Mixing
Integrability of a monotonic function and Darboux sum.
$begingroup$
I have two questions on integrability.
(a) Prove that $f(x)$ is integrable on $[a,b]$ if $f(x)$ is monotonic on the interval.
(b) Define $f(x) = 1$ if $x$ rational, $0$ otherwise. Prove $f(x)$ is not integrable on $[0,1].$
For (a), there is a proof on my text book but it uses something called mesh which I dont understand. Is there any other simple proof for this?
For (b), the function is clearly not continuous on the interval which implies that it is not integrable. But how should I start the proof??
tnaks!!
real-analysis
asked Dec 19 '13 at 6:04
eChung00eChung00
1,11831929
$endgroup$
add a comment |
$begingroup$
I have two questions on integrability.
(a) Prove that $f(x)$ is integrable on $[a,b]$ if $f(x)$ is monotonic on the interval.
(b) Define $f(x) = 1$ if $x$ rational, $0$ otherwise. Prove $f(x)$ is not integrable on $[0,1].$
For (a), there is a proof on my text book but it uses something called mesh which I dont understand. Is there any other simple proof for this?
For (b), the function is clearly not continuous on the interval which implies that it is not integrable. But how should I start the proof??
tnaks!!
real-analysis
asked Dec 19 '13 at 6:04
eChung00eChung00
1,11831929
$endgroup$
add a comment |
$begingroup$
I have two questions on integrability.
(a) Prove that $f(x)$ is integrable on $[a,b]$ if $f(x)$ is monotonic on the interval.
(b) Define $f(x) = 1$ if $x$ rational, $0$ otherwise. Prove $f(x)$ is not integrable on $[0,1].$
For (a), there is a proof on my text book but it uses something called mesh which I dont understand. Is there any other simple proof for this?
For (b), the function is clearly not continuous on the interval which implies that it is not integrable. But how should I start the proof??
tnaks!!
real-analysis
asked Dec 19 '13 at 6:04
eChung00eChung00
1,11831929
$endgroup$
I have two questions on integrability.
(a) Prove that $f(x)$ is integrable on $[a,b]$ if $f(x)$ is monotonic on the interval.
(b) Define $f(x) = 1$ if $x$ rational, $0$ otherwise. Prove $f(x)$ is not integrable on $[0,1].$
For (a), there is a proof on my text book but it uses something called mesh which I dont understand. Is there any other simple proof for this?
For (b), the function is clearly not continuous on the interval which implies that it is not integrable. But how should I start the proof??
tnaks!!
real-analysis
real-analysis
asked Dec 19 '13 at 6:04
eChung00eChung00
1,11831929
asked Dec 19 '13 at 6:04
eChung00eChung00
1,11831929
asked Dec 19 '13 at 6:04
eChung00eChung00
1,11831929
asked Dec 19 '13 at 6:04
eChung00eChung00
1,11831929
asked Dec 19 '13 at 6:04
eChung00eChung00
1,11831929
1,11831929
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The standard proof of (a) goes by noticing that since $f$ is monotone, given any partition $P$, if $[x_{i-1},x_i]$ is one of the intervals in the partition, the largest value of $f$ on the interval is one of $f(x_{i-1})$ and $f(x_i)$, and the least value is the other one: If $f$ is increasing, the largest value is $f(x_i)$. If $f$ is decreasing, it is the other one. This means that if we pick a partition where all the intervals have the same length $(b-a)/n$, $U(f,P)-L(f,P)$ forms a telescoping sum, and it is easy to see that it converges to $0$ as $ntoinfty$. In case there is confusion, $U(f,P)$ denotes here the upper Riemann sum, and $L(f,P)$ the lower Riemann sum, that is, if $P={x_0,dots,x_n}$, then $$L(f,P)=sum_{i=1}^ninf{f(x)mid x_{i-1}le xle x_i}cdot (x_i-x_{i-1}),$$ and $$U(f,P)=sum_{i=1}^nsup{f(x)mid x_{i-1}le xle x_i}cdot (x_i-x_{i-1}).$$
In detail, if $f$ is increasing and $P$ is the partition into $n$ pieces of the same size, $$ U(f,P)-L(f,P)=frac{b-a}nsum_{i=1}^n (f(x_i)-f(x_{i-1}))=frac{b-a}n(f(b)-f(a))to0$$ as $ntoinfty$. If $f$ is decreasing, the argument is the same, but the sum is now $(b-a)(f(a)-f(b))/n$.
The standard proof of (b) goes by noticing that $f$ is discontinuous everywhere. This is much stronger than just "not continuous", as you say. But this uses the Lebesgue criterion ($f$ is integrable iff it is continuous a.e., and bounded), which seems like an overkill. Easier to note that $L(f,P)=0$ and $U(f,P)=1$ for any partition $P$.
The point in both cases is that $f$ is integrable iff for any $epsilon>0$ there is $P$ such that $U(f,P)-L(f,P)<epsilon$.
By the way, (a) is fairly classical. Although it is not stated this way, the proof above is essentially the one given by Newton in Lemma II of Book I of his Philosophiæ Naturalis Principia Mathematica, see here. What we did algebraically above with a telescoping sum, Newton does geometrically: See the figure accompanying Lemma II, and note that the rectangles $aKbl,bLcm$, etc, correspond to the differences $(M_1-m_1)(b-a)/n$, $(M_2-m_2)(b-a)/2$, etc., in the argument above, where $M_i= sup{f(x)mid x_{i-1}le xle x_i}$ and $m_i=inf{f(x)mid x_{i-1}le xle x_i}$. Newton then notes than translating these rectangles so they all line up below $aKbl$ gives us the area of the rectangle $aABl$, that is, $(f(b)-f(a))(b-a)/n$. (David Bressoud also draws attention to this argument by Newton in the Introduction to his nice A radical approach to Lebesgue's theory of integration.)
One could suspect that the (Riemann) integrability of continuous functions could be verified just building on this argument, as it appears reasonable that any continuous function is piecewise monotonic -- Dirichlet seems to have believed that, as it is part of his argument that the Fourier series of a continuous function converges pointwise to the function. Nowadays, however, we know that this is false (and Dirichlet's theorem is false as well in full generality).
The first correct proof of integrability of continuous functions is due to Cauchy, in his 1823 book, where the definite integral is defined for the first time not in terms of antiderivatives, but rather in terms of a particular version of Riemann sum. His argument, naturally, proceeds along different lines than (a) above.
edited Apr 13 '17 at 12:20
Community♦
1
answered Dec 19 '13 at 6:16
Andrés E. CaicedoAndrés E. Caicedo
65.5k8158249
$endgroup$
add a comment |
$begingroup$
Assume that $f(x)$ is monotonic increasing on the interval. Choose, for each $n$, a uniform partition, $P_n$ on $[a,b]$ which is a finite sequence of real numbers such that
$$a=x_0<x_<cdots<x_n=b.$$
Fix $jin {1,2,cdots,n}$ and $,xin I_j,$ then
$$m_jleq f(t_j)leq M_j,;;text{where};;t_jin I_j,$$
Since $f$ is increasing,
begin{align}m_j=inf{f(x):;x_{j-1}leq x< x_{j} }=f(x_{j-1}) ;;text{and};; M_j=sup{f(x):;x_{j-1}leq x< x_{j} }leq f(x_{j}).end{align}
This means that
begin{align} &L(f,P_n)= sum^{n}_{j=1}m_j(x_{j}-x_{j-1}) = sum^{n}_{j=1}f(x_{j-1})(x_{j}-x_{j-1}) ,\&U(f,P_n)= sum^{n}_{j=1}M_j(x_{j}-x_{j-1})leq sum^{n}_{j=1}f(x_{j})(x_{j}-x_{j-1}).end{align}
Let $epsilon>0$ be given. Choose a partition, $P_epsilon$, such that $maxlimits_{1leq jleq n}(x_j-x_{j-1})<dfrac{epsilon}{(f(b)-f(a)+1)}$, then
begin{align}U(f,P_epsilon)-L(f,P_epsilon)&leq sum^{n}_{j=1}left[f(x_{j}-f(x_{j-1}) right](x_{j}-x_{j-1})\&leq maxlimits_{1leq jleq n}(x_j-x_{j-1})sum^{n}_{j=1}left[f(x_{j}-f(x_{j-1}) right]\&<dfrac{epsilon left(f(b)-f(a) right)}{(f(b)-f(a)+1)}\&<epsilonend{align}
Thus, $f$ is integrable on $[a,b]$. The same thing can be done for monotonically decreasing function on $[a,b]$.
answered Jan 15 at 19:39
Omojola MichealOmojola Micheal
1,889324
$endgroup$
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%2f612635%2fintegrability-of-a-monotonic-function-and-darboux-sum%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
$begingroup$
The standard proof of (a) goes by noticing that since $f$ is monotone, given any partition $P$, if $[x_{i-1},x_i]$ is one of the intervals in the partition, the largest value of $f$ on the interval is one of $f(x_{i-1})$ and $f(x_i)$, and the least value is the other one: If $f$ is increasing, the largest value is $f(x_i)$. If $f$ is decreasing, it is the other one. This means that if we pick a partition where all the intervals have the same length $(b-a)/n$, $U(f,P)-L(f,P)$ forms a telescoping sum, and it is easy to see that it converges to $0$ as $ntoinfty$. In case there is confusion, $U(f,P)$ denotes here the upper Riemann sum, and $L(f,P)$ the lower Riemann sum, that is, if $P={x_0,dots,x_n}$, then $$L(f,P)=sum_{i=1}^ninf{f(x)mid x_{i-1}le xle x_i}cdot (x_i-x_{i-1}),$$ and $$U(f,P)=sum_{i=1}^nsup{f(x)mid x_{i-1}le xle x_i}cdot (x_i-x_{i-1}).$$
In detail, if $f$ is increasing and $P$ is the partition into $n$ pieces of the same size, $$ U(f,P)-L(f,P)=frac{b-a}nsum_{i=1}^n (f(x_i)-f(x_{i-1}))=frac{b-a}n(f(b)-f(a))to0$$ as $ntoinfty$. If $f$ is decreasing, the argument is the same, but the sum is now $(b-a)(f(a)-f(b))/n$.
The standard proof of (b) goes by noticing that $f$ is discontinuous everywhere. This is much stronger than just "not continuous", as you say. But this uses the Lebesgue criterion ($f$ is integrable iff it is continuous a.e., and bounded), which seems like an overkill. Easier to note that $L(f,P)=0$ and $U(f,P)=1$ for any partition $P$.
The point in both cases is that $f$ is integrable iff for any $epsilon>0$ there is $P$ such that $U(f,P)-L(f,P)<epsilon$.
By the way, (a) is fairly classical. Although it is not stated this way, the proof above is essentially the one given by Newton in Lemma II of Book I of his Philosophiæ Naturalis Principia Mathematica, see here. What we did algebraically above with a telescoping sum, Newton does geometrically: See the figure accompanying Lemma II, and note that the rectangles $aKbl,bLcm$, etc, correspond to the differences $(M_1-m_1)(b-a)/n$, $(M_2-m_2)(b-a)/2$, etc., in the argument above, where $M_i= sup{f(x)mid x_{i-1}le xle x_i}$ and $m_i=inf{f(x)mid x_{i-1}le xle x_i}$. Newton then notes than translating these rectangles so they all line up below $aKbl$ gives us the area of the rectangle $aABl$, that is, $(f(b)-f(a))(b-a)/n$. (David Bressoud also draws attention to this argument by Newton in the Introduction to his nice A radical approach to Lebesgue's theory of integration.)
One could suspect that the (Riemann) integrability of continuous functions could be verified just building on this argument, as it appears reasonable that any continuous function is piecewise monotonic -- Dirichlet seems to have believed that, as it is part of his argument that the Fourier series of a continuous function converges pointwise to the function. Nowadays, however, we know that this is false (and Dirichlet's theorem is false as well in full generality).
The first correct proof of integrability of continuous functions is due to Cauchy, in his 1823 book, where the definite integral is defined for the first time not in terms of antiderivatives, but rather in terms of a particular version of Riemann sum. His argument, naturally, proceeds along different lines than (a) above.
edited Apr 13 '17 at 12:20
Community♦
1
answered Dec 19 '13 at 6:16
Andrés E. CaicedoAndrés E. Caicedo
65.5k8158249
$endgroup$
add a comment |
$begingroup$
The standard proof of (a) goes by noticing that since $f$ is monotone, given any partition $P$, if $[x_{i-1},x_i]$ is one of the intervals in the partition, the largest value of $f$ on the interval is one of $f(x_{i-1})$ and $f(x_i)$, and the least value is the other one: If $f$ is increasing, the largest value is $f(x_i)$. If $f$ is decreasing, it is the other one. This means that if we pick a partition where all the intervals have the same length $(b-a)/n$, $U(f,P)-L(f,P)$ forms a telescoping sum, and it is easy to see that it converges to $0$ as $ntoinfty$. In case there is confusion, $U(f,P)$ denotes here the upper Riemann sum, and $L(f,P)$ the lower Riemann sum, that is, if $P={x_0,dots,x_n}$, then $$L(f,P)=sum_{i=1}^ninf{f(x)mid x_{i-1}le xle x_i}cdot (x_i-x_{i-1}),$$ and $$U(f,P)=sum_{i=1}^nsup{f(x)mid x_{i-1}le xle x_i}cdot (x_i-x_{i-1}).$$
In detail, if $f$ is increasing and $P$ is the partition into $n$ pieces of the same size, $$ U(f,P)-L(f,P)=frac{b-a}nsum_{i=1}^n (f(x_i)-f(x_{i-1}))=frac{b-a}n(f(b)-f(a))to0$$ as $ntoinfty$. If $f$ is decreasing, the argument is the same, but the sum is now $(b-a)(f(a)-f(b))/n$.
The standard proof of (b) goes by noticing that $f$ is discontinuous everywhere. This is much stronger than just "not continuous", as you say. But this uses the Lebesgue criterion ($f$ is integrable iff it is continuous a.e., and bounded), which seems like an overkill. Easier to note that $L(f,P)=0$ and $U(f,P)=1$ for any partition $P$.
The point in both cases is that $f$ is integrable iff for any $epsilon>0$ there is $P$ such that $U(f,P)-L(f,P)<epsilon$.
By the way, (a) is fairly classical. Although it is not stated this way, the proof above is essentially the one given by Newton in Lemma II of Book I of his Philosophiæ Naturalis Principia Mathematica, see here. What we did algebraically above with a telescoping sum, Newton does geometrically: See the figure accompanying Lemma II, and note that the rectangles $aKbl,bLcm$, etc, correspond to the differences $(M_1-m_1)(b-a)/n$, $(M_2-m_2)(b-a)/2$, etc., in the argument above, where $M_i= sup{f(x)mid x_{i-1}le xle x_i}$ and $m_i=inf{f(x)mid x_{i-1}le xle x_i}$. Newton then notes than translating these rectangles so they all line up below $aKbl$ gives us the area of the rectangle $aABl$, that is, $(f(b)-f(a))(b-a)/n$. (David Bressoud also draws attention to this argument by Newton in the Introduction to his nice A radical approach to Lebesgue's theory of integration.)
One could suspect that the (Riemann) integrability of continuous functions could be verified just building on this argument, as it appears reasonable that any continuous function is piecewise monotonic -- Dirichlet seems to have believed that, as it is part of his argument that the Fourier series of a continuous function converges pointwise to the function. Nowadays, however, we know that this is false (and Dirichlet's theorem is false as well in full generality).
The first correct proof of integrability of continuous functions is due to Cauchy, in his 1823 book, where the definite integral is defined for the first time not in terms of antiderivatives, but rather in terms of a particular version of Riemann sum. His argument, naturally, proceeds along different lines than (a) above.
edited Apr 13 '17 at 12:20
Community♦
1
answered Dec 19 '13 at 6:16
Andrés E. CaicedoAndrés E. Caicedo
65.5k8158249
$endgroup$
add a comment |
$begingroup$
The standard proof of (a) goes by noticing that since $f$ is monotone, given any partition $P$, if $[x_{i-1},x_i]$ is one of the intervals in the partition, the largest value of $f$ on the interval is one of $f(x_{i-1})$ and $f(x_i)$, and the least value is the other one: If $f$ is increasing, the largest value is $f(x_i)$. If $f$ is decreasing, it is the other one. This means that if we pick a partition where all the intervals have the same length $(b-a)/n$, $U(f,P)-L(f,P)$ forms a telescoping sum, and it is easy to see that it converges to $0$ as $ntoinfty$. In case there is confusion, $U(f,P)$ denotes here the upper Riemann sum, and $L(f,P)$ the lower Riemann sum, that is, if $P={x_0,dots,x_n}$, then $$L(f,P)=sum_{i=1}^ninf{f(x)mid x_{i-1}le xle x_i}cdot (x_i-x_{i-1}),$$ and $$U(f,P)=sum_{i=1}^nsup{f(x)mid x_{i-1}le xle x_i}cdot (x_i-x_{i-1}).$$
In detail, if $f$ is increasing and $P$ is the partition into $n$ pieces of the same size, $$ U(f,P)-L(f,P)=frac{b-a}nsum_{i=1}^n (f(x_i)-f(x_{i-1}))=frac{b-a}n(f(b)-f(a))to0$$ as $ntoinfty$. If $f$ is decreasing, the argument is the same, but the sum is now $(b-a)(f(a)-f(b))/n$.
The standard proof of (b) goes by noticing that $f$ is discontinuous everywhere. This is much stronger than just "not continuous", as you say. But this uses the Lebesgue criterion ($f$ is integrable iff it is continuous a.e., and bounded), which seems like an overkill. Easier to note that $L(f,P)=0$ and $U(f,P)=1$ for any partition $P$.
The point in both cases is that $f$ is integrable iff for any $epsilon>0$ there is $P$ such that $U(f,P)-L(f,P)<epsilon$.
By the way, (a) is fairly classical. Although it is not stated this way, the proof above is essentially the one given by Newton in Lemma II of Book I of his Philosophiæ Naturalis Principia Mathematica, see here. What we did algebraically above with a telescoping sum, Newton does geometrically: See the figure accompanying Lemma II, and note that the rectangles $aKbl,bLcm$, etc, correspond to the differences $(M_1-m_1)(b-a)/n$, $(M_2-m_2)(b-a)/2$, etc., in the argument above, where $M_i= sup{f(x)mid x_{i-1}le xle x_i}$ and $m_i=inf{f(x)mid x_{i-1}le xle x_i}$. Newton then notes than translating these rectangles so they all line up below $aKbl$ gives us the area of the rectangle $aABl$, that is, $(f(b)-f(a))(b-a)/n$. (David Bressoud also draws attention to this argument by Newton in the Introduction to his nice A radical approach to Lebesgue's theory of integration.)
One could suspect that the (Riemann) integrability of continuous functions could be verified just building on this argument, as it appears reasonable that any continuous function is piecewise monotonic -- Dirichlet seems to have believed that, as it is part of his argument that the Fourier series of a continuous function converges pointwise to the function. Nowadays, however, we know that this is false (and Dirichlet's theorem is false as well in full generality).
The first correct proof of integrability of continuous functions is due to Cauchy, in his 1823 book, where the definite integral is defined for the first time not in terms of antiderivatives, but rather in terms of a particular version of Riemann sum. His argument, naturally, proceeds along different lines than (a) above.
edited Apr 13 '17 at 12:20
Community♦
1
answered Dec 19 '13 at 6:16
Andrés E. CaicedoAndrés E. Caicedo
65.5k8158249
$endgroup$
The standard proof of (a) goes by noticing that since $f$ is monotone, given any partition $P$, if $[x_{i-1},x_i]$ is one of the intervals in the partition, the largest value of $f$ on the interval is one of $f(x_{i-1})$ and $f(x_i)$, and the least value is the other one: If $f$ is increasing, the largest value is $f(x_i)$. If $f$ is decreasing, it is the other one. This means that if we pick a partition where all the intervals have the same length $(b-a)/n$, $U(f,P)-L(f,P)$ forms a telescoping sum, and it is easy to see that it converges to $0$ as $ntoinfty$. In case there is confusion, $U(f,P)$ denotes here the upper Riemann sum, and $L(f,P)$ the lower Riemann sum, that is, if $P={x_0,dots,x_n}$, then $$L(f,P)=sum_{i=1}^ninf{f(x)mid x_{i-1}le xle x_i}cdot (x_i-x_{i-1}),$$ and $$U(f,P)=sum_{i=1}^nsup{f(x)mid x_{i-1}le xle x_i}cdot (x_i-x_{i-1}).$$
In detail, if $f$ is increasing and $P$ is the partition into $n$ pieces of the same size, $$ U(f,P)-L(f,P)=frac{b-a}nsum_{i=1}^n (f(x_i)-f(x_{i-1}))=frac{b-a}n(f(b)-f(a))to0$$ as $ntoinfty$. If $f$ is decreasing, the argument is the same, but the sum is now $(b-a)(f(a)-f(b))/n$.
The standard proof of (b) goes by noticing that $f$ is discontinuous everywhere. This is much stronger than just "not continuous", as you say. But this uses the Lebesgue criterion ($f$ is integrable iff it is continuous a.e., and bounded), which seems like an overkill. Easier to note that $L(f,P)=0$ and $U(f,P)=1$ for any partition $P$.
The point in both cases is that $f$ is integrable iff for any $epsilon>0$ there is $P$ such that $U(f,P)-L(f,P)<epsilon$.
By the way, (a) is fairly classical. Although it is not stated this way, the proof above is essentially the one given by Newton in Lemma II of Book I of his Philosophiæ Naturalis Principia Mathematica, see here. What we did algebraically above with a telescoping sum, Newton does geometrically: See the figure accompanying Lemma II, and note that the rectangles $aKbl,bLcm$, etc, correspond to the differences $(M_1-m_1)(b-a)/n$, $(M_2-m_2)(b-a)/2$, etc., in the argument above, where $M_i= sup{f(x)mid x_{i-1}le xle x_i}$ and $m_i=inf{f(x)mid x_{i-1}le xle x_i}$. Newton then notes than translating these rectangles so they all line up below $aKbl$ gives us the area of the rectangle $aABl$, that is, $(f(b)-f(a))(b-a)/n$. (David Bressoud also draws attention to this argument by Newton in the Introduction to his nice A radical approach to Lebesgue's theory of integration.)
One could suspect that the (Riemann) integrability of continuous functions could be verified just building on this argument, as it appears reasonable that any continuous function is piecewise monotonic -- Dirichlet seems to have believed that, as it is part of his argument that the Fourier series of a continuous function converges pointwise to the function. Nowadays, however, we know that this is false (and Dirichlet's theorem is false as well in full generality).
The first correct proof of integrability of continuous functions is due to Cauchy, in his 1823 book, where the definite integral is defined for the first time not in terms of antiderivatives, but rather in terms of a particular version of Riemann sum. His argument, naturally, proceeds along different lines than (a) above.
edited Apr 13 '17 at 12:20
Community♦
1
answered Dec 19 '13 at 6:16
Andrés E. CaicedoAndrés E. Caicedo
65.5k8158249
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
1
answered Dec 19 '13 at 6:16
Andrés E. CaicedoAndrés E. Caicedo
65.5k8158249
answered Dec 19 '13 at 6:16
Andrés E. CaicedoAndrés E. Caicedo
65.5k8158249
answered Dec 19 '13 at 6:16
Andrés E. CaicedoAndrés E. Caicedo
65.5k8158249
65.5k8158249
add a comment |
add a comment |
$begingroup$
Assume that $f(x)$ is monotonic increasing on the interval. Choose, for each $n$, a uniform partition, $P_n$ on $[a,b]$ which is a finite sequence of real numbers such that
$$a=x_0<x_<cdots<x_n=b.$$
Fix $jin {1,2,cdots,n}$ and $,xin I_j,$ then
$$m_jleq f(t_j)leq M_j,;;text{where};;t_jin I_j,$$
Since $f$ is increasing,
begin{align}m_j=inf{f(x):;x_{j-1}leq x< x_{j} }=f(x_{j-1}) ;;text{and};; M_j=sup{f(x):;x_{j-1}leq x< x_{j} }leq f(x_{j}).end{align}
This means that
begin{align} &L(f,P_n)= sum^{n}_{j=1}m_j(x_{j}-x_{j-1}) = sum^{n}_{j=1}f(x_{j-1})(x_{j}-x_{j-1}) ,\&U(f,P_n)= sum^{n}_{j=1}M_j(x_{j}-x_{j-1})leq sum^{n}_{j=1}f(x_{j})(x_{j}-x_{j-1}).end{align}
Let $epsilon>0$ be given. Choose a partition, $P_epsilon$, such that $maxlimits_{1leq jleq n}(x_j-x_{j-1})<dfrac{epsilon}{(f(b)-f(a)+1)}$, then
begin{align}U(f,P_epsilon)-L(f,P_epsilon)&leq sum^{n}_{j=1}left[f(x_{j}-f(x_{j-1}) right](x_{j}-x_{j-1})\&leq maxlimits_{1leq jleq n}(x_j-x_{j-1})sum^{n}_{j=1}left[f(x_{j}-f(x_{j-1}) right]\&<dfrac{epsilon left(f(b)-f(a) right)}{(f(b)-f(a)+1)}\&<epsilonend{align}
Thus, $f$ is integrable on $[a,b]$. The same thing can be done for monotonically decreasing function on $[a,b]$.
answered Jan 15 at 19:39
Omojola MichealOmojola Micheal
1,889324
$endgroup$
add a comment |
$begingroup$
Assume that $f(x)$ is monotonic increasing on the interval. Choose, for each $n$, a uniform partition, $P_n$ on $[a,b]$ which is a finite sequence of real numbers such that
$$a=x_0<x_<cdots<x_n=b.$$
Fix $jin {1,2,cdots,n}$ and $,xin I_j,$ then
$$m_jleq f(t_j)leq M_j,;;text{where};;t_jin I_j,$$
Since $f$ is increasing,
begin{align}m_j=inf{f(x):;x_{j-1}leq x< x_{j} }=f(x_{j-1}) ;;text{and};; M_j=sup{f(x):;x_{j-1}leq x< x_{j} }leq f(x_{j}).end{align}
This means that
begin{align} &L(f,P_n)= sum^{n}_{j=1}m_j(x_{j}-x_{j-1}) = sum^{n}_{j=1}f(x_{j-1})(x_{j}-x_{j-1}) ,\&U(f,P_n)= sum^{n}_{j=1}M_j(x_{j}-x_{j-1})leq sum^{n}_{j=1}f(x_{j})(x_{j}-x_{j-1}).end{align}
Let $epsilon>0$ be given. Choose a partition, $P_epsilon$, such that $maxlimits_{1leq jleq n}(x_j-x_{j-1})<dfrac{epsilon}{(f(b)-f(a)+1)}$, then
begin{align}U(f,P_epsilon)-L(f,P_epsilon)&leq sum^{n}_{j=1}left[f(x_{j}-f(x_{j-1}) right](x_{j}-x_{j-1})\&leq maxlimits_{1leq jleq n}(x_j-x_{j-1})sum^{n}_{j=1}left[f(x_{j}-f(x_{j-1}) right]\&<dfrac{epsilon left(f(b)-f(a) right)}{(f(b)-f(a)+1)}\&<epsilonend{align}
Thus, $f$ is integrable on $[a,b]$. The same thing can be done for monotonically decreasing function on $[a,b]$.
answered Jan 15 at 19:39
Omojola MichealOmojola Micheal
1,889324
$endgroup$
add a comment |
$begingroup$
Assume that $f(x)$ is monotonic increasing on the interval. Choose, for each $n$, a uniform partition, $P_n$ on $[a,b]$ which is a finite sequence of real numbers such that
$$a=x_0<x_<cdots<x_n=b.$$
Fix $jin {1,2,cdots,n}$ and $,xin I_j,$ then
$$m_jleq f(t_j)leq M_j,;;text{where};;t_jin I_j,$$
Since $f$ is increasing,
begin{align}m_j=inf{f(x):;x_{j-1}leq x< x_{j} }=f(x_{j-1}) ;;text{and};; M_j=sup{f(x):;x_{j-1}leq x< x_{j} }leq f(x_{j}).end{align}
This means that
begin{align} &L(f,P_n)= sum^{n}_{j=1}m_j(x_{j}-x_{j-1}) = sum^{n}_{j=1}f(x_{j-1})(x_{j}-x_{j-1}) ,\&U(f,P_n)= sum^{n}_{j=1}M_j(x_{j}-x_{j-1})leq sum^{n}_{j=1}f(x_{j})(x_{j}-x_{j-1}).end{align}
Let $epsilon>0$ be given. Choose a partition, $P_epsilon$, such that $maxlimits_{1leq jleq n}(x_j-x_{j-1})<dfrac{epsilon}{(f(b)-f(a)+1)}$, then
begin{align}U(f,P_epsilon)-L(f,P_epsilon)&leq sum^{n}_{j=1}left[f(x_{j}-f(x_{j-1}) right](x_{j}-x_{j-1})\&leq maxlimits_{1leq jleq n}(x_j-x_{j-1})sum^{n}_{j=1}left[f(x_{j}-f(x_{j-1}) right]\&<dfrac{epsilon left(f(b)-f(a) right)}{(f(b)-f(a)+1)}\&<epsilonend{align}
Thus, $f$ is integrable on $[a,b]$. The same thing can be done for monotonically decreasing function on $[a,b]$.
answered Jan 15 at 19:39
Omojola MichealOmojola Micheal
1,889324
$endgroup$
Assume that $f(x)$ is monotonic increasing on the interval. Choose, for each $n$, a uniform partition, $P_n$ on $[a,b]$ which is a finite sequence of real numbers such that
$$a=x_0<x_<cdots<x_n=b.$$
Fix $jin {1,2,cdots,n}$ and $,xin I_j,$ then
$$m_jleq f(t_j)leq M_j,;;text{where};;t_jin I_j,$$
Since $f$ is increasing,
begin{align}m_j=inf{f(x):;x_{j-1}leq x< x_{j} }=f(x_{j-1}) ;;text{and};; M_j=sup{f(x):;x_{j-1}leq x< x_{j} }leq f(x_{j}).end{align}
This means that
begin{align} &L(f,P_n)= sum^{n}_{j=1}m_j(x_{j}-x_{j-1}) = sum^{n}_{j=1}f(x_{j-1})(x_{j}-x_{j-1}) ,\&U(f,P_n)= sum^{n}_{j=1}M_j(x_{j}-x_{j-1})leq sum^{n}_{j=1}f(x_{j})(x_{j}-x_{j-1}).end{align}
Let $epsilon>0$ be given. Choose a partition, $P_epsilon$, such that $maxlimits_{1leq jleq n}(x_j-x_{j-1})<dfrac{epsilon}{(f(b)-f(a)+1)}$, then
begin{align}U(f,P_epsilon)-L(f,P_epsilon)&leq sum^{n}_{j=1}left[f(x_{j}-f(x_{j-1}) right](x_{j}-x_{j-1})\&leq maxlimits_{1leq jleq n}(x_j-x_{j-1})sum^{n}_{j=1}left[f(x_{j}-f(x_{j-1}) right]\&<dfrac{epsilon left(f(b)-f(a) right)}{(f(b)-f(a)+1)}\&<epsilonend{align}
Thus, $f$ is integrable on $[a,b]$. The same thing can be done for monotonically decreasing function on $[a,b]$.
answered Jan 15 at 19:39
Omojola MichealOmojola Micheal
1,889324
edited Jan 15 at 19:48
answered Jan 15 at 19:39
Omojola MichealOmojola Micheal
1,889324
answered Jan 15 at 19:39
Omojola MichealOmojola Micheal
1,889324
answered Jan 15 at 19:39
Omojola MichealOmojola Micheal
1,889324
1,889324
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%2f612635%2fintegrability-of-a-monotonic-function-and-darboux-sum%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
