Undefined Subintervals - Riemann Integrals
$begingroup$
I searched through stackexchange and multiple other PDFs but couldn't find an answer
I'm curious to know when talking about Riemann Integrals with respect to functions that are bounded on closed bounded intervals (which we've been studying).
If I take some arbitrary function $f$ which is bounded on the interval $[a,b]$ (where a and b are finite real numbers); what happens if the function $f$ is undefined on a sub interval of a particular partition in that closed bounded interval $[a,b]$? This sub interval can't have a lub or a glb since the sub interval doesn't contain any real numbers (in this example). Is it just completely ignored in the sum since there is no "area"? Please explain
Cheers
integration
edited Apr 21 '15 at 13:11
Simon S
23.3k63481
asked Apr 21 '15 at 13:01
charl1echarl1e
11
$endgroup$
add a comment |
$begingroup$
I searched through stackexchange and multiple other PDFs but couldn't find an answer
I'm curious to know when talking about Riemann Integrals with respect to functions that are bounded on closed bounded intervals (which we've been studying).
If I take some arbitrary function $f$ which is bounded on the interval $[a,b]$ (where a and b are finite real numbers); what happens if the function $f$ is undefined on a sub interval of a particular partition in that closed bounded interval $[a,b]$? This sub interval can't have a lub or a glb since the sub interval doesn't contain any real numbers (in this example). Is it just completely ignored in the sum since there is no "area"? Please explain
Cheers
integration
edited Apr 21 '15 at 13:11
Simon S
23.3k63481
asked Apr 21 '15 at 13:01
charl1echarl1e
11
$endgroup$
add a comment |
$begingroup$
I searched through stackexchange and multiple other PDFs but couldn't find an answer
I'm curious to know when talking about Riemann Integrals with respect to functions that are bounded on closed bounded intervals (which we've been studying).
If I take some arbitrary function $f$ which is bounded on the interval $[a,b]$ (where a and b are finite real numbers); what happens if the function $f$ is undefined on a sub interval of a particular partition in that closed bounded interval $[a,b]$? This sub interval can't have a lub or a glb since the sub interval doesn't contain any real numbers (in this example). Is it just completely ignored in the sum since there is no "area"? Please explain
Cheers
integration
edited Apr 21 '15 at 13:11
Simon S
23.3k63481
asked Apr 21 '15 at 13:01
charl1echarl1e
11
$endgroup$
I searched through stackexchange and multiple other PDFs but couldn't find an answer
I'm curious to know when talking about Riemann Integrals with respect to functions that are bounded on closed bounded intervals (which we've been studying).
If I take some arbitrary function $f$ which is bounded on the interval $[a,b]$ (where a and b are finite real numbers); what happens if the function $f$ is undefined on a sub interval of a particular partition in that closed bounded interval $[a,b]$? This sub interval can't have a lub or a glb since the sub interval doesn't contain any real numbers (in this example). Is it just completely ignored in the sum since there is no "area"? Please explain
Cheers
integration
integration
edited Apr 21 '15 at 13:11
Simon S
23.3k63481
asked Apr 21 '15 at 13:01
charl1echarl1e
11
edited Apr 21 '15 at 13:11
Simon S
23.3k63481
asked Apr 21 '15 at 13:01
charl1echarl1e
11
edited Apr 21 '15 at 13:11
Simon S
23.3k63481
edited Apr 21 '15 at 13:11
Simon S
23.3k63481
edited Apr 21 '15 at 13:11
Simon S
23.3k63481
23.3k63481
asked Apr 21 '15 at 13:01
charl1echarl1e
11
asked Apr 21 '15 at 13:01
charl1echarl1e
11
asked Apr 21 '15 at 13:01
charl1echarl1e
11
11
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Maybe, i'm not understanding your question correctly, but are you essentially asking if you have some function that is discontinuous in a partition how would you perform the Riemann Integral? I would imagine that the best approach would be to break up your function into however many discontinuities that may exist on the partition and perform the the Riemann Integral on those sub partitions that exist and then add them up.
answered Apr 21 '15 at 13:19
TophatTophat
20317
$endgroup$
$begingroup$
Yep, discontinuous. So you're saying that you ignore the subintervals in [a,b] where the function is discontinuous, and only sum the continuous subintervals (given some partition)? I don't really believe your answer, since the Riemann Sum is defined such that you're finding the glb or lub of a particular subinterval of [a,b] (depending on whether you're talking about lower or upper sums) which doesn't exist if the function is discontinuous (undefined) on the entire subinterval
$endgroup$
– charl1e
Apr 21 '15 at 13:34
$begingroup$
I don't really see the problem. If a function isn't defined on a certain sub-interval then how can you possibly find a glb or lub? It's the same idea of breaking up an integral of a discontinuous function on an interval.
$endgroup$
– Tophat
Apr 21 '15 at 13:49
$begingroup$
Exactly, you can't have a glb or lub, that's the problem. So given a partition, where the function is discontinuous over a single subinterval (for a given partition) is the sum still defined even though one of the terms in the sum doesn't have a glb or a lub?
$endgroup$
– charl1e
Apr 21 '15 at 22:17
$begingroup$
The Riemann sum is only defined on the intervals where it is not discontinuous. Thus if you want to take a Riemann sum over an entire interval and there exist discontinuous subintervals you have to break up the sum to not include those intervals.
$endgroup$
– Tophat
Apr 22 '15 at 13:56
$begingroup$
Perfect, cheers for the reply man
$endgroup$
– charl1e
Apr 23 '15 at 0:29
add a comment |
$begingroup$
For Riemann integrability, you need the function to be bounded in the interval, and the set of discontinuities to have a Lebesgue measure of 0.
You can extend this to undefined points by saying that the undefined point could be any value without changing the value of the Riemann integral, so the value at that point doesn't matter. This isn't the most rigorous thing you can say, but it can be said.
An entire sub-interval would have a Lebesgue measure greater than 0, because it would scale with the unit interval.
I wouldn't trust any definition of an integral that gives a value to a set with an undefined value. I think it makes the most sense to say that an integral over an undefined interval is undefined. Other than that, you might get away with calling it 0, or you might give it a value in a different unit that basically only gives you the measure of the undefined set under some obscure math.
answered Dec 13 '18 at 1:49
Prince_TatPrince_Tat
354
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
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%2f1244867%2fundefined-subintervals-riemann-integrals%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
$begingroup$
Maybe, i'm not understanding your question correctly, but are you essentially asking if you have some function that is discontinuous in a partition how would you perform the Riemann Integral? I would imagine that the best approach would be to break up your function into however many discontinuities that may exist on the partition and perform the the Riemann Integral on those sub partitions that exist and then add them up.
answered Apr 21 '15 at 13:19
TophatTophat
20317
$endgroup$
$begingroup$
Yep, discontinuous. So you're saying that you ignore the subintervals in [a,b] where the function is discontinuous, and only sum the continuous subintervals (given some partition)? I don't really believe your answer, since the Riemann Sum is defined such that you're finding the glb or lub of a particular subinterval of [a,b] (depending on whether you're talking about lower or upper sums) which doesn't exist if the function is discontinuous (undefined) on the entire subinterval
$endgroup$
– charl1e
Apr 21 '15 at 13:34
$begingroup$
I don't really see the problem. If a function isn't defined on a certain sub-interval then how can you possibly find a glb or lub? It's the same idea of breaking up an integral of a discontinuous function on an interval.
$endgroup$
– Tophat
Apr 21 '15 at 13:49
$begingroup$
Exactly, you can't have a glb or lub, that's the problem. So given a partition, where the function is discontinuous over a single subinterval (for a given partition) is the sum still defined even though one of the terms in the sum doesn't have a glb or a lub?
$endgroup$
– charl1e
Apr 21 '15 at 22:17
$begingroup$
The Riemann sum is only defined on the intervals where it is not discontinuous. Thus if you want to take a Riemann sum over an entire interval and there exist discontinuous subintervals you have to break up the sum to not include those intervals.
$endgroup$
– Tophat
Apr 22 '15 at 13:56
$begingroup$
Perfect, cheers for the reply man
$endgroup$
– charl1e
Apr 23 '15 at 0:29
add a comment |
$begingroup$
Maybe, i'm not understanding your question correctly, but are you essentially asking if you have some function that is discontinuous in a partition how would you perform the Riemann Integral? I would imagine that the best approach would be to break up your function into however many discontinuities that may exist on the partition and perform the the Riemann Integral on those sub partitions that exist and then add them up.
answered Apr 21 '15 at 13:19
TophatTophat
20317
$endgroup$
$begingroup$
Yep, discontinuous. So you're saying that you ignore the subintervals in [a,b] where the function is discontinuous, and only sum the continuous subintervals (given some partition)? I don't really believe your answer, since the Riemann Sum is defined such that you're finding the glb or lub of a particular subinterval of [a,b] (depending on whether you're talking about lower or upper sums) which doesn't exist if the function is discontinuous (undefined) on the entire subinterval
$endgroup$
– charl1e
Apr 21 '15 at 13:34
$begingroup$
I don't really see the problem. If a function isn't defined on a certain sub-interval then how can you possibly find a glb or lub? It's the same idea of breaking up an integral of a discontinuous function on an interval.
$endgroup$
– Tophat
Apr 21 '15 at 13:49
$begingroup$
Exactly, you can't have a glb or lub, that's the problem. So given a partition, where the function is discontinuous over a single subinterval (for a given partition) is the sum still defined even though one of the terms in the sum doesn't have a glb or a lub?
$endgroup$
– charl1e
Apr 21 '15 at 22:17
$begingroup$
The Riemann sum is only defined on the intervals where it is not discontinuous. Thus if you want to take a Riemann sum over an entire interval and there exist discontinuous subintervals you have to break up the sum to not include those intervals.
$endgroup$
– Tophat
Apr 22 '15 at 13:56
$begingroup$
Perfect, cheers for the reply man
$endgroup$
– charl1e
Apr 23 '15 at 0:29
add a comment |
$begingroup$
Maybe, i'm not understanding your question correctly, but are you essentially asking if you have some function that is discontinuous in a partition how would you perform the Riemann Integral? I would imagine that the best approach would be to break up your function into however many discontinuities that may exist on the partition and perform the the Riemann Integral on those sub partitions that exist and then add them up.
answered Apr 21 '15 at 13:19
TophatTophat
20317
$endgroup$
Maybe, i'm not understanding your question correctly, but are you essentially asking if you have some function that is discontinuous in a partition how would you perform the Riemann Integral? I would imagine that the best approach would be to break up your function into however many discontinuities that may exist on the partition and perform the the Riemann Integral on those sub partitions that exist and then add them up.
answered Apr 21 '15 at 13:19
TophatTophat
20317
answered Apr 21 '15 at 13:19
TophatTophat
20317
answered Apr 21 '15 at 13:19
TophatTophat
20317
answered Apr 21 '15 at 13:19
TophatTophat
20317
20317
$begingroup$
Yep, discontinuous. So you're saying that you ignore the subintervals in [a,b] where the function is discontinuous, and only sum the continuous subintervals (given some partition)? I don't really believe your answer, since the Riemann Sum is defined such that you're finding the glb or lub of a particular subinterval of [a,b] (depending on whether you're talking about lower or upper sums) which doesn't exist if the function is discontinuous (undefined) on the entire subinterval
$endgroup$
– charl1e
Apr 21 '15 at 13:34
$begingroup$
I don't really see the problem. If a function isn't defined on a certain sub-interval then how can you possibly find a glb or lub? It's the same idea of breaking up an integral of a discontinuous function on an interval.
$endgroup$
– Tophat
Apr 21 '15 at 13:49
$begingroup$
Exactly, you can't have a glb or lub, that's the problem. So given a partition, where the function is discontinuous over a single subinterval (for a given partition) is the sum still defined even though one of the terms in the sum doesn't have a glb or a lub?
$endgroup$
– charl1e
Apr 21 '15 at 22:17
$begingroup$
The Riemann sum is only defined on the intervals where it is not discontinuous. Thus if you want to take a Riemann sum over an entire interval and there exist discontinuous subintervals you have to break up the sum to not include those intervals.
$endgroup$
– Tophat
Apr 22 '15 at 13:56
$begingroup$
Perfect, cheers for the reply man
$endgroup$
– charl1e
Apr 23 '15 at 0:29
add a comment |
$begingroup$
Yep, discontinuous. So you're saying that you ignore the subintervals in [a,b] where the function is discontinuous, and only sum the continuous subintervals (given some partition)? I don't really believe your answer, since the Riemann Sum is defined such that you're finding the glb or lub of a particular subinterval of [a,b] (depending on whether you're talking about lower or upper sums) which doesn't exist if the function is discontinuous (undefined) on the entire subinterval
$endgroup$
– charl1e
Apr 21 '15 at 13:34
$begingroup$
I don't really see the problem. If a function isn't defined on a certain sub-interval then how can you possibly find a glb or lub? It's the same idea of breaking up an integral of a discontinuous function on an interval.
$endgroup$
– Tophat
Apr 21 '15 at 13:49
$begingroup$
Exactly, you can't have a glb or lub, that's the problem. So given a partition, where the function is discontinuous over a single subinterval (for a given partition) is the sum still defined even though one of the terms in the sum doesn't have a glb or a lub?
$endgroup$
– charl1e
Apr 21 '15 at 22:17
$begingroup$
The Riemann sum is only defined on the intervals where it is not discontinuous. Thus if you want to take a Riemann sum over an entire interval and there exist discontinuous subintervals you have to break up the sum to not include those intervals.
$endgroup$
– Tophat
Apr 22 '15 at 13:56
$begingroup$
Perfect, cheers for the reply man
$endgroup$
– charl1e
Apr 23 '15 at 0:29
$begingroup$
Yep, discontinuous. So you're saying that you ignore the subintervals in [a,b] where the function is discontinuous, and only sum the continuous subintervals (given some partition)? I don't really believe your answer, since the Riemann Sum is defined such that you're finding the glb or lub of a particular subinterval of [a,b] (depending on whether you're talking about lower or upper sums) which doesn't exist if the function is discontinuous (undefined) on the entire subinterval
$endgroup$
– charl1e
Apr 21 '15 at 13:34
$begingroup$
Yep, discontinuous. So you're saying that you ignore the subintervals in [a,b] where the function is discontinuous, and only sum the continuous subintervals (given some partition)? I don't really believe your answer, since the Riemann Sum is defined such that you're finding the glb or lub of a particular subinterval of [a,b] (depending on whether you're talking about lower or upper sums) which doesn't exist if the function is discontinuous (undefined) on the entire subinterval
$endgroup$
– charl1e
Apr 21 '15 at 13:34
$begingroup$
I don't really see the problem. If a function isn't defined on a certain sub-interval then how can you possibly find a glb or lub? It's the same idea of breaking up an integral of a discontinuous function on an interval.
$endgroup$
– Tophat
Apr 21 '15 at 13:49
$begingroup$
I don't really see the problem. If a function isn't defined on a certain sub-interval then how can you possibly find a glb or lub? It's the same idea of breaking up an integral of a discontinuous function on an interval.
$endgroup$
– Tophat
Apr 21 '15 at 13:49
$begingroup$
Exactly, you can't have a glb or lub, that's the problem. So given a partition, where the function is discontinuous over a single subinterval (for a given partition) is the sum still defined even though one of the terms in the sum doesn't have a glb or a lub?
$endgroup$
– charl1e
Apr 21 '15 at 22:17
$begingroup$
Exactly, you can't have a glb or lub, that's the problem. So given a partition, where the function is discontinuous over a single subinterval (for a given partition) is the sum still defined even though one of the terms in the sum doesn't have a glb or a lub?
$endgroup$
– charl1e
Apr 21 '15 at 22:17
$begingroup$
The Riemann sum is only defined on the intervals where it is not discontinuous. Thus if you want to take a Riemann sum over an entire interval and there exist discontinuous subintervals you have to break up the sum to not include those intervals.
$endgroup$
– Tophat
Apr 22 '15 at 13:56
$begingroup$
The Riemann sum is only defined on the intervals where it is not discontinuous. Thus if you want to take a Riemann sum over an entire interval and there exist discontinuous subintervals you have to break up the sum to not include those intervals.
$endgroup$
– Tophat
Apr 22 '15 at 13:56
$begingroup$
Perfect, cheers for the reply man
$endgroup$
– charl1e
Apr 23 '15 at 0:29
$begingroup$
Perfect, cheers for the reply man
$endgroup$
– charl1e
Apr 23 '15 at 0:29
add a comment |
$begingroup$
For Riemann integrability, you need the function to be bounded in the interval, and the set of discontinuities to have a Lebesgue measure of 0.
You can extend this to undefined points by saying that the undefined point could be any value without changing the value of the Riemann integral, so the value at that point doesn't matter. This isn't the most rigorous thing you can say, but it can be said.
An entire sub-interval would have a Lebesgue measure greater than 0, because it would scale with the unit interval.
I wouldn't trust any definition of an integral that gives a value to a set with an undefined value. I think it makes the most sense to say that an integral over an undefined interval is undefined. Other than that, you might get away with calling it 0, or you might give it a value in a different unit that basically only gives you the measure of the undefined set under some obscure math.
answered Dec 13 '18 at 1:49
Prince_TatPrince_Tat
354
$endgroup$
add a comment |
$begingroup$
For Riemann integrability, you need the function to be bounded in the interval, and the set of discontinuities to have a Lebesgue measure of 0.
You can extend this to undefined points by saying that the undefined point could be any value without changing the value of the Riemann integral, so the value at that point doesn't matter. This isn't the most rigorous thing you can say, but it can be said.
An entire sub-interval would have a Lebesgue measure greater than 0, because it would scale with the unit interval.
I wouldn't trust any definition of an integral that gives a value to a set with an undefined value. I think it makes the most sense to say that an integral over an undefined interval is undefined. Other than that, you might get away with calling it 0, or you might give it a value in a different unit that basically only gives you the measure of the undefined set under some obscure math.
answered Dec 13 '18 at 1:49
Prince_TatPrince_Tat
354
$endgroup$
add a comment |
$begingroup$
For Riemann integrability, you need the function to be bounded in the interval, and the set of discontinuities to have a Lebesgue measure of 0.
You can extend this to undefined points by saying that the undefined point could be any value without changing the value of the Riemann integral, so the value at that point doesn't matter. This isn't the most rigorous thing you can say, but it can be said.
An entire sub-interval would have a Lebesgue measure greater than 0, because it would scale with the unit interval.
I wouldn't trust any definition of an integral that gives a value to a set with an undefined value. I think it makes the most sense to say that an integral over an undefined interval is undefined. Other than that, you might get away with calling it 0, or you might give it a value in a different unit that basically only gives you the measure of the undefined set under some obscure math.
answered Dec 13 '18 at 1:49
Prince_TatPrince_Tat
354
$endgroup$
For Riemann integrability, you need the function to be bounded in the interval, and the set of discontinuities to have a Lebesgue measure of 0.
You can extend this to undefined points by saying that the undefined point could be any value without changing the value of the Riemann integral, so the value at that point doesn't matter. This isn't the most rigorous thing you can say, but it can be said.
An entire sub-interval would have a Lebesgue measure greater than 0, because it would scale with the unit interval.
I wouldn't trust any definition of an integral that gives a value to a set with an undefined value. I think it makes the most sense to say that an integral over an undefined interval is undefined. Other than that, you might get away with calling it 0, or you might give it a value in a different unit that basically only gives you the measure of the undefined set under some obscure math.
answered Dec 13 '18 at 1:49
Prince_TatPrince_Tat
354
answered Dec 13 '18 at 1:49
Prince_TatPrince_Tat
354
answered Dec 13 '18 at 1:49
Prince_TatPrince_Tat
354
answered Dec 13 '18 at 1:49
Prince_TatPrince_Tat
354
354
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%2f1244867%2fundefined-subintervals-riemann-integrals%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
