. Any help would be really appreciated.
real-analysis measure-theory riemann-integration
asked Dec 8 '18 at 3:17
DadadaveDadadave
9118
$endgroup$
I started to self-study measure theory and as a review the book Im using(Rana) begin explaining some theorems on Riemann Integrals and im trying to understand to the fullest the intuition and proofs of the next theorem:
Let $f$ be a Riemann integrble function on $[a,b]$, then there exists a sequence of partitions $(pi_n)_n$ such that $pi_nsubset pi_{n+1}$ for all $ngeq 1$.
Im not even sure how to start, besides noticing that since f is Riemann integrable $int_{a}^{b}f(x)dx=sup{L(pi,f)}=inf{U(pi,f)}$ where $pi$ is a partition of $[a,b]$. Then we can choose $(pi_{n}^{1})_{n},(pi_{n}^{2})_{n}$ such that $lim$ $U(pi_{n}^{1},f)$=$lim$ $L(pi_{n}^{2},f)=int_{a}^{b}f(x)dx$ as n tends to $infty$ and such that $lim$ $U(pi_{n},f)=$$lim$ $L(pi_{n},f)=int_{a}^{b}f(x)dx$.
After that Im stuck. Any help would be really appreciated.
real-analysis measure-theory riemann-integration
real-analysis measure-theory riemann-integration
asked Dec 8 '18 at 3:17
DadadaveDadadave
9118
asked Dec 8 '18 at 3:17
DadadaveDadadave
9118
edited Dec 8 '18 at 17:01
Dadadave
asked Dec 8 '18 at 3:17
DadadaveDadadave
9118
asked Dec 8 '18 at 3:17
DadadaveDadadave
9118
asked Dec 8 '18 at 3:17
DadadaveDadadave
9118
9118
closed as unclear what you're asking by Andrés E. Caicedo, Brahadeesh, Paramanand Singh, A. Pongrácz, RRL Dec 8 '18 at 23:48
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Andrés E. Caicedo, Brahadeesh, Paramanand Singh, A. Pongrácz, RRL Dec 8 '18 at 23:48
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
Think about what happens to $L(pi, f)$ when you take a finer or a coarser partition and similarly for $U(pi, f)$.
$endgroup$
– twnly
Dec 8 '18 at 4:21
4
$begingroup$
What exactly is the full statement of the theorem? As written, there is no connection between $f$ and the sequence of partitions, and RRL's answer below seems to be the best interpretation of your question. I went through the first chapter of the book but I can't find any related theorem or exercise, so a more precise citation would also help.
$endgroup$
– Brahadeesh
Dec 8 '18 at 11:42
1
$begingroup$
Your question is incomplete. Do you mean to say that there is a sequence $pi_n$ such that $pi_nsubset pi_{n+1}$ and $lim_{ntoinfty} L(pi_n, f) =lim_{ntoinfty} U(pi_n, f) =int_{a} ^{b} f(x) , dx$?
$endgroup$
– Paramanand Singh
Dec 8 '18 at 15:22
$begingroup$
@ParamanandSingh Yes, thats the full question but I thought that there were separated statements, maybe is because of my english but I thought that the first part of the problem was to show that if f is Riemann integrable then I could find a sequence of partitions s.t $ pi_{n} subset pi_{n+1}$.
$endgroup$
– Dadadave
Dec 8 '18 at 16:32
1
$begingroup$
Ok then please edit the question accordingly. I have given a close vote with the reason "unclear what you are asking". Perhaps the close vote can be retracted after your edit.
$endgroup$
– Paramanand Singh
Dec 8 '18 at 16:40
|
show 2 more comments
1
$begingroup$
Think about what happens to $L(pi, f)$ when you take a finer or a coarser partition and similarly for $U(pi, f)$.
$endgroup$
– twnly
Dec 8 '18 at 4:21
4
$begingroup$
What exactly is the full statement of the theorem? As written, there is no connection between $f$ and the sequence of partitions, and RRL's answer below seems to be the best interpretation of your question. I went through the first chapter of the book but I can't find any related theorem or exercise, so a more precise citation would also help.
$endgroup$
– Brahadeesh
Dec 8 '18 at 11:42
1
$begingroup$
Your question is incomplete. Do you mean to say that there is a sequence $pi_n$ such that $pi_nsubset pi_{n+1}$ and $lim_{ntoinfty} L(pi_n, f) =lim_{ntoinfty} U(pi_n, f) =int_{a} ^{b} f(x) , dx$?
$endgroup$
– Paramanand Singh
Dec 8 '18 at 15:22
$begingroup$
@ParamanandSingh Yes, thats the full question but I thought that there were separated statements, maybe is because of my english but I thought that the first part of the problem was to show that if f is Riemann integrable then I could find a sequence of partitions s.t $ pi_{n} subset pi_{n+1}$.
$endgroup$
– Dadadave
Dec 8 '18 at 16:32
1
$begingroup$
Ok then please edit the question accordingly. I have given a close vote with the reason "unclear what you are asking". Perhaps the close vote can be retracted after your edit.
$endgroup$
– Paramanand Singh
Dec 8 '18 at 16:40
1
1
$begingroup$
Think about what happens to $L(pi, f)$ when you take a finer or a coarser partition and similarly for $U(pi, f)$.
$endgroup$
– twnly
Dec 8 '18 at 4:21
$begingroup$
Think about what happens to $L(pi, f)$ when you take a finer or a coarser partition and similarly for $U(pi, f)$.
$endgroup$
– twnly
Dec 8 '18 at 4:21
4
4
$begingroup$
What exactly is the full statement of the theorem? As written, there is no connection between $f$ and the sequence of partitions, and RRL's answer below seems to be the best interpretation of your question. I went through the first chapter of the book but I can't find any related theorem or exercise, so a more precise citation would also help.
$endgroup$
– Brahadeesh
Dec 8 '18 at 11:42
$begingroup$
What exactly is the full statement of the theorem? As written, there is no connection between $f$ and the sequence of partitions, and RRL's answer below seems to be the best interpretation of your question. I went through the first chapter of the book but I can't find any related theorem or exercise, so a more precise citation would also help.
$endgroup$
– Brahadeesh
Dec 8 '18 at 11:42
1
1
$begingroup$
Your question is incomplete. Do you mean to say that there is a sequence $pi_n$ such that $pi_nsubset pi_{n+1}$ and $lim_{ntoinfty} L(pi_n, f) =lim_{ntoinfty} U(pi_n, f) =int_{a} ^{b} f(x) , dx$?
$endgroup$
– Paramanand Singh
Dec 8 '18 at 15:22
$begingroup$
Your question is incomplete. Do you mean to say that there is a sequence $pi_n$ such that $pi_nsubset pi_{n+1}$ and $lim_{ntoinfty} L(pi_n, f) =lim_{ntoinfty} U(pi_n, f) =int_{a} ^{b} f(x) , dx$?
$endgroup$
– Paramanand Singh
Dec 8 '18 at 15:22
$begingroup$
@ParamanandSingh Yes, thats the full question but I thought that there were separated statements, maybe is because of my english but I thought that the first part of the problem was to show that if f is Riemann integrable then I could find a sequence of partitions s.t $ pi_{n} subset pi_{n+1}$.
$endgroup$
– Dadadave
Dec 8 '18 at 16:32
$begingroup$
@ParamanandSingh Yes, thats the full question but I thought that there were separated statements, maybe is because of my english but I thought that the first part of the problem was to show that if f is Riemann integrable then I could find a sequence of partitions s.t $ pi_{n} subset pi_{n+1}$.
$endgroup$
– Dadadave
Dec 8 '18 at 16:32
1
1
$begingroup$
Ok then please edit the question accordingly. I have given a close vote with the reason "unclear what you are asking". Perhaps the close vote can be retracted after your edit.
$endgroup$
– Paramanand Singh
Dec 8 '18 at 16:40
$begingroup$
Ok then please edit the question accordingly. I have given a close vote with the reason "unclear what you are asking". Perhaps the close vote can be retracted after your edit.
$endgroup$
– Paramanand Singh
Dec 8 '18 at 16:40
|
show 2 more comments
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$begingroup$
Think about what happens to $L(pi, f)$ when you take a finer or a coarser partition and similarly for $U(pi, f)$.
$endgroup$
– twnly
Dec 8 '18 at 4:21
4
$begingroup$
What exactly is the full statement of the theorem? As written, there is no connection between $f$ and the sequence of partitions, and RRL's answer below seems to be the best interpretation of your question. I went through the first chapter of the book but I can't find any related theorem or exercise, so a more precise citation would also help.
$endgroup$
– Brahadeesh
Dec 8 '18 at 11:42
1
$begingroup$
Your question is incomplete. Do you mean to say that there is a sequence $pi_n$ such that $pi_nsubset pi_{n+1}$ and $lim_{ntoinfty} L(pi_n, f) =lim_{ntoinfty} U(pi_n, f) =int_{a} ^{b} f(x) , dx$?
$endgroup$
– Paramanand Singh
Dec 8 '18 at 15:22
$begingroup$
@ParamanandSingh Yes, thats the full question but I thought that there were separated statements, maybe is because of my english but I thought that the first part of the problem was to show that if f is Riemann integrable then I could find a sequence of partitions s.t $ pi_{n} subset pi_{n+1}$.
$endgroup$
– Dadadave
Dec 8 '18 at 16:32
1
$begingroup$
Ok then please edit the question accordingly. I have given a close vote with the reason "unclear what you are asking". Perhaps the close vote can be retracted after your edit.
$endgroup$
– Paramanand Singh
Dec 8 '18 at 16:40