Prove that $f in R(F_s)$ on the interval $[a, b]$ and that $int f dF_s = f(s)$
$begingroup$
Let $a<s<b$ and let $f:[a, b] to mathbb{R}$ be a bounded function that is continous at the point s. Define
$F_s(x) = begin{cases}
0, & text{if $a le x lt s$} \
1, & text{if $s le x le b$}
end{cases}$
Prove that $f in R(F_s)$ on the interval $[a, b]$ and that
$int fdF_s = f(s)$.
My attempt to the proof:
Consider the partition,$P$ of the interval $[a, b]$ where $a=x_o le x_1 le ... le x_{i-1} le x_i le ... le x_n = b$.
Let $M_i = supf(x) and m_i = inf f(x)$ for $x_{i-1} le x le x_i$.
Given $epsilon gt 0$, and $P$. $fin R(F_s)$ if $U(P,f, F_s)-L(P,f,F_s) lt epsilon$.
We know $F(x)$ is continuous at $s in [a,b]$. Let there exist subinterval $[x_{i-1}, x_i]$ in $[a, b]$ such that $s in (x_{i-1}, x_i)$.
Let $F(x_i) - F(x_{i-1}) lt frac{epsilon}{2mM}$. Let $M_i - m_i lt 2M$. Then $sum_{i=0}^{n}(M_i - m_i) lt 2mM$ because there are exactly $m$ terms in this sum, as there are exactly $m$ of the intervals in $[x_{i-1}, x_i]$.
$U(P,f, F_s) - L(P,f, F_s) = (F(x_i)-F(x_{i-1}))sum_{i=0}^{n}(M_i - m_i) lt 2mM frac{epsilon}{2mM} = epsilon$.
Therefore, $f in R(F_s)$.
For the second part of the proof, we need to show: $int fdF_s = f(s)$.
Am I doing this proof right so far? could someone help me with the rest?
Thanks in advance.
proof-verification metric-spaces riemann-integration
asked Dec 6 '18 at 22:17
ISuckAtMathPleaseHELPMEISuckAtMathPleaseHELPME
1549
$endgroup$
add a comment |
$begingroup$
Let $a<s<b$ and let $f:[a, b] to mathbb{R}$ be a bounded function that is continous at the point s. Define
$F_s(x) = begin{cases}
0, & text{if $a le x lt s$} \
1, & text{if $s le x le b$}
end{cases}$
Prove that $f in R(F_s)$ on the interval $[a, b]$ and that
$int fdF_s = f(s)$.
My attempt to the proof:
Consider the partition,$P$ of the interval $[a, b]$ where $a=x_o le x_1 le ... le x_{i-1} le x_i le ... le x_n = b$.
Let $M_i = supf(x) and m_i = inf f(x)$ for $x_{i-1} le x le x_i$.
Given $epsilon gt 0$, and $P$. $fin R(F_s)$ if $U(P,f, F_s)-L(P,f,F_s) lt epsilon$.
We know $F(x)$ is continuous at $s in [a,b]$. Let there exist subinterval $[x_{i-1}, x_i]$ in $[a, b]$ such that $s in (x_{i-1}, x_i)$.
Let $F(x_i) - F(x_{i-1}) lt frac{epsilon}{2mM}$. Let $M_i - m_i lt 2M$. Then $sum_{i=0}^{n}(M_i - m_i) lt 2mM$ because there are exactly $m$ terms in this sum, as there are exactly $m$ of the intervals in $[x_{i-1}, x_i]$.
$U(P,f, F_s) - L(P,f, F_s) = (F(x_i)-F(x_{i-1}))sum_{i=0}^{n}(M_i - m_i) lt 2mM frac{epsilon}{2mM} = epsilon$.
Therefore, $f in R(F_s)$.
For the second part of the proof, we need to show: $int fdF_s = f(s)$.
Am I doing this proof right so far? could someone help me with the rest?
Thanks in advance.
proof-verification metric-spaces riemann-integration
asked Dec 6 '18 at 22:17
ISuckAtMathPleaseHELPMEISuckAtMathPleaseHELPME
1549
$endgroup$
add a comment |
$begingroup$
Let $a<s<b$ and let $f:[a, b] to mathbb{R}$ be a bounded function that is continous at the point s. Define
$F_s(x) = begin{cases}
0, & text{if $a le x lt s$} \
1, & text{if $s le x le b$}
end{cases}$
Prove that $f in R(F_s)$ on the interval $[a, b]$ and that
$int fdF_s = f(s)$.
My attempt to the proof:
Consider the partition,$P$ of the interval $[a, b]$ where $a=x_o le x_1 le ... le x_{i-1} le x_i le ... le x_n = b$.
Let $M_i = supf(x) and m_i = inf f(x)$ for $x_{i-1} le x le x_i$.
Given $epsilon gt 0$, and $P$. $fin R(F_s)$ if $U(P,f, F_s)-L(P,f,F_s) lt epsilon$.
We know $F(x)$ is continuous at $s in [a,b]$. Let there exist subinterval $[x_{i-1}, x_i]$ in $[a, b]$ such that $s in (x_{i-1}, x_i)$.
Let $F(x_i) - F(x_{i-1}) lt frac{epsilon}{2mM}$. Let $M_i - m_i lt 2M$. Then $sum_{i=0}^{n}(M_i - m_i) lt 2mM$ because there are exactly $m$ terms in this sum, as there are exactly $m$ of the intervals in $[x_{i-1}, x_i]$.
$U(P,f, F_s) - L(P,f, F_s) = (F(x_i)-F(x_{i-1}))sum_{i=0}^{n}(M_i - m_i) lt 2mM frac{epsilon}{2mM} = epsilon$.
Therefore, $f in R(F_s)$.
For the second part of the proof, we need to show: $int fdF_s = f(s)$.
Am I doing this proof right so far? could someone help me with the rest?
Thanks in advance.
proof-verification metric-spaces riemann-integration
asked Dec 6 '18 at 22:17
ISuckAtMathPleaseHELPMEISuckAtMathPleaseHELPME
1549
$endgroup$
Let $a<s<b$ and let $f:[a, b] to mathbb{R}$ be a bounded function that is continous at the point s. Define
$F_s(x) = begin{cases}
0, & text{if $a le x lt s$} \
1, & text{if $s le x le b$}
end{cases}$
Prove that $f in R(F_s)$ on the interval $[a, b]$ and that
$int fdF_s = f(s)$.
My attempt to the proof:
Consider the partition,$P$ of the interval $[a, b]$ where $a=x_o le x_1 le ... le x_{i-1} le x_i le ... le x_n = b$.
Let $M_i = supf(x) and m_i = inf f(x)$ for $x_{i-1} le x le x_i$.
Given $epsilon gt 0$, and $P$. $fin R(F_s)$ if $U(P,f, F_s)-L(P,f,F_s) lt epsilon$.
We know $F(x)$ is continuous at $s in [a,b]$. Let there exist subinterval $[x_{i-1}, x_i]$ in $[a, b]$ such that $s in (x_{i-1}, x_i)$.
Let $F(x_i) - F(x_{i-1}) lt frac{epsilon}{2mM}$. Let $M_i - m_i lt 2M$. Then $sum_{i=0}^{n}(M_i - m_i) lt 2mM$ because there are exactly $m$ terms in this sum, as there are exactly $m$ of the intervals in $[x_{i-1}, x_i]$.
$U(P,f, F_s) - L(P,f, F_s) = (F(x_i)-F(x_{i-1}))sum_{i=0}^{n}(M_i - m_i) lt 2mM frac{epsilon}{2mM} = epsilon$.
Therefore, $f in R(F_s)$.
For the second part of the proof, we need to show: $int fdF_s = f(s)$.
Am I doing this proof right so far? could someone help me with the rest?
Thanks in advance.
proof-verification metric-spaces riemann-integration
proof-verification metric-spaces riemann-integration
asked Dec 6 '18 at 22:17
ISuckAtMathPleaseHELPMEISuckAtMathPleaseHELPME
1549
asked Dec 6 '18 at 22:17
ISuckAtMathPleaseHELPMEISuckAtMathPleaseHELPME
1549
edited Dec 6 '18 at 22:34
ISuckAtMathPleaseHELPME
asked Dec 6 '18 at 22:17
ISuckAtMathPleaseHELPMEISuckAtMathPleaseHELPME
1549
asked Dec 6 '18 at 22:17
ISuckAtMathPleaseHELPMEISuckAtMathPleaseHELPME
1549
asked Dec 6 '18 at 22:17
ISuckAtMathPleaseHELPMEISuckAtMathPleaseHELPME
1549
1549
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Everything looks good. You've proven that $f$ is Riemann-Lebesgue integrable with respect to $F_s$. Now pick any given sequence of partitions that allow the integral to converge $Pi_k$ and focus on the partition points just before and after $s$. Their contribution is $f(x_r)(F(x_r)-F(x_{r-1}))=f(x_r)cdot 1$. Now just use continuity of $f$ at $s$ and the definition of a sequence of partitions to conclude the integral converges to $f(s)$.
answered Dec 6 '18 at 23:00
Alex R.Alex R.
24.9k12452
$endgroup$
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%2f3029128%2fprove-that-f-in-rf-s-on-the-interval-a-b-and-that-int-f-df-s-fs%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
$begingroup$
Everything looks good. You've proven that $f$ is Riemann-Lebesgue integrable with respect to $F_s$. Now pick any given sequence of partitions that allow the integral to converge $Pi_k$ and focus on the partition points just before and after $s$. Their contribution is $f(x_r)(F(x_r)-F(x_{r-1}))=f(x_r)cdot 1$. Now just use continuity of $f$ at $s$ and the definition of a sequence of partitions to conclude the integral converges to $f(s)$.
answered Dec 6 '18 at 23:00
Alex R.Alex R.
24.9k12452
$endgroup$
add a comment |
$begingroup$
Everything looks good. You've proven that $f$ is Riemann-Lebesgue integrable with respect to $F_s$. Now pick any given sequence of partitions that allow the integral to converge $Pi_k$ and focus on the partition points just before and after $s$. Their contribution is $f(x_r)(F(x_r)-F(x_{r-1}))=f(x_r)cdot 1$. Now just use continuity of $f$ at $s$ and the definition of a sequence of partitions to conclude the integral converges to $f(s)$.
answered Dec 6 '18 at 23:00
Alex R.Alex R.
24.9k12452
$endgroup$
add a comment |
$begingroup$
Everything looks good. You've proven that $f$ is Riemann-Lebesgue integrable with respect to $F_s$. Now pick any given sequence of partitions that allow the integral to converge $Pi_k$ and focus on the partition points just before and after $s$. Their contribution is $f(x_r)(F(x_r)-F(x_{r-1}))=f(x_r)cdot 1$. Now just use continuity of $f$ at $s$ and the definition of a sequence of partitions to conclude the integral converges to $f(s)$.
answered Dec 6 '18 at 23:00
Alex R.Alex R.
24.9k12452
$endgroup$
Everything looks good. You've proven that $f$ is Riemann-Lebesgue integrable with respect to $F_s$. Now pick any given sequence of partitions that allow the integral to converge $Pi_k$ and focus on the partition points just before and after $s$. Their contribution is $f(x_r)(F(x_r)-F(x_{r-1}))=f(x_r)cdot 1$. Now just use continuity of $f$ at $s$ and the definition of a sequence of partitions to conclude the integral converges to $f(s)$.
answered Dec 6 '18 at 23:00
Alex R.Alex R.
24.9k12452
answered Dec 6 '18 at 23:00
Alex R.Alex R.
24.9k12452
answered Dec 6 '18 at 23:00
Alex R.Alex R.
24.9k12452
answered Dec 6 '18 at 23:00
Alex R.Alex R.
24.9k12452
24.9k12452
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%2f3029128%2fprove-that-f-in-rf-s-on-the-interval-a-b-and-that-int-f-df-s-fs%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
