Conditions for applying the second fundamental theorem of calculus with gauge integrals
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I was thinking about this question while walking home today and can't seem to prove or come up with a counterexample myself.
Let $f:[a,b]rightarrowmathbb{R}$ be a continuous function, $f(x),$ differentiable everywhere on $(a,b)$, except for one point $xi$ in $(a,b)$ where $f'(xi)$ does not exist. Since $f$ is continuous everywhere and differentiable everywhere on $(a,b)$ except at $xi$, $f'$ is gauge integrable on $[a,b]$. Am I correct that $int_a^bf'(x);mathrm{d}x=f(b)-f(a)$ despite that nasty point $xi$ in $(a,b)$? And could this value be interpreted as the "signed" area under the curve from high school calculus?
real-analysis gauge-integral
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add a comment |
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I was thinking about this question while walking home today and can't seem to prove or come up with a counterexample myself.
Let $f:[a,b]rightarrowmathbb{R}$ be a continuous function, $f(x),$ differentiable everywhere on $(a,b)$, except for one point $xi$ in $(a,b)$ where $f'(xi)$ does not exist. Since $f$ is continuous everywhere and differentiable everywhere on $(a,b)$ except at $xi$, $f'$ is gauge integrable on $[a,b]$. Am I correct that $int_a^bf'(x);mathrm{d}x=f(b)-f(a)$ despite that nasty point $xi$ in $(a,b)$? And could this value be interpreted as the "signed" area under the curve from high school calculus?
real-analysis gauge-integral
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What makes you thing $f'$ is integrable?
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– Kavi Rama Murthy
Dec 13 '18 at 23:52
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I thought that $f'$ is gauge integrable because $f$ is differentiable everywhere except at a countable number of points?
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– user626213
Dec 14 '18 at 0:42
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I do not know what 'gauge integrable' means but I think $f'$ need not be integrable in any sense. Note that $f'$ need not be bounded.
$endgroup$
– Kavi Rama Murthy
Dec 14 '18 at 5:32
add a comment |
$begingroup$
I was thinking about this question while walking home today and can't seem to prove or come up with a counterexample myself.
Let $f:[a,b]rightarrowmathbb{R}$ be a continuous function, $f(x),$ differentiable everywhere on $(a,b)$, except for one point $xi$ in $(a,b)$ where $f'(xi)$ does not exist. Since $f$ is continuous everywhere and differentiable everywhere on $(a,b)$ except at $xi$, $f'$ is gauge integrable on $[a,b]$. Am I correct that $int_a^bf'(x);mathrm{d}x=f(b)-f(a)$ despite that nasty point $xi$ in $(a,b)$? And could this value be interpreted as the "signed" area under the curve from high school calculus?
real-analysis gauge-integral
$endgroup$
I was thinking about this question while walking home today and can't seem to prove or come up with a counterexample myself.
Let $f:[a,b]rightarrowmathbb{R}$ be a continuous function, $f(x),$ differentiable everywhere on $(a,b)$, except for one point $xi$ in $(a,b)$ where $f'(xi)$ does not exist. Since $f$ is continuous everywhere and differentiable everywhere on $(a,b)$ except at $xi$, $f'$ is gauge integrable on $[a,b]$. Am I correct that $int_a^bf'(x);mathrm{d}x=f(b)-f(a)$ despite that nasty point $xi$ in $(a,b)$? And could this value be interpreted as the "signed" area under the curve from high school calculus?
real-analysis gauge-integral
real-analysis gauge-integral
asked Dec 13 '18 at 21:54
user626213
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What makes you thing $f'$ is integrable?
$endgroup$
– Kavi Rama Murthy
Dec 13 '18 at 23:52
$begingroup$
I thought that $f'$ is gauge integrable because $f$ is differentiable everywhere except at a countable number of points?
$endgroup$
– user626213
Dec 14 '18 at 0:42
$begingroup$
I do not know what 'gauge integrable' means but I think $f'$ need not be integrable in any sense. Note that $f'$ need not be bounded.
$endgroup$
– Kavi Rama Murthy
Dec 14 '18 at 5:32
add a comment |
$begingroup$
What makes you thing $f'$ is integrable?
$endgroup$
– Kavi Rama Murthy
Dec 13 '18 at 23:52
$begingroup$
I thought that $f'$ is gauge integrable because $f$ is differentiable everywhere except at a countable number of points?
$endgroup$
– user626213
Dec 14 '18 at 0:42
$begingroup$
I do not know what 'gauge integrable' means but I think $f'$ need not be integrable in any sense. Note that $f'$ need not be bounded.
$endgroup$
– Kavi Rama Murthy
Dec 14 '18 at 5:32
$begingroup$
What makes you thing $f'$ is integrable?
$endgroup$
– Kavi Rama Murthy
Dec 13 '18 at 23:52
$begingroup$
What makes you thing $f'$ is integrable?
$endgroup$
– Kavi Rama Murthy
Dec 13 '18 at 23:52
$begingroup$
I thought that $f'$ is gauge integrable because $f$ is differentiable everywhere except at a countable number of points?
$endgroup$
– user626213
Dec 14 '18 at 0:42
$begingroup$
I thought that $f'$ is gauge integrable because $f$ is differentiable everywhere except at a countable number of points?
$endgroup$
– user626213
Dec 14 '18 at 0:42
$begingroup$
I do not know what 'gauge integrable' means but I think $f'$ need not be integrable in any sense. Note that $f'$ need not be bounded.
$endgroup$
– Kavi Rama Murthy
Dec 14 '18 at 5:32
$begingroup$
I do not know what 'gauge integrable' means but I think $f'$ need not be integrable in any sense. Note that $f'$ need not be bounded.
$endgroup$
– Kavi Rama Murthy
Dec 14 '18 at 5:32
add a comment |
1 Answer
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If $f(x)=xsin (frac 1 x)$ ($f(0)=0$) then $f$ satisfies your hypothesis but $f$ does not satisfy the Fundamental Theorem of Calculus . If it did satisfy this on each interval contained in $[-1,1]$ the it would be absolutely continuous and hence of bounded variation on $[-1,1]$ but it is not of bounded variation.
answered Dec 13 '18 at 23:57
Kavi Rama MurthyKavi Rama Murthy
72.9k53170
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add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $f(x)=xsin (frac 1 x)$ ($f(0)=0$) then $f$ satisfies your hypothesis but $f$ does not satisfy the Fundamental Theorem of Calculus . If it did satisfy this on each interval contained in $[-1,1]$ the it would be absolutely continuous and hence of bounded variation on $[-1,1]$ but it is not of bounded variation.
answered Dec 13 '18 at 23:57
Kavi Rama MurthyKavi Rama Murthy
72.9k53170
$endgroup$
add a comment |
$begingroup$
If $f(x)=xsin (frac 1 x)$ ($f(0)=0$) then $f$ satisfies your hypothesis but $f$ does not satisfy the Fundamental Theorem of Calculus . If it did satisfy this on each interval contained in $[-1,1]$ the it would be absolutely continuous and hence of bounded variation on $[-1,1]$ but it is not of bounded variation.
answered Dec 13 '18 at 23:57
Kavi Rama MurthyKavi Rama Murthy
72.9k53170
$endgroup$
add a comment |
$begingroup$
If $f(x)=xsin (frac 1 x)$ ($f(0)=0$) then $f$ satisfies your hypothesis but $f$ does not satisfy the Fundamental Theorem of Calculus . If it did satisfy this on each interval contained in $[-1,1]$ the it would be absolutely continuous and hence of bounded variation on $[-1,1]$ but it is not of bounded variation.
answered Dec 13 '18 at 23:57
Kavi Rama MurthyKavi Rama Murthy
72.9k53170
$endgroup$
If $f(x)=xsin (frac 1 x)$ ($f(0)=0$) then $f$ satisfies your hypothesis but $f$ does not satisfy the Fundamental Theorem of Calculus . If it did satisfy this on each interval contained in $[-1,1]$ the it would be absolutely continuous and hence of bounded variation on $[-1,1]$ but it is not of bounded variation.
answered Dec 13 '18 at 23:57
Kavi Rama MurthyKavi Rama Murthy
72.9k53170
answered Dec 13 '18 at 23:57
Kavi Rama MurthyKavi Rama Murthy
72.9k53170
answered Dec 13 '18 at 23:57
Kavi Rama MurthyKavi Rama Murthy
72.9k53170
answered Dec 13 '18 at 23:57
Kavi Rama MurthyKavi Rama Murthy
72.9k53170
72.9k53170
add a comment |
add a comment |
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$begingroup$
What makes you thing $f'$ is integrable?
$endgroup$
– Kavi Rama Murthy
Dec 13 '18 at 23:52
$begingroup$
I thought that $f'$ is gauge integrable because $f$ is differentiable everywhere except at a countable number of points?
$endgroup$
– user626213
Dec 14 '18 at 0:42
$begingroup$
I do not know what 'gauge integrable' means but I think $f'$ need not be integrable in any sense. Note that $f'$ need not be bounded.
$endgroup$
– Kavi Rama Murthy
Dec 14 '18 at 5:32