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Proving Riemann integrability using sequences of Riemann sums
I am trying to prove the following:
Suppose $ f:[a,b]rightarrowmathbb{R} $ is bounded. Then $ f $ is Riemann integrable if and only if for each sequence of marked partitions ${P_n}$ with ${mu(P_n)}rightarrow0$, the sequence ${S(P_n,f)}$ is convergent
,where $mu(P)$ is the mesh of partition $P$ and $S(P,f)$ is the Riemann sum of $f$ over partition $P$.
My attempt at a solution:
Suppose for each sequence of marked partitions ${P}$ with ${mu(P_n)}rightarrow0,$ $ {S(P_n,f)}$ converges.
Let $epsilon>0$ be given. Then there is an $Ainmathbb{R}$ and $Ninmathbb{N}$ such that when $n>N$, there exists $delta$ such that $mu(P_n)<deltaimplies |S(P_n,f)-A|<epsilon$
Then, by the theorem provided by leo below, the existence of $A$ implies that $f$ is Riemann integrable.
Now suppose $f$ is integrable. Then given $epsilon>0$, there exists $Ainmathbb{R}$ such that there exists $delta$ for which $mu(P)<deltaimplies |S(P,f)-A|<epsilon, forall P$.
Then for each sequence each sequence of marked partitions ${P}$ with ${mu(P_n)}rightarrow0,$ $mu(P_n)<delta$.
Then, $|S(P_n,f)-A|<epsilon$ which means that ${S(P_n,f)}$ converges to A. Also by the theorem below, $A=int f dt$
real-analysis
asked Feb 1 '12 at 2:03
Shafat Arbaz AlamShafat Arbaz Alam
122119
|
show 5 more comments
I am trying to prove the following:
Suppose $ f:[a,b]rightarrowmathbb{R} $ is bounded. Then $ f $ is Riemann integrable if and only if for each sequence of marked partitions ${P_n}$ with ${mu(P_n)}rightarrow0$, the sequence ${S(P_n,f)}$ is convergent
,where $mu(P)$ is the mesh of partition $P$ and $S(P,f)$ is the Riemann sum of $f$ over partition $P$.
My attempt at a solution:
Suppose for each sequence of marked partitions ${P}$ with ${mu(P_n)}rightarrow0,$ $ {S(P_n,f)}$ converges.
Let $epsilon>0$ be given. Then there is an $Ainmathbb{R}$ and $Ninmathbb{N}$ such that when $n>N$, there exists $delta$ such that $mu(P_n)<deltaimplies |S(P_n,f)-A|<epsilon$
Then, by the theorem provided by leo below, the existence of $A$ implies that $f$ is Riemann integrable.
Now suppose $f$ is integrable. Then given $epsilon>0$, there exists $Ainmathbb{R}$ such that there exists $delta$ for which $mu(P)<deltaimplies |S(P,f)-A|<epsilon, forall P$.
Then for each sequence each sequence of marked partitions ${P}$ with ${mu(P_n)}rightarrow0,$ $mu(P_n)<delta$.
Then, $|S(P_n,f)-A|<epsilon$ which means that ${S(P_n,f)}$ converges to A. Also by the theorem below, $A=int f dt$
real-analysis
asked Feb 1 '12 at 2:03
Shafat Arbaz AlamShafat Arbaz Alam
122119
1
What is $mu(P_n)$?
– leo
Feb 1 '12 at 2:13
This is a notation I have seen for the mesh of a partition; this is the length of the longest interval in the partition.
– ncmathsadist
Feb 1 '12 at 2:16
Sorry, clarified
– Shafat Arbaz Alam
Feb 1 '12 at 2:21
Use the definition that says: $f$ is integrable over $[a,b]$ if there exist a number $I$ such that for every $epsilongt 0$ exist $deltagt 0$ s.t. if $P$ is partition of $[a,b]$ with $mu(P)ltdelta$, then $$|S(P_n,f)-I|ltepsilon.$$ Now I'm tired. I'll post an answer tomorrow if nobody does.
– leo
Feb 1 '12 at 3:10
1
Converges is enough: if two sequences converge to some different limits, interlace them to get a divergent sequence, which contradicts the hypothesis.
– Did
Feb 25 '12 at 12:02
|
show 5 more comments
I am trying to prove the following:
Suppose $ f:[a,b]rightarrowmathbb{R} $ is bounded. Then $ f $ is Riemann integrable if and only if for each sequence of marked partitions ${P_n}$ with ${mu(P_n)}rightarrow0$, the sequence ${S(P_n,f)}$ is convergent
,where $mu(P)$ is the mesh of partition $P$ and $S(P,f)$ is the Riemann sum of $f$ over partition $P$.
My attempt at a solution:
Suppose for each sequence of marked partitions ${P}$ with ${mu(P_n)}rightarrow0,$ $ {S(P_n,f)}$ converges.
Let $epsilon>0$ be given. Then there is an $Ainmathbb{R}$ and $Ninmathbb{N}$ such that when $n>N$, there exists $delta$ such that $mu(P_n)<deltaimplies |S(P_n,f)-A|<epsilon$
Then, by the theorem provided by leo below, the existence of $A$ implies that $f$ is Riemann integrable.
Now suppose $f$ is integrable. Then given $epsilon>0$, there exists $Ainmathbb{R}$ such that there exists $delta$ for which $mu(P)<deltaimplies |S(P,f)-A|<epsilon, forall P$.
Then for each sequence each sequence of marked partitions ${P}$ with ${mu(P_n)}rightarrow0,$ $mu(P_n)<delta$.
Then, $|S(P_n,f)-A|<epsilon$ which means that ${S(P_n,f)}$ converges to A. Also by the theorem below, $A=int f dt$
real-analysis
asked Feb 1 '12 at 2:03
Shafat Arbaz AlamShafat Arbaz Alam
122119
I am trying to prove the following:
Suppose $ f:[a,b]rightarrowmathbb{R} $ is bounded. Then $ f $ is Riemann integrable if and only if for each sequence of marked partitions ${P_n}$ with ${mu(P_n)}rightarrow0$, the sequence ${S(P_n,f)}$ is convergent
,where $mu(P)$ is the mesh of partition $P$ and $S(P,f)$ is the Riemann sum of $f$ over partition $P$.
My attempt at a solution:
Suppose for each sequence of marked partitions ${P}$ with ${mu(P_n)}rightarrow0,$ $ {S(P_n,f)}$ converges.
Let $epsilon>0$ be given. Then there is an $Ainmathbb{R}$ and $Ninmathbb{N}$ such that when $n>N$, there exists $delta$ such that $mu(P_n)<deltaimplies |S(P_n,f)-A|<epsilon$
Then, by the theorem provided by leo below, the existence of $A$ implies that $f$ is Riemann integrable.
Now suppose $f$ is integrable. Then given $epsilon>0$, there exists $Ainmathbb{R}$ such that there exists $delta$ for which $mu(P)<deltaimplies |S(P,f)-A|<epsilon, forall P$.
Then for each sequence each sequence of marked partitions ${P}$ with ${mu(P_n)}rightarrow0,$ $mu(P_n)<delta$.
Then, $|S(P_n,f)-A|<epsilon$ which means that ${S(P_n,f)}$ converges to A. Also by the theorem below, $A=int f dt$
real-analysis
real-analysis
asked Feb 1 '12 at 2:03
Shafat Arbaz AlamShafat Arbaz Alam
122119
asked Feb 1 '12 at 2:03
Shafat Arbaz AlamShafat Arbaz Alam
122119
edited Feb 1 '12 at 23:16
Shafat Arbaz Alam
asked Feb 1 '12 at 2:03
Shafat Arbaz AlamShafat Arbaz Alam
122119
asked Feb 1 '12 at 2:03
Shafat Arbaz AlamShafat Arbaz Alam
122119
asked Feb 1 '12 at 2:03
Shafat Arbaz AlamShafat Arbaz Alam
122119
122119
1
What is $mu(P_n)$?
– leo
Feb 1 '12 at 2:13
This is a notation I have seen for the mesh of a partition; this is the length of the longest interval in the partition.
– ncmathsadist
Feb 1 '12 at 2:16
Sorry, clarified
– Shafat Arbaz Alam
Feb 1 '12 at 2:21
Use the definition that says: $f$ is integrable over $[a,b]$ if there exist a number $I$ such that for every $epsilongt 0$ exist $deltagt 0$ s.t. if $P$ is partition of $[a,b]$ with $mu(P)ltdelta$, then $$|S(P_n,f)-I|ltepsilon.$$ Now I'm tired. I'll post an answer tomorrow if nobody does.
– leo
Feb 1 '12 at 3:10
1
Converges is enough: if two sequences converge to some different limits, interlace them to get a divergent sequence, which contradicts the hypothesis.
– Did
Feb 25 '12 at 12:02
|
show 5 more comments
1
What is $mu(P_n)$?
– leo
Feb 1 '12 at 2:13
This is a notation I have seen for the mesh of a partition; this is the length of the longest interval in the partition.
– ncmathsadist
Feb 1 '12 at 2:16
Sorry, clarified
– Shafat Arbaz Alam
Feb 1 '12 at 2:21
Use the definition that says: $f$ is integrable over $[a,b]$ if there exist a number $I$ such that for every $epsilongt 0$ exist $deltagt 0$ s.t. if $P$ is partition of $[a,b]$ with $mu(P)ltdelta$, then $$|S(P_n,f)-I|ltepsilon.$$ Now I'm tired. I'll post an answer tomorrow if nobody does.
– leo
Feb 1 '12 at 3:10
1
Converges is enough: if two sequences converge to some different limits, interlace them to get a divergent sequence, which contradicts the hypothesis.
– Did
Feb 25 '12 at 12:02
1
1
What is $mu(P_n)$?
– leo
Feb 1 '12 at 2:13
What is $mu(P_n)$?
– leo
Feb 1 '12 at 2:13
This is a notation I have seen for the mesh of a partition; this is the length of the longest interval in the partition.
– ncmathsadist
Feb 1 '12 at 2:16
This is a notation I have seen for the mesh of a partition; this is the length of the longest interval in the partition.
– ncmathsadist
Feb 1 '12 at 2:16
Sorry, clarified
– Shafat Arbaz Alam
Feb 1 '12 at 2:21
Sorry, clarified
– Shafat Arbaz Alam
Feb 1 '12 at 2:21
Use the definition that says: $f$ is integrable over $[a,b]$ if there exist a number $I$ such that for every $epsilongt 0$ exist $deltagt 0$ s.t. if $P$ is partition of $[a,b]$ with $mu(P)ltdelta$, then $$|S(P_n,f)-I|ltepsilon.$$ Now I'm tired. I'll post an answer tomorrow if nobody does.
– leo
Feb 1 '12 at 3:10
Use the definition that says: $f$ is integrable over $[a,b]$ if there exist a number $I$ such that for every $epsilongt 0$ exist $deltagt 0$ s.t. if $P$ is partition of $[a,b]$ with $mu(P)ltdelta$, then $$|S(P_n,f)-I|ltepsilon.$$ Now I'm tired. I'll post an answer tomorrow if nobody does.
– leo
Feb 1 '12 at 3:10
1
1
Converges is enough: if two sequences converge to some different limits, interlace them to get a divergent sequence, which contradicts the hypothesis.
– Did
Feb 25 '12 at 12:02
Converges is enough: if two sequences converge to some different limits, interlace them to get a divergent sequence, which contradicts the hypothesis.
– Did
Feb 25 '12 at 12:02
|
show 5 more comments
1 Answer
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oldest
votes
Your statement should be an easy corollary of the Du bois-Reymond and Darboux integration theorem. The proof of the theorem is rather cumbersome. So here is a reference: the proof can be found in Analysis by Its History by Hairer and Wanner.
And by the way, you need to define what convergence as ${mu(P_n)}to 0$ and what marked partition mean.
answered May 28 '16 at 6:49
ArgyllArgyll
333213
add a comment |
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1 Answer
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Your statement should be an easy corollary of the Du bois-Reymond and Darboux integration theorem. The proof of the theorem is rather cumbersome. So here is a reference: the proof can be found in Analysis by Its History by Hairer and Wanner.
And by the way, you need to define what convergence as ${mu(P_n)}to 0$ and what marked partition mean.
answered May 28 '16 at 6:49
ArgyllArgyll
333213
add a comment |
Your statement should be an easy corollary of the Du bois-Reymond and Darboux integration theorem. The proof of the theorem is rather cumbersome. So here is a reference: the proof can be found in Analysis by Its History by Hairer and Wanner.
And by the way, you need to define what convergence as ${mu(P_n)}to 0$ and what marked partition mean.
answered May 28 '16 at 6:49
ArgyllArgyll
333213
add a comment |
Your statement should be an easy corollary of the Du bois-Reymond and Darboux integration theorem. The proof of the theorem is rather cumbersome. So here is a reference: the proof can be found in Analysis by Its History by Hairer and Wanner.
And by the way, you need to define what convergence as ${mu(P_n)}to 0$ and what marked partition mean.
answered May 28 '16 at 6:49
ArgyllArgyll
333213
Your statement should be an easy corollary of the Du bois-Reymond and Darboux integration theorem. The proof of the theorem is rather cumbersome. So here is a reference: the proof can be found in Analysis by Its History by Hairer and Wanner.
And by the way, you need to define what convergence as ${mu(P_n)}to 0$ and what marked partition mean.
answered May 28 '16 at 6:49
ArgyllArgyll
333213
edited May 28 '16 at 7:11
answered May 28 '16 at 6:49
ArgyllArgyll
333213
answered May 28 '16 at 6:49
ArgyllArgyll
333213
answered May 28 '16 at 6:49
ArgyllArgyll
333213
333213
add a comment |
add a comment |
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1
What is $mu(P_n)$?
– leo
Feb 1 '12 at 2:13
This is a notation I have seen for the mesh of a partition; this is the length of the longest interval in the partition.
– ncmathsadist
Feb 1 '12 at 2:16
Sorry, clarified
– Shafat Arbaz Alam
Feb 1 '12 at 2:21
Use the definition that says: $f$ is integrable over $[a,b]$ if there exist a number $I$ such that for every $epsilongt 0$ exist $deltagt 0$ s.t. if $P$ is partition of $[a,b]$ with $mu(P)ltdelta$, then $$|S(P_n,f)-I|ltepsilon.$$ Now I'm tired. I'll post an answer tomorrow if nobody does.
– leo
Feb 1 '12 at 3:10
1
Converges is enough: if two sequences converge to some different limits, interlace them to get a divergent sequence, which contradicts the hypothesis.
– Did
Feb 25 '12 at 12:02