Riemann-integral for a function that has infinitely many discontinuity points
$begingroup$
The question is following:
Let $f:[0,1]rightarrow mathbb{R}.$
$f(x)=x,$ if $x=1/n, ninmathbb{N}$
$f(x)=0,$ otherwise.
Is $f$ Riemann-integrable? If it is, what is its value?
I know that the basic idea of Riemann-integral is to find two step-functions $h$ and $g$ which both integrate to the same value and $h(x) leq f(x) leq g(x)$. This time I can't find the upper step-function. However I guess that this function is Riemann-integrable as it's discontinuous only on $1/n$. Also I'm guessing that it's value is $0$. Am I correct and how could I proof it?
riemann-integration
asked Jan 22 at 19:21
jtejte
336
$endgroup$
add a comment |
$begingroup$
The question is following:
Let $f:[0,1]rightarrow mathbb{R}.$
$f(x)=x,$ if $x=1/n, ninmathbb{N}$
$f(x)=0,$ otherwise.
Is $f$ Riemann-integrable? If it is, what is its value?
I know that the basic idea of Riemann-integral is to find two step-functions $h$ and $g$ which both integrate to the same value and $h(x) leq f(x) leq g(x)$. This time I can't find the upper step-function. However I guess that this function is Riemann-integrable as it's discontinuous only on $1/n$. Also I'm guessing that it's value is $0$. Am I correct and how could I proof it?
riemann-integration
asked Jan 22 at 19:21
jtejte
336
$endgroup$
$begingroup$
There is a theorem that says that a bounded function is Riemann integrable if and only if the set of its discontinuities has measure zero. In particular, if a bounded function has countably many discontinuities, it is Riemann integrable.
$endgroup$
– uniquesolution
Jan 22 at 19:30
$begingroup$
Okay so this tells me that I'm correct with my guess that this function is Riemann-integrable and it's value is 0. However I'm stuck trying to proof it
$endgroup$
– jte
Jan 22 at 19:31
$begingroup$
There is a different approach that uses more machinery, which I'm not sure is allowed for you. But, there is a theorem that says if $f$ is Riemann integrable, then its Riemann integral equals its Lebesgue integral. Your function satisfies the hypothesis of theorem. Using the Lebesgue integral, it is very easy to calculate that it is 0.
$endgroup$
– Nicholas Roberts
Jan 22 at 20:03
$begingroup$
Sadly I have to use the definition of Riemann's integral for this :( I might check that out if I can't come up with a solution otherwise.
$endgroup$
– jte
Jan 22 at 20:05
$begingroup$
Ok. The theorem I have referenced is Theorem 2.28 in Folland's Real Analysis, page 57. Good luck!
$endgroup$
– Nicholas Roberts
Jan 22 at 20:06
add a comment |
$begingroup$
The question is following:
Let $f:[0,1]rightarrow mathbb{R}.$
$f(x)=x,$ if $x=1/n, ninmathbb{N}$
$f(x)=0,$ otherwise.
Is $f$ Riemann-integrable? If it is, what is its value?
I know that the basic idea of Riemann-integral is to find two step-functions $h$ and $g$ which both integrate to the same value and $h(x) leq f(x) leq g(x)$. This time I can't find the upper step-function. However I guess that this function is Riemann-integrable as it's discontinuous only on $1/n$. Also I'm guessing that it's value is $0$. Am I correct and how could I proof it?
riemann-integration
asked Jan 22 at 19:21
jtejte
336
$endgroup$
The question is following:
Let $f:[0,1]rightarrow mathbb{R}.$
$f(x)=x,$ if $x=1/n, ninmathbb{N}$
$f(x)=0,$ otherwise.
Is $f$ Riemann-integrable? If it is, what is its value?
I know that the basic idea of Riemann-integral is to find two step-functions $h$ and $g$ which both integrate to the same value and $h(x) leq f(x) leq g(x)$. This time I can't find the upper step-function. However I guess that this function is Riemann-integrable as it's discontinuous only on $1/n$. Also I'm guessing that it's value is $0$. Am I correct and how could I proof it?
riemann-integration
riemann-integration
asked Jan 22 at 19:21
jtejte
336
asked Jan 22 at 19:21
jtejte
336
asked Jan 22 at 19:21
jtejte
336
asked Jan 22 at 19:21
jtejte
336
asked Jan 22 at 19:21
jtejte
336
336
$begingroup$
There is a theorem that says that a bounded function is Riemann integrable if and only if the set of its discontinuities has measure zero. In particular, if a bounded function has countably many discontinuities, it is Riemann integrable.
$endgroup$
– uniquesolution
Jan 22 at 19:30
$begingroup$
Okay so this tells me that I'm correct with my guess that this function is Riemann-integrable and it's value is 0. However I'm stuck trying to proof it
$endgroup$
– jte
Jan 22 at 19:31
$begingroup$
There is a different approach that uses more machinery, which I'm not sure is allowed for you. But, there is a theorem that says if $f$ is Riemann integrable, then its Riemann integral equals its Lebesgue integral. Your function satisfies the hypothesis of theorem. Using the Lebesgue integral, it is very easy to calculate that it is 0.
$endgroup$
– Nicholas Roberts
Jan 22 at 20:03
$begingroup$
Sadly I have to use the definition of Riemann's integral for this :( I might check that out if I can't come up with a solution otherwise.
$endgroup$
– jte
Jan 22 at 20:05
$begingroup$
Ok. The theorem I have referenced is Theorem 2.28 in Folland's Real Analysis, page 57. Good luck!
$endgroup$
– Nicholas Roberts
Jan 22 at 20:06
add a comment |
$begingroup$
There is a theorem that says that a bounded function is Riemann integrable if and only if the set of its discontinuities has measure zero. In particular, if a bounded function has countably many discontinuities, it is Riemann integrable.
$endgroup$
– uniquesolution
Jan 22 at 19:30
$begingroup$
Okay so this tells me that I'm correct with my guess that this function is Riemann-integrable and it's value is 0. However I'm stuck trying to proof it
$endgroup$
– jte
Jan 22 at 19:31
$begingroup$
There is a different approach that uses more machinery, which I'm not sure is allowed for you. But, there is a theorem that says if $f$ is Riemann integrable, then its Riemann integral equals its Lebesgue integral. Your function satisfies the hypothesis of theorem. Using the Lebesgue integral, it is very easy to calculate that it is 0.
$endgroup$
– Nicholas Roberts
Jan 22 at 20:03
$begingroup$
Sadly I have to use the definition of Riemann's integral for this :( I might check that out if I can't come up with a solution otherwise.
$endgroup$
– jte
Jan 22 at 20:05
$begingroup$
Ok. The theorem I have referenced is Theorem 2.28 in Folland's Real Analysis, page 57. Good luck!
$endgroup$
– Nicholas Roberts
Jan 22 at 20:06
$begingroup$
There is a theorem that says that a bounded function is Riemann integrable if and only if the set of its discontinuities has measure zero. In particular, if a bounded function has countably many discontinuities, it is Riemann integrable.
$endgroup$
– uniquesolution
Jan 22 at 19:30
$begingroup$
There is a theorem that says that a bounded function is Riemann integrable if and only if the set of its discontinuities has measure zero. In particular, if a bounded function has countably many discontinuities, it is Riemann integrable.
$endgroup$
– uniquesolution
Jan 22 at 19:30
$begingroup$
Okay so this tells me that I'm correct with my guess that this function is Riemann-integrable and it's value is 0. However I'm stuck trying to proof it
$endgroup$
– jte
Jan 22 at 19:31
$begingroup$
Okay so this tells me that I'm correct with my guess that this function is Riemann-integrable and it's value is 0. However I'm stuck trying to proof it
$endgroup$
– jte
Jan 22 at 19:31
$begingroup$
There is a different approach that uses more machinery, which I'm not sure is allowed for you. But, there is a theorem that says if $f$ is Riemann integrable, then its Riemann integral equals its Lebesgue integral. Your function satisfies the hypothesis of theorem. Using the Lebesgue integral, it is very easy to calculate that it is 0.
$endgroup$
– Nicholas Roberts
Jan 22 at 20:03
$begingroup$
There is a different approach that uses more machinery, which I'm not sure is allowed for you. But, there is a theorem that says if $f$ is Riemann integrable, then its Riemann integral equals its Lebesgue integral. Your function satisfies the hypothesis of theorem. Using the Lebesgue integral, it is very easy to calculate that it is 0.
$endgroup$
– Nicholas Roberts
Jan 22 at 20:03
$begingroup$
Sadly I have to use the definition of Riemann's integral for this :( I might check that out if I can't come up with a solution otherwise.
$endgroup$
– jte
Jan 22 at 20:05
$begingroup$
Sadly I have to use the definition of Riemann's integral for this :( I might check that out if I can't come up with a solution otherwise.
$endgroup$
– jte
Jan 22 at 20:05
$begingroup$
Ok. The theorem I have referenced is Theorem 2.28 in Folland's Real Analysis, page 57. Good luck!
$endgroup$
– Nicholas Roberts
Jan 22 at 20:06
$begingroup$
Ok. The theorem I have referenced is Theorem 2.28 in Folland's Real Analysis, page 57. Good luck!
$endgroup$
– Nicholas Roberts
Jan 22 at 20:06
add a comment |
1 Answer
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$begingroup$
To do this from scratch, let $epsilon>0$ and consider the partition $Q$, such that:
$1). epsilon$ is a partition point, so that if $n>1/epsilon, 1/nin [0,epsilon].$ Let $Nge 2$ be the smallest integer for which this happens
and
$2). $ there are $x_{k}in Q,$ such that $x_k<1/k<x_{k+1}$ and $x_{k+1}-x_k<epsilon/2^k$ for $k<N.$
Then, $U(f,Q)=epsilonsum^N_{k=1}frac{1}{k2^k}+frac{epsilon }{N}<epsilon.$ Now clearly $L(f,Q)=0$ so $U(f,Q)-L(f,Q)<epsilon.$
Since $L(f,Q)leunderline int fle overline int fle U(f,Q)$, we conclude that $overline int f- underline int f<epsilon$, and since $epsilon$ is arbitrary, $int f$ exists and is equal to $0.$
answered Jan 22 at 21:21
MatematletaMatematleta
11.4k2920
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add a comment |
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$begingroup$
To do this from scratch, let $epsilon>0$ and consider the partition $Q$, such that:
$1). epsilon$ is a partition point, so that if $n>1/epsilon, 1/nin [0,epsilon].$ Let $Nge 2$ be the smallest integer for which this happens
and
$2). $ there are $x_{k}in Q,$ such that $x_k<1/k<x_{k+1}$ and $x_{k+1}-x_k<epsilon/2^k$ for $k<N.$
Then, $U(f,Q)=epsilonsum^N_{k=1}frac{1}{k2^k}+frac{epsilon }{N}<epsilon.$ Now clearly $L(f,Q)=0$ so $U(f,Q)-L(f,Q)<epsilon.$
Since $L(f,Q)leunderline int fle overline int fle U(f,Q)$, we conclude that $overline int f- underline int f<epsilon$, and since $epsilon$ is arbitrary, $int f$ exists and is equal to $0.$
answered Jan 22 at 21:21
MatematletaMatematleta
11.4k2920
$endgroup$
add a comment |
$begingroup$
To do this from scratch, let $epsilon>0$ and consider the partition $Q$, such that:
$1). epsilon$ is a partition point, so that if $n>1/epsilon, 1/nin [0,epsilon].$ Let $Nge 2$ be the smallest integer for which this happens
and
$2). $ there are $x_{k}in Q,$ such that $x_k<1/k<x_{k+1}$ and $x_{k+1}-x_k<epsilon/2^k$ for $k<N.$
Then, $U(f,Q)=epsilonsum^N_{k=1}frac{1}{k2^k}+frac{epsilon }{N}<epsilon.$ Now clearly $L(f,Q)=0$ so $U(f,Q)-L(f,Q)<epsilon.$
Since $L(f,Q)leunderline int fle overline int fle U(f,Q)$, we conclude that $overline int f- underline int f<epsilon$, and since $epsilon$ is arbitrary, $int f$ exists and is equal to $0.$
answered Jan 22 at 21:21
MatematletaMatematleta
11.4k2920
$endgroup$
add a comment |
$begingroup$
To do this from scratch, let $epsilon>0$ and consider the partition $Q$, such that:
$1). epsilon$ is a partition point, so that if $n>1/epsilon, 1/nin [0,epsilon].$ Let $Nge 2$ be the smallest integer for which this happens
and
$2). $ there are $x_{k}in Q,$ such that $x_k<1/k<x_{k+1}$ and $x_{k+1}-x_k<epsilon/2^k$ for $k<N.$
Then, $U(f,Q)=epsilonsum^N_{k=1}frac{1}{k2^k}+frac{epsilon }{N}<epsilon.$ Now clearly $L(f,Q)=0$ so $U(f,Q)-L(f,Q)<epsilon.$
Since $L(f,Q)leunderline int fle overline int fle U(f,Q)$, we conclude that $overline int f- underline int f<epsilon$, and since $epsilon$ is arbitrary, $int f$ exists and is equal to $0.$
answered Jan 22 at 21:21
MatematletaMatematleta
11.4k2920
$endgroup$
To do this from scratch, let $epsilon>0$ and consider the partition $Q$, such that:
$1). epsilon$ is a partition point, so that if $n>1/epsilon, 1/nin [0,epsilon].$ Let $Nge 2$ be the smallest integer for which this happens
and
$2). $ there are $x_{k}in Q,$ such that $x_k<1/k<x_{k+1}$ and $x_{k+1}-x_k<epsilon/2^k$ for $k<N.$
Then, $U(f,Q)=epsilonsum^N_{k=1}frac{1}{k2^k}+frac{epsilon }{N}<epsilon.$ Now clearly $L(f,Q)=0$ so $U(f,Q)-L(f,Q)<epsilon.$
Since $L(f,Q)leunderline int fle overline int fle U(f,Q)$, we conclude that $overline int f- underline int f<epsilon$, and since $epsilon$ is arbitrary, $int f$ exists and is equal to $0.$
answered Jan 22 at 21:21
MatematletaMatematleta
11.4k2920
edited Jan 22 at 21:36
answered Jan 22 at 21:21
MatematletaMatematleta
11.4k2920
answered Jan 22 at 21:21
MatematletaMatematleta
11.4k2920
answered Jan 22 at 21:21
MatematletaMatematleta
11.4k2920
11.4k2920
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$begingroup$
There is a theorem that says that a bounded function is Riemann integrable if and only if the set of its discontinuities has measure zero. In particular, if a bounded function has countably many discontinuities, it is Riemann integrable.
$endgroup$
– uniquesolution
Jan 22 at 19:30
$begingroup$
Okay so this tells me that I'm correct with my guess that this function is Riemann-integrable and it's value is 0. However I'm stuck trying to proof it
$endgroup$
– jte
Jan 22 at 19:31
$begingroup$
There is a different approach that uses more machinery, which I'm not sure is allowed for you. But, there is a theorem that says if $f$ is Riemann integrable, then its Riemann integral equals its Lebesgue integral. Your function satisfies the hypothesis of theorem. Using the Lebesgue integral, it is very easy to calculate that it is 0.
$endgroup$
– Nicholas Roberts
Jan 22 at 20:03
$begingroup$
Sadly I have to use the definition of Riemann's integral for this :( I might check that out if I can't come up with a solution otherwise.
$endgroup$
– jte
Jan 22 at 20:05
$begingroup$
Ok. The theorem I have referenced is Theorem 2.28 in Folland's Real Analysis, page 57. Good luck!
$endgroup$
– Nicholas Roberts
Jan 22 at 20:06