Theorem 11.33 rudin
$begingroup$
Please I have a slight confusion with the notation used by Rudin in the following proof.
11.33 Theorem. If $f$ is Riemann integrable on $[a,b],$ then $f$ is Lebesgue integrable on $[a,b]$ with respect to the Lebesgue measure $m$ and
$$ int_a^b f dx = mathscr{R} int_a^b f dx $$
Where $mathscr{R} int$ denotes the Riemann integral, while $int$ denotes the Lebesgue integral.
Proof Suppose $f$ is bounded. Then there exists a sequence ${P_k}$ of partitions of $[a,b]$ such that $P_{k+1}$ is a refinement of $P_k$ for each $k$ and
$$ lim_{ktoinfty} L(P_k,f) = mathscr{R}underline{int_a^b} f dx, quad lim_{ktoinfty} U(P_k,f) = mathscr{R}overline{int_a^b} f dx.
$$
Where $L(P_k), U(P_k)$ are the upper and lower sums respectively. If $P_k={a=x_0<x_1<dots<x_n=b},$ these are defined as,
$$ L(P_k,f) = sum_{i=1}^n (x_i-x_{i-1})m_i, quad U(P_k,f) = sum_{i=1}^n (x_i-x_{i-1})M_i,$$
where $M_i = sup_{xin[x_{i-1},x_i]} f(x)$ and $m_i = inf_{xin[x_{i-1},x_i]} f(x).$
We then define functions $U,L$ as $U_k(a)=L_k(a)=f(a)$ and for each $x in(x_{i-1},x_i],$ $1leq i leq n,$ $U_k(x)=M_i$ and $L_k(x)=m_i.$ Then for all $xin [a,b],$
$$ L(P_k,f) = int_a^b L_k dx, quad U(P_k,f) = int_a^b U_k dx, $$
and $$L_1(x) leq L_2(x) leq dots leq f(x) leq dots leq U_2(x) leq U_1(x). $$
There the sequence of functions $L_k, U_k$ converge point-wise on $[a,b],$ so let $L, U$ be the limit functions respectively. Then $L$ and $U$ are bounded measurable functions on $[a,b]$ and for any $x in [a,b],$
$$ L(x) leq f(x) leq U(x), $$
and by the monotone convergence theorem,
$$
int_a^b L(x) dx = mathscr{R} underline{int_a^b} f dx, quad int_a^b U(x) dx = mathscr{R} overline{int_a^b} f dx.
$$
Since $f$ is Riemann integrable, the upper and lower Riemann integrals are equal. Since $L(x) leq U(x),$ it follows that $L(x) = U(x)$ almost everywhere on [a,b].
Then $L(x) = f(x) = U(x)$ almost everywhere on $[a,b],$ so $f$ is measurable and the result follows.
I do understand the idea of the proof but the definition of the partition $P_k={a=x_0<x_1<dots<x_n=b}$. Shouldn't the end point depend on $k$ and not $n$?
Edit: By the explanation of @Ben W, how then is $$L_1(x) leq L_2(x) leq dots leq f(x) leq dots leq U_2(x) leq U_1(x). $$ obtained.
lebesgue-integral riemann-integration
edited Jan 1 at 21:41
Namaste
1
$endgroup$
add a comment |
$begingroup$
Please I have a slight confusion with the notation used by Rudin in the following proof.
11.33 Theorem. If $f$ is Riemann integrable on $[a,b],$ then $f$ is Lebesgue integrable on $[a,b]$ with respect to the Lebesgue measure $m$ and
$$ int_a^b f dx = mathscr{R} int_a^b f dx $$
Where $mathscr{R} int$ denotes the Riemann integral, while $int$ denotes the Lebesgue integral.
Proof Suppose $f$ is bounded. Then there exists a sequence ${P_k}$ of partitions of $[a,b]$ such that $P_{k+1}$ is a refinement of $P_k$ for each $k$ and
$$ lim_{ktoinfty} L(P_k,f) = mathscr{R}underline{int_a^b} f dx, quad lim_{ktoinfty} U(P_k,f) = mathscr{R}overline{int_a^b} f dx.
$$
Where $L(P_k), U(P_k)$ are the upper and lower sums respectively. If $P_k={a=x_0<x_1<dots<x_n=b},$ these are defined as,
$$ L(P_k,f) = sum_{i=1}^n (x_i-x_{i-1})m_i, quad U(P_k,f) = sum_{i=1}^n (x_i-x_{i-1})M_i,$$
where $M_i = sup_{xin[x_{i-1},x_i]} f(x)$ and $m_i = inf_{xin[x_{i-1},x_i]} f(x).$
We then define functions $U,L$ as $U_k(a)=L_k(a)=f(a)$ and for each $x in(x_{i-1},x_i],$ $1leq i leq n,$ $U_k(x)=M_i$ and $L_k(x)=m_i.$ Then for all $xin [a,b],$
$$ L(P_k,f) = int_a^b L_k dx, quad U(P_k,f) = int_a^b U_k dx, $$
and $$L_1(x) leq L_2(x) leq dots leq f(x) leq dots leq U_2(x) leq U_1(x). $$
There the sequence of functions $L_k, U_k$ converge point-wise on $[a,b],$ so let $L, U$ be the limit functions respectively. Then $L$ and $U$ are bounded measurable functions on $[a,b]$ and for any $x in [a,b],$
$$ L(x) leq f(x) leq U(x), $$
and by the monotone convergence theorem,
$$
int_a^b L(x) dx = mathscr{R} underline{int_a^b} f dx, quad int_a^b U(x) dx = mathscr{R} overline{int_a^b} f dx.
$$
Since $f$ is Riemann integrable, the upper and lower Riemann integrals are equal. Since $L(x) leq U(x),$ it follows that $L(x) = U(x)$ almost everywhere on [a,b].
Then $L(x) = f(x) = U(x)$ almost everywhere on $[a,b],$ so $f$ is measurable and the result follows.
I do understand the idea of the proof but the definition of the partition $P_k={a=x_0<x_1<dots<x_n=b}$. Shouldn't the end point depend on $k$ and not $n$?
Edit: By the explanation of @Ben W, how then is $$L_1(x) leq L_2(x) leq dots leq f(x) leq dots leq U_2(x) leq U_1(x). $$ obtained.
lebesgue-integral riemann-integration
edited Jan 1 at 21:41
Namaste
1
$endgroup$
add a comment |
$begingroup$
Please I have a slight confusion with the notation used by Rudin in the following proof.
11.33 Theorem. If $f$ is Riemann integrable on $[a,b],$ then $f$ is Lebesgue integrable on $[a,b]$ with respect to the Lebesgue measure $m$ and
$$ int_a^b f dx = mathscr{R} int_a^b f dx $$
Where $mathscr{R} int$ denotes the Riemann integral, while $int$ denotes the Lebesgue integral.
Proof Suppose $f$ is bounded. Then there exists a sequence ${P_k}$ of partitions of $[a,b]$ such that $P_{k+1}$ is a refinement of $P_k$ for each $k$ and
$$ lim_{ktoinfty} L(P_k,f) = mathscr{R}underline{int_a^b} f dx, quad lim_{ktoinfty} U(P_k,f) = mathscr{R}overline{int_a^b} f dx.
$$
Where $L(P_k), U(P_k)$ are the upper and lower sums respectively. If $P_k={a=x_0<x_1<dots<x_n=b},$ these are defined as,
$$ L(P_k,f) = sum_{i=1}^n (x_i-x_{i-1})m_i, quad U(P_k,f) = sum_{i=1}^n (x_i-x_{i-1})M_i,$$
where $M_i = sup_{xin[x_{i-1},x_i]} f(x)$ and $m_i = inf_{xin[x_{i-1},x_i]} f(x).$
We then define functions $U,L$ as $U_k(a)=L_k(a)=f(a)$ and for each $x in(x_{i-1},x_i],$ $1leq i leq n,$ $U_k(x)=M_i$ and $L_k(x)=m_i.$ Then for all $xin [a,b],$
$$ L(P_k,f) = int_a^b L_k dx, quad U(P_k,f) = int_a^b U_k dx, $$
and $$L_1(x) leq L_2(x) leq dots leq f(x) leq dots leq U_2(x) leq U_1(x). $$
There the sequence of functions $L_k, U_k$ converge point-wise on $[a,b],$ so let $L, U$ be the limit functions respectively. Then $L$ and $U$ are bounded measurable functions on $[a,b]$ and for any $x in [a,b],$
$$ L(x) leq f(x) leq U(x), $$
and by the monotone convergence theorem,
$$
int_a^b L(x) dx = mathscr{R} underline{int_a^b} f dx, quad int_a^b U(x) dx = mathscr{R} overline{int_a^b} f dx.
$$
Since $f$ is Riemann integrable, the upper and lower Riemann integrals are equal. Since $L(x) leq U(x),$ it follows that $L(x) = U(x)$ almost everywhere on [a,b].
Then $L(x) = f(x) = U(x)$ almost everywhere on $[a,b],$ so $f$ is measurable and the result follows.
I do understand the idea of the proof but the definition of the partition $P_k={a=x_0<x_1<dots<x_n=b}$. Shouldn't the end point depend on $k$ and not $n$?
Edit: By the explanation of @Ben W, how then is $$L_1(x) leq L_2(x) leq dots leq f(x) leq dots leq U_2(x) leq U_1(x). $$ obtained.
lebesgue-integral riemann-integration
edited Jan 1 at 21:41
Namaste
1
$endgroup$
Please I have a slight confusion with the notation used by Rudin in the following proof.
11.33 Theorem. If $f$ is Riemann integrable on $[a,b],$ then $f$ is Lebesgue integrable on $[a,b]$ with respect to the Lebesgue measure $m$ and
$$ int_a^b f dx = mathscr{R} int_a^b f dx $$
Where $mathscr{R} int$ denotes the Riemann integral, while $int$ denotes the Lebesgue integral.
Proof Suppose $f$ is bounded. Then there exists a sequence ${P_k}$ of partitions of $[a,b]$ such that $P_{k+1}$ is a refinement of $P_k$ for each $k$ and
$$ lim_{ktoinfty} L(P_k,f) = mathscr{R}underline{int_a^b} f dx, quad lim_{ktoinfty} U(P_k,f) = mathscr{R}overline{int_a^b} f dx.
$$
Where $L(P_k), U(P_k)$ are the upper and lower sums respectively. If $P_k={a=x_0<x_1<dots<x_n=b},$ these are defined as,
$$ L(P_k,f) = sum_{i=1}^n (x_i-x_{i-1})m_i, quad U(P_k,f) = sum_{i=1}^n (x_i-x_{i-1})M_i,$$
where $M_i = sup_{xin[x_{i-1},x_i]} f(x)$ and $m_i = inf_{xin[x_{i-1},x_i]} f(x).$
We then define functions $U,L$ as $U_k(a)=L_k(a)=f(a)$ and for each $x in(x_{i-1},x_i],$ $1leq i leq n,$ $U_k(x)=M_i$ and $L_k(x)=m_i.$ Then for all $xin [a,b],$
$$ L(P_k,f) = int_a^b L_k dx, quad U(P_k,f) = int_a^b U_k dx, $$
and $$L_1(x) leq L_2(x) leq dots leq f(x) leq dots leq U_2(x) leq U_1(x). $$
There the sequence of functions $L_k, U_k$ converge point-wise on $[a,b],$ so let $L, U$ be the limit functions respectively. Then $L$ and $U$ are bounded measurable functions on $[a,b]$ and for any $x in [a,b],$
$$ L(x) leq f(x) leq U(x), $$
and by the monotone convergence theorem,
$$
int_a^b L(x) dx = mathscr{R} underline{int_a^b} f dx, quad int_a^b U(x) dx = mathscr{R} overline{int_a^b} f dx.
$$
Since $f$ is Riemann integrable, the upper and lower Riemann integrals are equal. Since $L(x) leq U(x),$ it follows that $L(x) = U(x)$ almost everywhere on [a,b].
Then $L(x) = f(x) = U(x)$ almost everywhere on $[a,b],$ so $f$ is measurable and the result follows.
I do understand the idea of the proof but the definition of the partition $P_k={a=x_0<x_1<dots<x_n=b}$. Shouldn't the end point depend on $k$ and not $n$?
Edit: By the explanation of @Ben W, how then is $$L_1(x) leq L_2(x) leq dots leq f(x) leq dots leq U_2(x) leq U_1(x). $$ obtained.
lebesgue-integral riemann-integration
lebesgue-integral riemann-integration
edited Jan 1 at 21:41
Namaste
1
edited Jan 1 at 21:41
Namaste
1
edited Jan 1 at 21:41
Namaste
1
edited Jan 1 at 21:41
Namaste
1
edited Jan 1 at 21:41
Namaste
1
1
asked Jan 1 at 21:00
user397197
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For each $k$ there is $n(k)$ such that
$$P_k={x_0<x_1<cdots<x_{n(k)}}$$
However, it is more convenient just to write $n$ instead of $n(k)$.
answered Jan 1 at 21:03
Ben WBen W
2,321615
$endgroup$
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%2f3058872%2ftheorem-11-33-rudin%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For each $k$ there is $n(k)$ such that
$$P_k={x_0<x_1<cdots<x_{n(k)}}$$
However, it is more convenient just to write $n$ instead of $n(k)$.
answered Jan 1 at 21:03
Ben WBen W
2,321615
$endgroup$
add a comment |
$begingroup$
For each $k$ there is $n(k)$ such that
$$P_k={x_0<x_1<cdots<x_{n(k)}}$$
However, it is more convenient just to write $n$ instead of $n(k)$.
answered Jan 1 at 21:03
Ben WBen W
2,321615
$endgroup$
add a comment |
$begingroup$
For each $k$ there is $n(k)$ such that
$$P_k={x_0<x_1<cdots<x_{n(k)}}$$
However, it is more convenient just to write $n$ instead of $n(k)$.
answered Jan 1 at 21:03
Ben WBen W
2,321615
$endgroup$
For each $k$ there is $n(k)$ such that
$$P_k={x_0<x_1<cdots<x_{n(k)}}$$
However, it is more convenient just to write $n$ instead of $n(k)$.
answered Jan 1 at 21:03
Ben WBen W
2,321615
answered Jan 1 at 21:03
Ben WBen W
2,321615
answered Jan 1 at 21:03
Ben WBen W
2,321615
answered Jan 1 at 21:03
Ben WBen W
2,321615
2,321615
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%2f3058872%2ftheorem-11-33-rudin%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
