What kind of integral or analysis is capable of dealing with cantor's function?
$begingroup$
According to Lebesgue’s characterization of Riemann integrable functions (see: http://www.math.ru.nl/~mueger/Lebesgue.pdf)
cantor funciton on $[0,1]$ (https://en.wikipedia.org/wiki/Cantor_function) is riemann integrable, and since it's bounded. They should be equal. However, since $f$ (the cantor function) is singular on $[0,1]$, $f'=0$ almost everywhere. Thus remmann and lebesgue both equal to $0$.
So both lebesgue and riemann integralbe can not deal with cantor's function or Weierstrass function on $[0,1]$(https://en.wikipedia.org/wiki/Weierstrass_function).
My question was what kind of analysis techniques can be used to deal with cantor funciton, or even Weierstrass function?
definition lebesgue-integral
asked Dec 12 '18 at 4:26
user9976437user9976437
759
$endgroup$
add a comment |
$begingroup$
According to Lebesgue’s characterization of Riemann integrable functions (see: http://www.math.ru.nl/~mueger/Lebesgue.pdf)
cantor funciton on $[0,1]$ (https://en.wikipedia.org/wiki/Cantor_function) is riemann integrable, and since it's bounded. They should be equal. However, since $f$ (the cantor function) is singular on $[0,1]$, $f'=0$ almost everywhere. Thus remmann and lebesgue both equal to $0$.
So both lebesgue and riemann integralbe can not deal with cantor's function or Weierstrass function on $[0,1]$(https://en.wikipedia.org/wiki/Weierstrass_function).
My question was what kind of analysis techniques can be used to deal with cantor funciton, or even Weierstrass function?
definition lebesgue-integral
asked Dec 12 '18 at 4:26
user9976437user9976437
759
$endgroup$
1
$begingroup$
Technically, the Cantor function can be differentiated in the sense of distribution theory, but this is in some sense "cheating".
$endgroup$
– Ian
Dec 12 '18 at 4:37
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Are you talking about integral of $f$ or $f'$? Integral of $f$ is not zero.
$endgroup$
– Kavi Rama Murthy
Dec 12 '18 at 5:33
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@KaviRamaMurthy Two cases, 1 $f'$ exists almost everywhere, but obviously "did not work", second case, $f'$ does not exists at all.
$endgroup$
– user9976437
Dec 12 '18 at 5:38
$begingroup$
What do you mean by 'deal with'? The integral of the Cantor function is ${1 over 2}$ not $0$.
$endgroup$
– copper.hat
Dec 12 '18 at 6:45
$begingroup$
@copper.hat They are talking about the failure of (the hypotheses of) the fundamental theorem of calculus. The first part is quite nonsensical however (the Cantor function itself is indeed both Riemann and Lebesgue integrable and the integrals are the same).
$endgroup$
– Ian
Dec 12 '18 at 9:48
add a comment |
$begingroup$
According to Lebesgue’s characterization of Riemann integrable functions (see: http://www.math.ru.nl/~mueger/Lebesgue.pdf)
cantor funciton on $[0,1]$ (https://en.wikipedia.org/wiki/Cantor_function) is riemann integrable, and since it's bounded. They should be equal. However, since $f$ (the cantor function) is singular on $[0,1]$, $f'=0$ almost everywhere. Thus remmann and lebesgue both equal to $0$.
So both lebesgue and riemann integralbe can not deal with cantor's function or Weierstrass function on $[0,1]$(https://en.wikipedia.org/wiki/Weierstrass_function).
My question was what kind of analysis techniques can be used to deal with cantor funciton, or even Weierstrass function?
definition lebesgue-integral
asked Dec 12 '18 at 4:26
user9976437user9976437
759
$endgroup$
According to Lebesgue’s characterization of Riemann integrable functions (see: http://www.math.ru.nl/~mueger/Lebesgue.pdf)
cantor funciton on $[0,1]$ (https://en.wikipedia.org/wiki/Cantor_function) is riemann integrable, and since it's bounded. They should be equal. However, since $f$ (the cantor function) is singular on $[0,1]$, $f'=0$ almost everywhere. Thus remmann and lebesgue both equal to $0$.
So both lebesgue and riemann integralbe can not deal with cantor's function or Weierstrass function on $[0,1]$(https://en.wikipedia.org/wiki/Weierstrass_function).
My question was what kind of analysis techniques can be used to deal with cantor funciton, or even Weierstrass function?
definition lebesgue-integral
definition lebesgue-integral
asked Dec 12 '18 at 4:26
user9976437user9976437
759
asked Dec 12 '18 at 4:26
user9976437user9976437
759
asked Dec 12 '18 at 4:26
user9976437user9976437
759
asked Dec 12 '18 at 4:26
user9976437user9976437
759
asked Dec 12 '18 at 4:26
user9976437user9976437
759
759
1
$begingroup$
Technically, the Cantor function can be differentiated in the sense of distribution theory, but this is in some sense "cheating".
$endgroup$
– Ian
Dec 12 '18 at 4:37
$begingroup$
Are you talking about integral of $f$ or $f'$? Integral of $f$ is not zero.
$endgroup$
– Kavi Rama Murthy
Dec 12 '18 at 5:33
$begingroup$
@KaviRamaMurthy Two cases, 1 $f'$ exists almost everywhere, but obviously "did not work", second case, $f'$ does not exists at all.
$endgroup$
– user9976437
Dec 12 '18 at 5:38
$begingroup$
What do you mean by 'deal with'? The integral of the Cantor function is ${1 over 2}$ not $0$.
$endgroup$
– copper.hat
Dec 12 '18 at 6:45
$begingroup$
@copper.hat They are talking about the failure of (the hypotheses of) the fundamental theorem of calculus. The first part is quite nonsensical however (the Cantor function itself is indeed both Riemann and Lebesgue integrable and the integrals are the same).
$endgroup$
– Ian
Dec 12 '18 at 9:48
add a comment |
1
$begingroup$
Technically, the Cantor function can be differentiated in the sense of distribution theory, but this is in some sense "cheating".
$endgroup$
– Ian
Dec 12 '18 at 4:37
$begingroup$
Are you talking about integral of $f$ or $f'$? Integral of $f$ is not zero.
$endgroup$
– Kavi Rama Murthy
Dec 12 '18 at 5:33
$begingroup$
@KaviRamaMurthy Two cases, 1 $f'$ exists almost everywhere, but obviously "did not work", second case, $f'$ does not exists at all.
$endgroup$
– user9976437
Dec 12 '18 at 5:38
$begingroup$
What do you mean by 'deal with'? The integral of the Cantor function is ${1 over 2}$ not $0$.
$endgroup$
– copper.hat
Dec 12 '18 at 6:45
$begingroup$
@copper.hat They are talking about the failure of (the hypotheses of) the fundamental theorem of calculus. The first part is quite nonsensical however (the Cantor function itself is indeed both Riemann and Lebesgue integrable and the integrals are the same).
$endgroup$
– Ian
Dec 12 '18 at 9:48
1
1
$begingroup$
Technically, the Cantor function can be differentiated in the sense of distribution theory, but this is in some sense "cheating".
$endgroup$
– Ian
Dec 12 '18 at 4:37
$begingroup$
Technically, the Cantor function can be differentiated in the sense of distribution theory, but this is in some sense "cheating".
$endgroup$
– Ian
Dec 12 '18 at 4:37
$begingroup$
Are you talking about integral of $f$ or $f'$? Integral of $f$ is not zero.
$endgroup$
– Kavi Rama Murthy
Dec 12 '18 at 5:33
$begingroup$
Are you talking about integral of $f$ or $f'$? Integral of $f$ is not zero.
$endgroup$
– Kavi Rama Murthy
Dec 12 '18 at 5:33
$begingroup$
@KaviRamaMurthy Two cases, 1 $f'$ exists almost everywhere, but obviously "did not work", second case, $f'$ does not exists at all.
$endgroup$
– user9976437
Dec 12 '18 at 5:38
$begingroup$
@KaviRamaMurthy Two cases, 1 $f'$ exists almost everywhere, but obviously "did not work", second case, $f'$ does not exists at all.
$endgroup$
– user9976437
Dec 12 '18 at 5:38
$begingroup$
What do you mean by 'deal with'? The integral of the Cantor function is ${1 over 2}$ not $0$.
$endgroup$
– copper.hat
Dec 12 '18 at 6:45
$begingroup$
What do you mean by 'deal with'? The integral of the Cantor function is ${1 over 2}$ not $0$.
$endgroup$
– copper.hat
Dec 12 '18 at 6:45
$begingroup$
@copper.hat They are talking about the failure of (the hypotheses of) the fundamental theorem of calculus. The first part is quite nonsensical however (the Cantor function itself is indeed both Riemann and Lebesgue integrable and the integrals are the same).
$endgroup$
– Ian
Dec 12 '18 at 9:48
$begingroup$
@copper.hat They are talking about the failure of (the hypotheses of) the fundamental theorem of calculus. The first part is quite nonsensical however (the Cantor function itself is indeed both Riemann and Lebesgue integrable and the integrals are the same).
$endgroup$
– Ian
Dec 12 '18 at 9:48
add a comment |
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$begingroup$
Technically, the Cantor function can be differentiated in the sense of distribution theory, but this is in some sense "cheating".
$endgroup$
– Ian
Dec 12 '18 at 4:37
$begingroup$
Are you talking about integral of $f$ or $f'$? Integral of $f$ is not zero.
$endgroup$
– Kavi Rama Murthy
Dec 12 '18 at 5:33
$begingroup$
@KaviRamaMurthy Two cases, 1 $f'$ exists almost everywhere, but obviously "did not work", second case, $f'$ does not exists at all.
$endgroup$
– user9976437
Dec 12 '18 at 5:38
$begingroup$
What do you mean by 'deal with'? The integral of the Cantor function is ${1 over 2}$ not $0$.
$endgroup$
– copper.hat
Dec 12 '18 at 6:45
$begingroup$
@copper.hat They are talking about the failure of (the hypotheses of) the fundamental theorem of calculus. The first part is quite nonsensical however (the Cantor function itself is indeed both Riemann and Lebesgue integrable and the integrals are the same).
$endgroup$
– Ian
Dec 12 '18 at 9:48