Why is the upper riemann integral the infimum of all upper sums? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Proof that Darboux upper/lower sums always converge to the Darboux upper/lower integral as the partition gets thinnerProving that $ f: [a,b] to BbbR $ is Riemann-integrable using an $ epsilon $-$ delta $ definition.How to prove that a function is Riemann integrable if and only if it is Darboux integrable?If the Darboux sums converge to the integral, does the parameter of the partition tend to zero?Some basic properties of Riemann integrable functionsLet $finmathfrak R[0,1],tau_n=0,frac1n,ldots,frac nn$, & $lim_ntoinftyU[f;tau_n]=lim_ntoinftyL[f;tau_n]=A$. Prove $int_0^1f=A$What does Infimum of Upper Sum and Supremum of Lower Sums mean?How to evaluate this Riemann integral using this definition?Method for evaluating Darboux integrals by a sequence of partitions?Prove that for any $epsilon > 0, exists delta > 0,$ if $||P|| < delta $, then $|L(f,P) - I|<epsilon $ , and $|U(f,P) - I|<epsilon $Converging of Riemann Sums with different partitionsnondecreasing f prove integrable with upper and lower sumsIs there always a partition that gives the Riemann Integral?A function with more than one number between the lower and upper riemann sums?Prove the Riemann integral $int_a^b x^2 dx = fracb^3-a^33$ by using the mean value theoremEvaluate Riemann integral $int_a^b e^x dx$ using upper and lower integral definitions and theoremsProve that the lower sum is the infimum of all Riemann sums over a closed, bounded partitionHow to show that the upper and lower Riemann integrals of a function, say $f(x) = -2x$, are equal?Textbook Definition of Riemann Integral: Supremum, Infimum, Lower/Upper BoundsShow that $f(x)=0,;0leq x<1/2,; f(x)=1,;1/2leq xleq 1$ is Riemann integrable over $[0,1]$ and find its value.
Can anything be seen from the center of the Boötes void? How dark would it be?
Is CEO the "profession" with the most psychopaths?
Find 108 by using 3,4,6
Did Mueller's report provide an evidentiary basis for the claim of Russian govt election interference via social media?
A term for a woman complaining about things/begging in a cute/childish way
How to get all distinct words within a set of lines?
MLE of the unknown radius
Maximum summed subsequences with non-adjacent items
Do I really need to have a message in a novel to appeal to readers?
Most bit efficient text communication method?
Using audio cues to encourage good posture
Put R under double integral
Project Euler #1 in C++
How fail-safe is nr as stop bytes?
Is grep documentation about ignoring case wrong, since it doesn't ignore case in filenames?
What order were files/directories outputted in dir?
Why does it sometimes sound good to play a grace note as a lead in to a note in a melody?
Sum letters are not two different
Importance of からだ in this sentence
Did Deadpool rescue all of the X-Force?
Is it fair for a professor to grade us on the possession of past papers?
Why do early math courses focus on the cross sections of a cone and not on other 3D objects?
What is the appropriate index architecture when forced to implement IsDeleted (soft deletes)?
Strange behavior of Object.defineProperty() in JavaScript
Why is the upper riemann integral the infimum of all upper sums?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Proof that Darboux upper/lower sums always converge to the Darboux upper/lower integral as the partition gets thinnerProving that $ f: [a,b] to BbbR $ is Riemann-integrable using an $ epsilon $-$ delta $ definition.How to prove that a function is Riemann integrable if and only if it is Darboux integrable?If the Darboux sums converge to the integral, does the parameter of the partition tend to zero?Some basic properties of Riemann integrable functionsLet $finmathfrak R[0,1],tau_n=0,frac1n,ldots,frac nn$, & $lim_ntoinftyU[f;tau_n]=lim_ntoinftyL[f;tau_n]=A$. Prove $int_0^1f=A$What does Infimum of Upper Sum and Supremum of Lower Sums mean?How to evaluate this Riemann integral using this definition?Method for evaluating Darboux integrals by a sequence of partitions?Prove that for any $epsilon > 0, exists delta > 0,$ if $||P|| < delta $, then $|L(f,P) - I|<epsilon $ , and $|U(f,P) - I|<epsilon $Converging of Riemann Sums with different partitionsnondecreasing f prove integrable with upper and lower sumsIs there always a partition that gives the Riemann Integral?A function with more than one number between the lower and upper riemann sums?Prove the Riemann integral $int_a^b x^2 dx = fracb^3-a^33$ by using the mean value theoremEvaluate Riemann integral $int_a^b e^x dx$ using upper and lower integral definitions and theoremsProve that the lower sum is the infimum of all Riemann sums over a closed, bounded partitionHow to show that the upper and lower Riemann integrals of a function, say $f(x) = -2x$, are equal?Textbook Definition of Riemann Integral: Supremum, Infimum, Lower/Upper BoundsShow that $f(x)=0,;0leq x<1/2,; f(x)=1,;1/2leq xleq 1$ is Riemann integrable over $[0,1]$ and find its value.
$begingroup$
I was reading the theory of Riemann integration when I cam across the following ,
If $f$ is bounded on $[a,b]$, and $P = x_0,x_1,x_2.......x_n$ is a partition of $[a,b]$, let $$M_j = sup_x_j-1leq xleq x_jf(x)$$ The upper sum of f over P is $$S(P) = sum_j=1^n M_j(x_j-x_j-1)$$ and the upper integral of $f$ over $[a,b]$, denoted by $$int_a^b^- f(x)dx$$ is the infimum of all upper sums.
The theorem similarly goes on to state the result for lower sums.
My doubt is : I do not understand how is $$int_a^b^- f(x)dx$$ the infimum of all upper sums. I understand that if we refine the partiton P , then the upper sum would decrease, so it may be a lower limit for all the upper sums computed on the refinements of P ( but still being the lower limit does not prove that it is the inifmum ) and what about those partitions for which P itslef is the refinement of ?
How do I know that it will be a lower limit for those , let alone a infimum ?
real-analysis riemann-integration
$endgroup$
add a comment |
$begingroup$
I was reading the theory of Riemann integration when I cam across the following ,
If $f$ is bounded on $[a,b]$, and $P = x_0,x_1,x_2.......x_n$ is a partition of $[a,b]$, let $$M_j = sup_x_j-1leq xleq x_jf(x)$$ The upper sum of f over P is $$S(P) = sum_j=1^n M_j(x_j-x_j-1)$$ and the upper integral of $f$ over $[a,b]$, denoted by $$int_a^b^- f(x)dx$$ is the infimum of all upper sums.
The theorem similarly goes on to state the result for lower sums.
My doubt is : I do not understand how is $$int_a^b^- f(x)dx$$ the infimum of all upper sums. I understand that if we refine the partiton P , then the upper sum would decrease, so it may be a lower limit for all the upper sums computed on the refinements of P ( but still being the lower limit does not prove that it is the inifmum ) and what about those partitions for which P itslef is the refinement of ?
How do I know that it will be a lower limit for those , let alone a infimum ?
real-analysis riemann-integration
$endgroup$
$begingroup$
Do you have a definition of the upper integral other than "the infimum of all upper sums"? From what you've written, it sounds like it's just the definition.
$endgroup$
– Eric Wofsey
Dec 6 '16 at 0:41
$begingroup$
It's not a theorem; it's a definition.
$endgroup$
– zhw.
Dec 6 '16 at 0:42
$begingroup$
@zhw , But is it right to claim that we could just define that that particular limit which is what the integral basically is equals the infimum ?
$endgroup$
– Noob101
Dec 6 '16 at 0:52
$begingroup$
@EricWofsey , I wanted to know that is the upper integral defined in this way a riemann integral too , that is if the partition goes to less than delta , then will my upper sum converge to it ?
$endgroup$
– Noob101
Dec 6 '16 at 0:54
$begingroup$
Nice but slightly ambiguous question. +1
$endgroup$
– Paramanand Singh
Dec 7 '16 at 13:17
add a comment |
$begingroup$
I was reading the theory of Riemann integration when I cam across the following ,
If $f$ is bounded on $[a,b]$, and $P = x_0,x_1,x_2.......x_n$ is a partition of $[a,b]$, let $$M_j = sup_x_j-1leq xleq x_jf(x)$$ The upper sum of f over P is $$S(P) = sum_j=1^n M_j(x_j-x_j-1)$$ and the upper integral of $f$ over $[a,b]$, denoted by $$int_a^b^- f(x)dx$$ is the infimum of all upper sums.
The theorem similarly goes on to state the result for lower sums.
My doubt is : I do not understand how is $$int_a^b^- f(x)dx$$ the infimum of all upper sums. I understand that if we refine the partiton P , then the upper sum would decrease, so it may be a lower limit for all the upper sums computed on the refinements of P ( but still being the lower limit does not prove that it is the inifmum ) and what about those partitions for which P itslef is the refinement of ?
How do I know that it will be a lower limit for those , let alone a infimum ?
real-analysis riemann-integration
$endgroup$
I was reading the theory of Riemann integration when I cam across the following ,
If $f$ is bounded on $[a,b]$, and $P = x_0,x_1,x_2.......x_n$ is a partition of $[a,b]$, let $$M_j = sup_x_j-1leq xleq x_jf(x)$$ The upper sum of f over P is $$S(P) = sum_j=1^n M_j(x_j-x_j-1)$$ and the upper integral of $f$ over $[a,b]$, denoted by $$int_a^b^- f(x)dx$$ is the infimum of all upper sums.
The theorem similarly goes on to state the result for lower sums.
My doubt is : I do not understand how is $$int_a^b^- f(x)dx$$ the infimum of all upper sums. I understand that if we refine the partiton P , then the upper sum would decrease, so it may be a lower limit for all the upper sums computed on the refinements of P ( but still being the lower limit does not prove that it is the inifmum ) and what about those partitions for which P itslef is the refinement of ?
How do I know that it will be a lower limit for those , let alone a infimum ?
real-analysis riemann-integration
real-analysis riemann-integration
edited Dec 7 '16 at 12:32
Paramanand Singh
51.5k560170
51.5k560170
asked Dec 6 '16 at 0:39
Noob101Noob101
431218
431218
$begingroup$
Do you have a definition of the upper integral other than "the infimum of all upper sums"? From what you've written, it sounds like it's just the definition.
$endgroup$
– Eric Wofsey
Dec 6 '16 at 0:41
$begingroup$
It's not a theorem; it's a definition.
$endgroup$
– zhw.
Dec 6 '16 at 0:42
$begingroup$
@zhw , But is it right to claim that we could just define that that particular limit which is what the integral basically is equals the infimum ?
$endgroup$
– Noob101
Dec 6 '16 at 0:52
$begingroup$
@EricWofsey , I wanted to know that is the upper integral defined in this way a riemann integral too , that is if the partition goes to less than delta , then will my upper sum converge to it ?
$endgroup$
– Noob101
Dec 6 '16 at 0:54
$begingroup$
Nice but slightly ambiguous question. +1
$endgroup$
– Paramanand Singh
Dec 7 '16 at 13:17
add a comment |
$begingroup$
Do you have a definition of the upper integral other than "the infimum of all upper sums"? From what you've written, it sounds like it's just the definition.
$endgroup$
– Eric Wofsey
Dec 6 '16 at 0:41
$begingroup$
It's not a theorem; it's a definition.
$endgroup$
– zhw.
Dec 6 '16 at 0:42
$begingroup$
@zhw , But is it right to claim that we could just define that that particular limit which is what the integral basically is equals the infimum ?
$endgroup$
– Noob101
Dec 6 '16 at 0:52
$begingroup$
@EricWofsey , I wanted to know that is the upper integral defined in this way a riemann integral too , that is if the partition goes to less than delta , then will my upper sum converge to it ?
$endgroup$
– Noob101
Dec 6 '16 at 0:54
$begingroup$
Nice but slightly ambiguous question. +1
$endgroup$
– Paramanand Singh
Dec 7 '16 at 13:17
$begingroup$
Do you have a definition of the upper integral other than "the infimum of all upper sums"? From what you've written, it sounds like it's just the definition.
$endgroup$
– Eric Wofsey
Dec 6 '16 at 0:41
$begingroup$
Do you have a definition of the upper integral other than "the infimum of all upper sums"? From what you've written, it sounds like it's just the definition.
$endgroup$
– Eric Wofsey
Dec 6 '16 at 0:41
$begingroup$
It's not a theorem; it's a definition.
$endgroup$
– zhw.
Dec 6 '16 at 0:42
$begingroup$
It's not a theorem; it's a definition.
$endgroup$
– zhw.
Dec 6 '16 at 0:42
$begingroup$
@zhw , But is it right to claim that we could just define that that particular limit which is what the integral basically is equals the infimum ?
$endgroup$
– Noob101
Dec 6 '16 at 0:52
$begingroup$
@zhw , But is it right to claim that we could just define that that particular limit which is what the integral basically is equals the infimum ?
$endgroup$
– Noob101
Dec 6 '16 at 0:52
$begingroup$
@EricWofsey , I wanted to know that is the upper integral defined in this way a riemann integral too , that is if the partition goes to less than delta , then will my upper sum converge to it ?
$endgroup$
– Noob101
Dec 6 '16 at 0:54
$begingroup$
@EricWofsey , I wanted to know that is the upper integral defined in this way a riemann integral too , that is if the partition goes to less than delta , then will my upper sum converge to it ?
$endgroup$
– Noob101
Dec 6 '16 at 0:54
$begingroup$
Nice but slightly ambiguous question. +1
$endgroup$
– Paramanand Singh
Dec 7 '16 at 13:17
$begingroup$
Nice but slightly ambiguous question. +1
$endgroup$
– Paramanand Singh
Dec 7 '16 at 13:17
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Your question does have some ambiguity. From the wording of your question and comments it appears that you want to know:
Does the limit of upper sums (with respect to partitions getting finer and finer) equal the infimum of all upper sums?
First of all note that when we are dealing with limits of things dependent on a partition of an interval then there are two ways in which the limit operation can be defined:
1) Limit via refinement of a partition: Let $P = x_0, x_1, x_2,ldots, x_n $ be a partition of $[a, b]$ where $$a =x_0 < x_1 < x_2 < cdots < x_n = b$$ A partition $P'$ of $[a, b]$ is said to be a refinement of $P$ (or finer than $P$) if $P subseteq P'$.
Let $mathcalP[a, b]$ denote the collection of all partitions of $[a, b]$ and let $F:mathcalP[a, b] to mathbbR$ be a function. A number $L$ is said to be the limit of $F$ (via refinement) if for every $epsilon > 0$ there is a partition $P_epsilonin mathcalP[a, b]$ such that $|F(P) - L| < epsilon$ for all $P in mathcalP[a, b]$ with $P_epsilon subseteq P$.
2) Limit as norm of parititon tends to $0$: If $P = a = x_0, x_1, x_2, ldots, x_n = b$ is a partition of $[a, b]$ then the norm $||P||$ of partition $P$ is defined as $||P|| = max_i = 1^n(x_i - x_i - 1)$.
Let $mathcalP[a, b]$ denote the collection of all partitions of $[a, b]$ and let $F: mathcalP[a, b] to mathbbR$ be a function. A number $L$ is said to be limit of $F$ as norm of partition tends to $0$ if for every $epsilon > 0$ there is a $delta > 0$ such that $|F(P) - L| < epsilon$ for all $Pin mathcalP[a, b]$ with $||P|| < delta$. This is written as $lim_ to 0F(P) = L$.
Note that for a given function $F:mathcalP[a, b] to mathbbR$ the limiting behavior of $F$ can be different according to these two definitions given above. In fact if $F(P) to L$ as $||P||to 0$ then $F(P) to L$ via refinement but the converse may not hold in general. This is because of the fact that refinement of a partition leads to a decrease in the norm, but decreasing the norm of a partition does not necessarily lead to a refinement.
Now let $f$ be a function defined and bounded on $[a, b]$ and let $P = x_0, x_1, x_2, ldots x_n$ be a partition of $[a, b]$. Let $M_k = sup,f(x), x in [x_k - 1, x_k]$ and let $mathcalP[a, b]$ denote the collection of all partitions of $[a, b]$. We define the upper sum function $S:mathcalP[a, b] to mathbbR$ by $$S(P) = sum_k = 1^nM_k(x_k - x_k - 1)$$ It is easy to prove that if $m = inf,f(x), x in [a, b]$ then $S(P) geq m(b - a)$ for all $P in mathcalP[a, b]$ and further if $P, P' in mathcalP[a, b]$ are such that $P subseteq P'$ then $S(P') leq S(P)$. It follows that $J = inf,S(P), P in mathcalP[a, b]$ exists.
Your question can now be worded more concretely into one of the following two forms:
Does $S(P) to J$ via refinement?
or
Does $lim_ to 0S(P) = J$?
The answer to the first question is obviously "yes" and you should be able to prove this using the definition of limit via refinement given above.
The answer to second question is also "yes" but it is difficult to prove. We first prove the result for a non-negative function $f$. Let $epsilon > 0$ be given. Since $J = inf,S(P), P in mathcalP[a, b]$, there is a partition $P_epsilon in mathcalP[a, b]$ such that $$J leq S(P_epsilon) < J + fracepsilon2tag1$$ Let $P_epsilon = x_0', x_1', x_2', ldots, x_N'$ and let $M = sup,f(x), x in [a, b] + 1$. Let $delta = epsilon / (2MN)$ and consider a partition $P = x_0, x_1, x_2, ldots, x_n$ with $||P|| < delta$.
We can write $$S(P) = sum_k = 1^nM_k(x_k - x_k - 1) = S_1 + S_2tag2$$ where $S_1$ is the sum corresponding to the index $k$ for which $[x_k - 1, x_k]$ does not contain any point of $P_epsilon$ and $S_2$ is the sum corresponding to other values of index $k$. Clearly for $S_1$ the interval $[x_k - 1, x_k]$ lies wholly in one of the intervals $[x_j - 1', x_j']$ made by $P_epsilon$ and hence $S_1 leq S(P_epsilon)$ (note that $f$ is non-negative). For $S_2$ we can see that the number of such indexes $k$ is no more than $N$ and hence $S_2 < MNdelta = epsilon / 2$ (note that $f$ is non-negative here). It follows that $$J leq S(P) = S_1 + S_2 < S(P_epsilon) + fracepsilon2 < J + epsilontag3$$ for all $P in mathcalP[a, b]$ with $||P|| < delta$. It follows that $S(P) to J$ as $||P|| to 0$.
Extension to a general function $f$ can be achieved by writing $f(x) = g(x) + m$ where $m = inf,f(x), x in [a, b]$ and noting that $g$ is non-negative.
Note: The limit of a Riemann sum is based on the two definitions given above but there is a slight complication. A Riemann sum depends not only on a partition but also on choice of tags corresponding to a partition. Formally one can view a Riemann sum not as a function from $mathcalP [a, b] $ to $mathbbR $ but rather as a relation from $mathcalP [a, b] $ to $mathbb R $ such that it relates every partition of $[a, b] $ to one or more real numbers.
$endgroup$
$begingroup$
I guess I should start referring back to this as the definitive answer on the $|P| to 0$ versus refinement question (+1).
$endgroup$
– RRL
Dec 3 '18 at 1:02
$begingroup$
@RRL: thanks for your encouraging words!
$endgroup$
– Paramanand Singh
Dec 3 '18 at 1:05
add a comment |
$begingroup$
You are having a fundamental misunderstanding on this topic for some reason. We have a bounded function. We define the upper integral. No question that it exists. We define the lower integral. Again, no question that it exists. We then define what it means for a bounded function to be Riemann integrable (RI): The uppper integral equals the lower integral. Plenty of questions about when this happens. The theory of the Riemann integral is all about when we are lucky enough to have $f$ RI, and about the value of the integral when it exists. For example, there is the theorem that if $f$ is continuous on $[a,b],$ then $f$ is RI on $[a,b].$ There is the FTC. A beautiful result of Lebesgue gives a necessary and sufficient condition: $f$ is RI iff $f$ is continuous a.e. All of these results go back to the definition.
$endgroup$
add a comment |
Your Answer
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2045722%2fwhy-is-the-upper-riemann-integral-the-infimum-of-all-upper-sums%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your question does have some ambiguity. From the wording of your question and comments it appears that you want to know:
Does the limit of upper sums (with respect to partitions getting finer and finer) equal the infimum of all upper sums?
First of all note that when we are dealing with limits of things dependent on a partition of an interval then there are two ways in which the limit operation can be defined:
1) Limit via refinement of a partition: Let $P = x_0, x_1, x_2,ldots, x_n $ be a partition of $[a, b]$ where $$a =x_0 < x_1 < x_2 < cdots < x_n = b$$ A partition $P'$ of $[a, b]$ is said to be a refinement of $P$ (or finer than $P$) if $P subseteq P'$.
Let $mathcalP[a, b]$ denote the collection of all partitions of $[a, b]$ and let $F:mathcalP[a, b] to mathbbR$ be a function. A number $L$ is said to be the limit of $F$ (via refinement) if for every $epsilon > 0$ there is a partition $P_epsilonin mathcalP[a, b]$ such that $|F(P) - L| < epsilon$ for all $P in mathcalP[a, b]$ with $P_epsilon subseteq P$.
2) Limit as norm of parititon tends to $0$: If $P = a = x_0, x_1, x_2, ldots, x_n = b$ is a partition of $[a, b]$ then the norm $||P||$ of partition $P$ is defined as $||P|| = max_i = 1^n(x_i - x_i - 1)$.
Let $mathcalP[a, b]$ denote the collection of all partitions of $[a, b]$ and let $F: mathcalP[a, b] to mathbbR$ be a function. A number $L$ is said to be limit of $F$ as norm of partition tends to $0$ if for every $epsilon > 0$ there is a $delta > 0$ such that $|F(P) - L| < epsilon$ for all $Pin mathcalP[a, b]$ with $||P|| < delta$. This is written as $lim_ to 0F(P) = L$.
Note that for a given function $F:mathcalP[a, b] to mathbbR$ the limiting behavior of $F$ can be different according to these two definitions given above. In fact if $F(P) to L$ as $||P||to 0$ then $F(P) to L$ via refinement but the converse may not hold in general. This is because of the fact that refinement of a partition leads to a decrease in the norm, but decreasing the norm of a partition does not necessarily lead to a refinement.
Now let $f$ be a function defined and bounded on $[a, b]$ and let $P = x_0, x_1, x_2, ldots x_n$ be a partition of $[a, b]$. Let $M_k = sup,f(x), x in [x_k - 1, x_k]$ and let $mathcalP[a, b]$ denote the collection of all partitions of $[a, b]$. We define the upper sum function $S:mathcalP[a, b] to mathbbR$ by $$S(P) = sum_k = 1^nM_k(x_k - x_k - 1)$$ It is easy to prove that if $m = inf,f(x), x in [a, b]$ then $S(P) geq m(b - a)$ for all $P in mathcalP[a, b]$ and further if $P, P' in mathcalP[a, b]$ are such that $P subseteq P'$ then $S(P') leq S(P)$. It follows that $J = inf,S(P), P in mathcalP[a, b]$ exists.
Your question can now be worded more concretely into one of the following two forms:
Does $S(P) to J$ via refinement?
or
Does $lim_ to 0S(P) = J$?
The answer to the first question is obviously "yes" and you should be able to prove this using the definition of limit via refinement given above.
The answer to second question is also "yes" but it is difficult to prove. We first prove the result for a non-negative function $f$. Let $epsilon > 0$ be given. Since $J = inf,S(P), P in mathcalP[a, b]$, there is a partition $P_epsilon in mathcalP[a, b]$ such that $$J leq S(P_epsilon) < J + fracepsilon2tag1$$ Let $P_epsilon = x_0', x_1', x_2', ldots, x_N'$ and let $M = sup,f(x), x in [a, b] + 1$. Let $delta = epsilon / (2MN)$ and consider a partition $P = x_0, x_1, x_2, ldots, x_n$ with $||P|| < delta$.
We can write $$S(P) = sum_k = 1^nM_k(x_k - x_k - 1) = S_1 + S_2tag2$$ where $S_1$ is the sum corresponding to the index $k$ for which $[x_k - 1, x_k]$ does not contain any point of $P_epsilon$ and $S_2$ is the sum corresponding to other values of index $k$. Clearly for $S_1$ the interval $[x_k - 1, x_k]$ lies wholly in one of the intervals $[x_j - 1', x_j']$ made by $P_epsilon$ and hence $S_1 leq S(P_epsilon)$ (note that $f$ is non-negative). For $S_2$ we can see that the number of such indexes $k$ is no more than $N$ and hence $S_2 < MNdelta = epsilon / 2$ (note that $f$ is non-negative here). It follows that $$J leq S(P) = S_1 + S_2 < S(P_epsilon) + fracepsilon2 < J + epsilontag3$$ for all $P in mathcalP[a, b]$ with $||P|| < delta$. It follows that $S(P) to J$ as $||P|| to 0$.
Extension to a general function $f$ can be achieved by writing $f(x) = g(x) + m$ where $m = inf,f(x), x in [a, b]$ and noting that $g$ is non-negative.
Note: The limit of a Riemann sum is based on the two definitions given above but there is a slight complication. A Riemann sum depends not only on a partition but also on choice of tags corresponding to a partition. Formally one can view a Riemann sum not as a function from $mathcalP [a, b] $ to $mathbbR $ but rather as a relation from $mathcalP [a, b] $ to $mathbb R $ such that it relates every partition of $[a, b] $ to one or more real numbers.
$endgroup$
$begingroup$
I guess I should start referring back to this as the definitive answer on the $|P| to 0$ versus refinement question (+1).
$endgroup$
– RRL
Dec 3 '18 at 1:02
$begingroup$
@RRL: thanks for your encouraging words!
$endgroup$
– Paramanand Singh
Dec 3 '18 at 1:05
add a comment |
$begingroup$
Your question does have some ambiguity. From the wording of your question and comments it appears that you want to know:
Does the limit of upper sums (with respect to partitions getting finer and finer) equal the infimum of all upper sums?
First of all note that when we are dealing with limits of things dependent on a partition of an interval then there are two ways in which the limit operation can be defined:
1) Limit via refinement of a partition: Let $P = x_0, x_1, x_2,ldots, x_n $ be a partition of $[a, b]$ where $$a =x_0 < x_1 < x_2 < cdots < x_n = b$$ A partition $P'$ of $[a, b]$ is said to be a refinement of $P$ (or finer than $P$) if $P subseteq P'$.
Let $mathcalP[a, b]$ denote the collection of all partitions of $[a, b]$ and let $F:mathcalP[a, b] to mathbbR$ be a function. A number $L$ is said to be the limit of $F$ (via refinement) if for every $epsilon > 0$ there is a partition $P_epsilonin mathcalP[a, b]$ such that $|F(P) - L| < epsilon$ for all $P in mathcalP[a, b]$ with $P_epsilon subseteq P$.
2) Limit as norm of parititon tends to $0$: If $P = a = x_0, x_1, x_2, ldots, x_n = b$ is a partition of $[a, b]$ then the norm $||P||$ of partition $P$ is defined as $||P|| = max_i = 1^n(x_i - x_i - 1)$.
Let $mathcalP[a, b]$ denote the collection of all partitions of $[a, b]$ and let $F: mathcalP[a, b] to mathbbR$ be a function. A number $L$ is said to be limit of $F$ as norm of partition tends to $0$ if for every $epsilon > 0$ there is a $delta > 0$ such that $|F(P) - L| < epsilon$ for all $Pin mathcalP[a, b]$ with $||P|| < delta$. This is written as $lim_ to 0F(P) = L$.
Note that for a given function $F:mathcalP[a, b] to mathbbR$ the limiting behavior of $F$ can be different according to these two definitions given above. In fact if $F(P) to L$ as $||P||to 0$ then $F(P) to L$ via refinement but the converse may not hold in general. This is because of the fact that refinement of a partition leads to a decrease in the norm, but decreasing the norm of a partition does not necessarily lead to a refinement.
Now let $f$ be a function defined and bounded on $[a, b]$ and let $P = x_0, x_1, x_2, ldots x_n$ be a partition of $[a, b]$. Let $M_k = sup,f(x), x in [x_k - 1, x_k]$ and let $mathcalP[a, b]$ denote the collection of all partitions of $[a, b]$. We define the upper sum function $S:mathcalP[a, b] to mathbbR$ by $$S(P) = sum_k = 1^nM_k(x_k - x_k - 1)$$ It is easy to prove that if $m = inf,f(x), x in [a, b]$ then $S(P) geq m(b - a)$ for all $P in mathcalP[a, b]$ and further if $P, P' in mathcalP[a, b]$ are such that $P subseteq P'$ then $S(P') leq S(P)$. It follows that $J = inf,S(P), P in mathcalP[a, b]$ exists.
Your question can now be worded more concretely into one of the following two forms:
Does $S(P) to J$ via refinement?
or
Does $lim_ to 0S(P) = J$?
The answer to the first question is obviously "yes" and you should be able to prove this using the definition of limit via refinement given above.
The answer to second question is also "yes" but it is difficult to prove. We first prove the result for a non-negative function $f$. Let $epsilon > 0$ be given. Since $J = inf,S(P), P in mathcalP[a, b]$, there is a partition $P_epsilon in mathcalP[a, b]$ such that $$J leq S(P_epsilon) < J + fracepsilon2tag1$$ Let $P_epsilon = x_0', x_1', x_2', ldots, x_N'$ and let $M = sup,f(x), x in [a, b] + 1$. Let $delta = epsilon / (2MN)$ and consider a partition $P = x_0, x_1, x_2, ldots, x_n$ with $||P|| < delta$.
We can write $$S(P) = sum_k = 1^nM_k(x_k - x_k - 1) = S_1 + S_2tag2$$ where $S_1$ is the sum corresponding to the index $k$ for which $[x_k - 1, x_k]$ does not contain any point of $P_epsilon$ and $S_2$ is the sum corresponding to other values of index $k$. Clearly for $S_1$ the interval $[x_k - 1, x_k]$ lies wholly in one of the intervals $[x_j - 1', x_j']$ made by $P_epsilon$ and hence $S_1 leq S(P_epsilon)$ (note that $f$ is non-negative). For $S_2$ we can see that the number of such indexes $k$ is no more than $N$ and hence $S_2 < MNdelta = epsilon / 2$ (note that $f$ is non-negative here). It follows that $$J leq S(P) = S_1 + S_2 < S(P_epsilon) + fracepsilon2 < J + epsilontag3$$ for all $P in mathcalP[a, b]$ with $||P|| < delta$. It follows that $S(P) to J$ as $||P|| to 0$.
Extension to a general function $f$ can be achieved by writing $f(x) = g(x) + m$ where $m = inf,f(x), x in [a, b]$ and noting that $g$ is non-negative.
Note: The limit of a Riemann sum is based on the two definitions given above but there is a slight complication. A Riemann sum depends not only on a partition but also on choice of tags corresponding to a partition. Formally one can view a Riemann sum not as a function from $mathcalP [a, b] $ to $mathbbR $ but rather as a relation from $mathcalP [a, b] $ to $mathbb R $ such that it relates every partition of $[a, b] $ to one or more real numbers.
$endgroup$
$begingroup$
I guess I should start referring back to this as the definitive answer on the $|P| to 0$ versus refinement question (+1).
$endgroup$
– RRL
Dec 3 '18 at 1:02
$begingroup$
@RRL: thanks for your encouraging words!
$endgroup$
– Paramanand Singh
Dec 3 '18 at 1:05
add a comment |
$begingroup$
Your question does have some ambiguity. From the wording of your question and comments it appears that you want to know:
Does the limit of upper sums (with respect to partitions getting finer and finer) equal the infimum of all upper sums?
First of all note that when we are dealing with limits of things dependent on a partition of an interval then there are two ways in which the limit operation can be defined:
1) Limit via refinement of a partition: Let $P = x_0, x_1, x_2,ldots, x_n $ be a partition of $[a, b]$ where $$a =x_0 < x_1 < x_2 < cdots < x_n = b$$ A partition $P'$ of $[a, b]$ is said to be a refinement of $P$ (or finer than $P$) if $P subseteq P'$.
Let $mathcalP[a, b]$ denote the collection of all partitions of $[a, b]$ and let $F:mathcalP[a, b] to mathbbR$ be a function. A number $L$ is said to be the limit of $F$ (via refinement) if for every $epsilon > 0$ there is a partition $P_epsilonin mathcalP[a, b]$ such that $|F(P) - L| < epsilon$ for all $P in mathcalP[a, b]$ with $P_epsilon subseteq P$.
2) Limit as norm of parititon tends to $0$: If $P = a = x_0, x_1, x_2, ldots, x_n = b$ is a partition of $[a, b]$ then the norm $||P||$ of partition $P$ is defined as $||P|| = max_i = 1^n(x_i - x_i - 1)$.
Let $mathcalP[a, b]$ denote the collection of all partitions of $[a, b]$ and let $F: mathcalP[a, b] to mathbbR$ be a function. A number $L$ is said to be limit of $F$ as norm of partition tends to $0$ if for every $epsilon > 0$ there is a $delta > 0$ such that $|F(P) - L| < epsilon$ for all $Pin mathcalP[a, b]$ with $||P|| < delta$. This is written as $lim_ to 0F(P) = L$.
Note that for a given function $F:mathcalP[a, b] to mathbbR$ the limiting behavior of $F$ can be different according to these two definitions given above. In fact if $F(P) to L$ as $||P||to 0$ then $F(P) to L$ via refinement but the converse may not hold in general. This is because of the fact that refinement of a partition leads to a decrease in the norm, but decreasing the norm of a partition does not necessarily lead to a refinement.
Now let $f$ be a function defined and bounded on $[a, b]$ and let $P = x_0, x_1, x_2, ldots x_n$ be a partition of $[a, b]$. Let $M_k = sup,f(x), x in [x_k - 1, x_k]$ and let $mathcalP[a, b]$ denote the collection of all partitions of $[a, b]$. We define the upper sum function $S:mathcalP[a, b] to mathbbR$ by $$S(P) = sum_k = 1^nM_k(x_k - x_k - 1)$$ It is easy to prove that if $m = inf,f(x), x in [a, b]$ then $S(P) geq m(b - a)$ for all $P in mathcalP[a, b]$ and further if $P, P' in mathcalP[a, b]$ are such that $P subseteq P'$ then $S(P') leq S(P)$. It follows that $J = inf,S(P), P in mathcalP[a, b]$ exists.
Your question can now be worded more concretely into one of the following two forms:
Does $S(P) to J$ via refinement?
or
Does $lim_ to 0S(P) = J$?
The answer to the first question is obviously "yes" and you should be able to prove this using the definition of limit via refinement given above.
The answer to second question is also "yes" but it is difficult to prove. We first prove the result for a non-negative function $f$. Let $epsilon > 0$ be given. Since $J = inf,S(P), P in mathcalP[a, b]$, there is a partition $P_epsilon in mathcalP[a, b]$ such that $$J leq S(P_epsilon) < J + fracepsilon2tag1$$ Let $P_epsilon = x_0', x_1', x_2', ldots, x_N'$ and let $M = sup,f(x), x in [a, b] + 1$. Let $delta = epsilon / (2MN)$ and consider a partition $P = x_0, x_1, x_2, ldots, x_n$ with $||P|| < delta$.
We can write $$S(P) = sum_k = 1^nM_k(x_k - x_k - 1) = S_1 + S_2tag2$$ where $S_1$ is the sum corresponding to the index $k$ for which $[x_k - 1, x_k]$ does not contain any point of $P_epsilon$ and $S_2$ is the sum corresponding to other values of index $k$. Clearly for $S_1$ the interval $[x_k - 1, x_k]$ lies wholly in one of the intervals $[x_j - 1', x_j']$ made by $P_epsilon$ and hence $S_1 leq S(P_epsilon)$ (note that $f$ is non-negative). For $S_2$ we can see that the number of such indexes $k$ is no more than $N$ and hence $S_2 < MNdelta = epsilon / 2$ (note that $f$ is non-negative here). It follows that $$J leq S(P) = S_1 + S_2 < S(P_epsilon) + fracepsilon2 < J + epsilontag3$$ for all $P in mathcalP[a, b]$ with $||P|| < delta$. It follows that $S(P) to J$ as $||P|| to 0$.
Extension to a general function $f$ can be achieved by writing $f(x) = g(x) + m$ where $m = inf,f(x), x in [a, b]$ and noting that $g$ is non-negative.
Note: The limit of a Riemann sum is based on the two definitions given above but there is a slight complication. A Riemann sum depends not only on a partition but also on choice of tags corresponding to a partition. Formally one can view a Riemann sum not as a function from $mathcalP [a, b] $ to $mathbbR $ but rather as a relation from $mathcalP [a, b] $ to $mathbb R $ such that it relates every partition of $[a, b] $ to one or more real numbers.
$endgroup$
Your question does have some ambiguity. From the wording of your question and comments it appears that you want to know:
Does the limit of upper sums (with respect to partitions getting finer and finer) equal the infimum of all upper sums?
First of all note that when we are dealing with limits of things dependent on a partition of an interval then there are two ways in which the limit operation can be defined:
1) Limit via refinement of a partition: Let $P = x_0, x_1, x_2,ldots, x_n $ be a partition of $[a, b]$ where $$a =x_0 < x_1 < x_2 < cdots < x_n = b$$ A partition $P'$ of $[a, b]$ is said to be a refinement of $P$ (or finer than $P$) if $P subseteq P'$.
Let $mathcalP[a, b]$ denote the collection of all partitions of $[a, b]$ and let $F:mathcalP[a, b] to mathbbR$ be a function. A number $L$ is said to be the limit of $F$ (via refinement) if for every $epsilon > 0$ there is a partition $P_epsilonin mathcalP[a, b]$ such that $|F(P) - L| < epsilon$ for all $P in mathcalP[a, b]$ with $P_epsilon subseteq P$.
2) Limit as norm of parititon tends to $0$: If $P = a = x_0, x_1, x_2, ldots, x_n = b$ is a partition of $[a, b]$ then the norm $||P||$ of partition $P$ is defined as $||P|| = max_i = 1^n(x_i - x_i - 1)$.
Let $mathcalP[a, b]$ denote the collection of all partitions of $[a, b]$ and let $F: mathcalP[a, b] to mathbbR$ be a function. A number $L$ is said to be limit of $F$ as norm of partition tends to $0$ if for every $epsilon > 0$ there is a $delta > 0$ such that $|F(P) - L| < epsilon$ for all $Pin mathcalP[a, b]$ with $||P|| < delta$. This is written as $lim_ to 0F(P) = L$.
Note that for a given function $F:mathcalP[a, b] to mathbbR$ the limiting behavior of $F$ can be different according to these two definitions given above. In fact if $F(P) to L$ as $||P||to 0$ then $F(P) to L$ via refinement but the converse may not hold in general. This is because of the fact that refinement of a partition leads to a decrease in the norm, but decreasing the norm of a partition does not necessarily lead to a refinement.
Now let $f$ be a function defined and bounded on $[a, b]$ and let $P = x_0, x_1, x_2, ldots x_n$ be a partition of $[a, b]$. Let $M_k = sup,f(x), x in [x_k - 1, x_k]$ and let $mathcalP[a, b]$ denote the collection of all partitions of $[a, b]$. We define the upper sum function $S:mathcalP[a, b] to mathbbR$ by $$S(P) = sum_k = 1^nM_k(x_k - x_k - 1)$$ It is easy to prove that if $m = inf,f(x), x in [a, b]$ then $S(P) geq m(b - a)$ for all $P in mathcalP[a, b]$ and further if $P, P' in mathcalP[a, b]$ are such that $P subseteq P'$ then $S(P') leq S(P)$. It follows that $J = inf,S(P), P in mathcalP[a, b]$ exists.
Your question can now be worded more concretely into one of the following two forms:
Does $S(P) to J$ via refinement?
or
Does $lim_ to 0S(P) = J$?
The answer to the first question is obviously "yes" and you should be able to prove this using the definition of limit via refinement given above.
The answer to second question is also "yes" but it is difficult to prove. We first prove the result for a non-negative function $f$. Let $epsilon > 0$ be given. Since $J = inf,S(P), P in mathcalP[a, b]$, there is a partition $P_epsilon in mathcalP[a, b]$ such that $$J leq S(P_epsilon) < J + fracepsilon2tag1$$ Let $P_epsilon = x_0', x_1', x_2', ldots, x_N'$ and let $M = sup,f(x), x in [a, b] + 1$. Let $delta = epsilon / (2MN)$ and consider a partition $P = x_0, x_1, x_2, ldots, x_n$ with $||P|| < delta$.
We can write $$S(P) = sum_k = 1^nM_k(x_k - x_k - 1) = S_1 + S_2tag2$$ where $S_1$ is the sum corresponding to the index $k$ for which $[x_k - 1, x_k]$ does not contain any point of $P_epsilon$ and $S_2$ is the sum corresponding to other values of index $k$. Clearly for $S_1$ the interval $[x_k - 1, x_k]$ lies wholly in one of the intervals $[x_j - 1', x_j']$ made by $P_epsilon$ and hence $S_1 leq S(P_epsilon)$ (note that $f$ is non-negative). For $S_2$ we can see that the number of such indexes $k$ is no more than $N$ and hence $S_2 < MNdelta = epsilon / 2$ (note that $f$ is non-negative here). It follows that $$J leq S(P) = S_1 + S_2 < S(P_epsilon) + fracepsilon2 < J + epsilontag3$$ for all $P in mathcalP[a, b]$ with $||P|| < delta$. It follows that $S(P) to J$ as $||P|| to 0$.
Extension to a general function $f$ can be achieved by writing $f(x) = g(x) + m$ where $m = inf,f(x), x in [a, b]$ and noting that $g$ is non-negative.
Note: The limit of a Riemann sum is based on the two definitions given above but there is a slight complication. A Riemann sum depends not only on a partition but also on choice of tags corresponding to a partition. Formally one can view a Riemann sum not as a function from $mathcalP [a, b] $ to $mathbbR $ but rather as a relation from $mathcalP [a, b] $ to $mathbb R $ such that it relates every partition of $[a, b] $ to one or more real numbers.
edited Mar 27 at 18:38
answered Dec 7 '16 at 12:28
Paramanand SinghParamanand Singh
51.5k560170
51.5k560170
$begingroup$
I guess I should start referring back to this as the definitive answer on the $|P| to 0$ versus refinement question (+1).
$endgroup$
– RRL
Dec 3 '18 at 1:02
$begingroup$
@RRL: thanks for your encouraging words!
$endgroup$
– Paramanand Singh
Dec 3 '18 at 1:05
add a comment |
$begingroup$
I guess I should start referring back to this as the definitive answer on the $|P| to 0$ versus refinement question (+1).
$endgroup$
– RRL
Dec 3 '18 at 1:02
$begingroup$
@RRL: thanks for your encouraging words!
$endgroup$
– Paramanand Singh
Dec 3 '18 at 1:05
$begingroup$
I guess I should start referring back to this as the definitive answer on the $|P| to 0$ versus refinement question (+1).
$endgroup$
– RRL
Dec 3 '18 at 1:02
$begingroup$
I guess I should start referring back to this as the definitive answer on the $|P| to 0$ versus refinement question (+1).
$endgroup$
– RRL
Dec 3 '18 at 1:02
$begingroup$
@RRL: thanks for your encouraging words!
$endgroup$
– Paramanand Singh
Dec 3 '18 at 1:05
$begingroup$
@RRL: thanks for your encouraging words!
$endgroup$
– Paramanand Singh
Dec 3 '18 at 1:05
add a comment |
$begingroup$
You are having a fundamental misunderstanding on this topic for some reason. We have a bounded function. We define the upper integral. No question that it exists. We define the lower integral. Again, no question that it exists. We then define what it means for a bounded function to be Riemann integrable (RI): The uppper integral equals the lower integral. Plenty of questions about when this happens. The theory of the Riemann integral is all about when we are lucky enough to have $f$ RI, and about the value of the integral when it exists. For example, there is the theorem that if $f$ is continuous on $[a,b],$ then $f$ is RI on $[a,b].$ There is the FTC. A beautiful result of Lebesgue gives a necessary and sufficient condition: $f$ is RI iff $f$ is continuous a.e. All of these results go back to the definition.
$endgroup$
add a comment |
$begingroup$
You are having a fundamental misunderstanding on this topic for some reason. We have a bounded function. We define the upper integral. No question that it exists. We define the lower integral. Again, no question that it exists. We then define what it means for a bounded function to be Riemann integrable (RI): The uppper integral equals the lower integral. Plenty of questions about when this happens. The theory of the Riemann integral is all about when we are lucky enough to have $f$ RI, and about the value of the integral when it exists. For example, there is the theorem that if $f$ is continuous on $[a,b],$ then $f$ is RI on $[a,b].$ There is the FTC. A beautiful result of Lebesgue gives a necessary and sufficient condition: $f$ is RI iff $f$ is continuous a.e. All of these results go back to the definition.
$endgroup$
add a comment |
$begingroup$
You are having a fundamental misunderstanding on this topic for some reason. We have a bounded function. We define the upper integral. No question that it exists. We define the lower integral. Again, no question that it exists. We then define what it means for a bounded function to be Riemann integrable (RI): The uppper integral equals the lower integral. Plenty of questions about when this happens. The theory of the Riemann integral is all about when we are lucky enough to have $f$ RI, and about the value of the integral when it exists. For example, there is the theorem that if $f$ is continuous on $[a,b],$ then $f$ is RI on $[a,b].$ There is the FTC. A beautiful result of Lebesgue gives a necessary and sufficient condition: $f$ is RI iff $f$ is continuous a.e. All of these results go back to the definition.
$endgroup$
You are having a fundamental misunderstanding on this topic for some reason. We have a bounded function. We define the upper integral. No question that it exists. We define the lower integral. Again, no question that it exists. We then define what it means for a bounded function to be Riemann integrable (RI): The uppper integral equals the lower integral. Plenty of questions about when this happens. The theory of the Riemann integral is all about when we are lucky enough to have $f$ RI, and about the value of the integral when it exists. For example, there is the theorem that if $f$ is continuous on $[a,b],$ then $f$ is RI on $[a,b].$ There is the FTC. A beautiful result of Lebesgue gives a necessary and sufficient condition: $f$ is RI iff $f$ is continuous a.e. All of these results go back to the definition.
answered Dec 6 '16 at 3:31
zhw.zhw.
75.1k43275
75.1k43275
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2045722%2fwhy-is-the-upper-riemann-integral-the-infimum-of-all-upper-sums%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Do you have a definition of the upper integral other than "the infimum of all upper sums"? From what you've written, it sounds like it's just the definition.
$endgroup$
– Eric Wofsey
Dec 6 '16 at 0:41
$begingroup$
It's not a theorem; it's a definition.
$endgroup$
– zhw.
Dec 6 '16 at 0:42
$begingroup$
@zhw , But is it right to claim that we could just define that that particular limit which is what the integral basically is equals the infimum ?
$endgroup$
– Noob101
Dec 6 '16 at 0:52
$begingroup$
@EricWofsey , I wanted to know that is the upper integral defined in this way a riemann integral too , that is if the partition goes to less than delta , then will my upper sum converge to it ?
$endgroup$
– Noob101
Dec 6 '16 at 0:54
$begingroup$
Nice but slightly ambiguous question. +1
$endgroup$
– Paramanand Singh
Dec 7 '16 at 13:17