Showing that $f(x)=1$ if $x=frac{1}{n}$, $0$ otherwise on [0,1] is Riemann Integrable
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I have to show that the following function $f:[0,1]rightarrowmathbb{R}$ is Riemann Integrable:
$$f(x) =
left{
begin{array}{ll}
1 & mbox{if } x = frac{1}{n} \
0 & mbox{otherwise}
end{array}
right.$$
For the upper and lower Riemann sum I am using the following definitions:
$$S_{l}(f,V)=sum^{n}_{j=1}inf_{I(j)}(f)(x_j-x_{j-1})$$
With $I(j)$ denoting the interval $[x_{j-1},x_j$] and $V$ is a partition $V={0,x_1,...,1}$. The upper sum is defined with the supremum. I have shown that for any partition on $[0,1]$ the lower sum is $0$. But now I need to prove that for every $epsilon>0$ there is a partition $V$ such that $S_{u}(f,V)<epsilon$. Completing the proof is easy. I see that any partition on $[0,1]$ will only contain a limited number of points of the set ${frac{1}{n}:ninmathbb{N}}$. But I can't make the proof concrete. Could anybody help me out?
riemann-sum
asked Jun 20 '14 at 13:44
GehaktmolenGehaktmolen
17611
$endgroup$
add a comment |
$begingroup$
I have to show that the following function $f:[0,1]rightarrowmathbb{R}$ is Riemann Integrable:
$$f(x) =
left{
begin{array}{ll}
1 & mbox{if } x = frac{1}{n} \
0 & mbox{otherwise}
end{array}
right.$$
For the upper and lower Riemann sum I am using the following definitions:
$$S_{l}(f,V)=sum^{n}_{j=1}inf_{I(j)}(f)(x_j-x_{j-1})$$
With $I(j)$ denoting the interval $[x_{j-1},x_j$] and $V$ is a partition $V={0,x_1,...,1}$. The upper sum is defined with the supremum. I have shown that for any partition on $[0,1]$ the lower sum is $0$. But now I need to prove that for every $epsilon>0$ there is a partition $V$ such that $S_{u}(f,V)<epsilon$. Completing the proof is easy. I see that any partition on $[0,1]$ will only contain a limited number of points of the set ${frac{1}{n}:ninmathbb{N}}$. But I can't make the proof concrete. Could anybody help me out?
riemann-sum
asked Jun 20 '14 at 13:44
GehaktmolenGehaktmolen
17611
$endgroup$
add a comment |
$begingroup$
I have to show that the following function $f:[0,1]rightarrowmathbb{R}$ is Riemann Integrable:
$$f(x) =
left{
begin{array}{ll}
1 & mbox{if } x = frac{1}{n} \
0 & mbox{otherwise}
end{array}
right.$$
For the upper and lower Riemann sum I am using the following definitions:
$$S_{l}(f,V)=sum^{n}_{j=1}inf_{I(j)}(f)(x_j-x_{j-1})$$
With $I(j)$ denoting the interval $[x_{j-1},x_j$] and $V$ is a partition $V={0,x_1,...,1}$. The upper sum is defined with the supremum. I have shown that for any partition on $[0,1]$ the lower sum is $0$. But now I need to prove that for every $epsilon>0$ there is a partition $V$ such that $S_{u}(f,V)<epsilon$. Completing the proof is easy. I see that any partition on $[0,1]$ will only contain a limited number of points of the set ${frac{1}{n}:ninmathbb{N}}$. But I can't make the proof concrete. Could anybody help me out?
riemann-sum
asked Jun 20 '14 at 13:44
GehaktmolenGehaktmolen
17611
$endgroup$
I have to show that the following function $f:[0,1]rightarrowmathbb{R}$ is Riemann Integrable:
$$f(x) =
left{
begin{array}{ll}
1 & mbox{if } x = frac{1}{n} \
0 & mbox{otherwise}
end{array}
right.$$
For the upper and lower Riemann sum I am using the following definitions:
$$S_{l}(f,V)=sum^{n}_{j=1}inf_{I(j)}(f)(x_j-x_{j-1})$$
With $I(j)$ denoting the interval $[x_{j-1},x_j$] and $V$ is a partition $V={0,x_1,...,1}$. The upper sum is defined with the supremum. I have shown that for any partition on $[0,1]$ the lower sum is $0$. But now I need to prove that for every $epsilon>0$ there is a partition $V$ such that $S_{u}(f,V)<epsilon$. Completing the proof is easy. I see that any partition on $[0,1]$ will only contain a limited number of points of the set ${frac{1}{n}:ninmathbb{N}}$. But I can't make the proof concrete. Could anybody help me out?
riemann-sum
riemann-sum
asked Jun 20 '14 at 13:44
GehaktmolenGehaktmolen
17611
asked Jun 20 '14 at 13:44
GehaktmolenGehaktmolen
17611
asked Jun 20 '14 at 13:44
GehaktmolenGehaktmolen
17611
asked Jun 20 '14 at 13:44
GehaktmolenGehaktmolen
17611
asked Jun 20 '14 at 13:44
GehaktmolenGehaktmolen
17611
17611
add a comment |
add a comment |
2 Answers
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oldest
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$begingroup$
Try the following:
The set $F={xin [0,1]: f(x)>epsilon }$ is finite for every $epsilon>0$. Then you can form a partition such that if an interval contains some $xin F$ then it have no other. Finally you can choose the partition such that the sum of interval who contains some $xin F$ is $<epsilon$. Separate the interval wich cover $F$ and those which don't.
Can you continue from this?
answered Jun 20 '14 at 14:12
YTSYTS
2,260827
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add a comment |
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$ou have the right idea. It's all about the technicalities. One possible partition set-up is as follows.
For every $epsilon>0$, there exists an integer $N_0$ such that $1/N_0<epsilon$. Let $N=max{N_0,5}$.
Partition $[0,1]$ with $V={0,x_1, x_2,ldots,x_{4N-5}}$ where $x_1=frac{1}{2N}$, $x_{4N-5}=1$, $x_{2k+1}-x_{2k}=frac{1}{N^3}$, and $x_{2k}<frac{1}{2N-k}<x_{2k+1}$.
Then, $$S_u(f,V)= 1cdot frac{1}{2N} + frac{1}{N^3} cdot (2N-1)<frac{1}{2N}+frac{2N}{N^3}=frac{1}{2N}+frac{2}{N^2}$$.
Since $Nge 5$, $frac{2}{N^2}<frac{1}{2N}$. Therefore, $S_u(f,V)<frac{1}{N}<epsilon$.
answered Jun 20 '14 at 14:13
BadoeBadoe
385111
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add a comment |
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2 Answers
2
active
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2 Answers
2
active
oldest
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active
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active
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$begingroup$
Try the following:
The set $F={xin [0,1]: f(x)>epsilon }$ is finite for every $epsilon>0$. Then you can form a partition such that if an interval contains some $xin F$ then it have no other. Finally you can choose the partition such that the sum of interval who contains some $xin F$ is $<epsilon$. Separate the interval wich cover $F$ and those which don't.
Can you continue from this?
answered Jun 20 '14 at 14:12
YTSYTS
2,260827
$endgroup$
add a comment |
$begingroup$
Try the following:
The set $F={xin [0,1]: f(x)>epsilon }$ is finite for every $epsilon>0$. Then you can form a partition such that if an interval contains some $xin F$ then it have no other. Finally you can choose the partition such that the sum of interval who contains some $xin F$ is $<epsilon$. Separate the interval wich cover $F$ and those which don't.
Can you continue from this?
answered Jun 20 '14 at 14:12
YTSYTS
2,260827
$endgroup$
add a comment |
$begingroup$
Try the following:
The set $F={xin [0,1]: f(x)>epsilon }$ is finite for every $epsilon>0$. Then you can form a partition such that if an interval contains some $xin F$ then it have no other. Finally you can choose the partition such that the sum of interval who contains some $xin F$ is $<epsilon$. Separate the interval wich cover $F$ and those which don't.
Can you continue from this?
answered Jun 20 '14 at 14:12
YTSYTS
2,260827
$endgroup$
Try the following:
The set $F={xin [0,1]: f(x)>epsilon }$ is finite for every $epsilon>0$. Then you can form a partition such that if an interval contains some $xin F$ then it have no other. Finally you can choose the partition such that the sum of interval who contains some $xin F$ is $<epsilon$. Separate the interval wich cover $F$ and those which don't.
Can you continue from this?
answered Jun 20 '14 at 14:12
YTSYTS
2,260827
answered Jun 20 '14 at 14:12
YTSYTS
2,260827
answered Jun 20 '14 at 14:12
YTSYTS
2,260827
answered Jun 20 '14 at 14:12
YTSYTS
2,260827
2,260827
add a comment |
add a comment |
$begingroup$
$ou have the right idea. It's all about the technicalities. One possible partition set-up is as follows.
For every $epsilon>0$, there exists an integer $N_0$ such that $1/N_0<epsilon$. Let $N=max{N_0,5}$.
Partition $[0,1]$ with $V={0,x_1, x_2,ldots,x_{4N-5}}$ where $x_1=frac{1}{2N}$, $x_{4N-5}=1$, $x_{2k+1}-x_{2k}=frac{1}{N^3}$, and $x_{2k}<frac{1}{2N-k}<x_{2k+1}$.
Then, $$S_u(f,V)= 1cdot frac{1}{2N} + frac{1}{N^3} cdot (2N-1)<frac{1}{2N}+frac{2N}{N^3}=frac{1}{2N}+frac{2}{N^2}$$.
Since $Nge 5$, $frac{2}{N^2}<frac{1}{2N}$. Therefore, $S_u(f,V)<frac{1}{N}<epsilon$.
answered Jun 20 '14 at 14:13
BadoeBadoe
385111
$endgroup$
add a comment |
$begingroup$
$ou have the right idea. It's all about the technicalities. One possible partition set-up is as follows.
For every $epsilon>0$, there exists an integer $N_0$ such that $1/N_0<epsilon$. Let $N=max{N_0,5}$.
Partition $[0,1]$ with $V={0,x_1, x_2,ldots,x_{4N-5}}$ where $x_1=frac{1}{2N}$, $x_{4N-5}=1$, $x_{2k+1}-x_{2k}=frac{1}{N^3}$, and $x_{2k}<frac{1}{2N-k}<x_{2k+1}$.
Then, $$S_u(f,V)= 1cdot frac{1}{2N} + frac{1}{N^3} cdot (2N-1)<frac{1}{2N}+frac{2N}{N^3}=frac{1}{2N}+frac{2}{N^2}$$.
Since $Nge 5$, $frac{2}{N^2}<frac{1}{2N}$. Therefore, $S_u(f,V)<frac{1}{N}<epsilon$.
answered Jun 20 '14 at 14:13
BadoeBadoe
385111
$endgroup$
add a comment |
$begingroup$
$ou have the right idea. It's all about the technicalities. One possible partition set-up is as follows.
For every $epsilon>0$, there exists an integer $N_0$ such that $1/N_0<epsilon$. Let $N=max{N_0,5}$.
Partition $[0,1]$ with $V={0,x_1, x_2,ldots,x_{4N-5}}$ where $x_1=frac{1}{2N}$, $x_{4N-5}=1$, $x_{2k+1}-x_{2k}=frac{1}{N^3}$, and $x_{2k}<frac{1}{2N-k}<x_{2k+1}$.
Then, $$S_u(f,V)= 1cdot frac{1}{2N} + frac{1}{N^3} cdot (2N-1)<frac{1}{2N}+frac{2N}{N^3}=frac{1}{2N}+frac{2}{N^2}$$.
Since $Nge 5$, $frac{2}{N^2}<frac{1}{2N}$. Therefore, $S_u(f,V)<frac{1}{N}<epsilon$.
answered Jun 20 '14 at 14:13
BadoeBadoe
385111
$endgroup$
$ou have the right idea. It's all about the technicalities. One possible partition set-up is as follows.
For every $epsilon>0$, there exists an integer $N_0$ such that $1/N_0<epsilon$. Let $N=max{N_0,5}$.
Partition $[0,1]$ with $V={0,x_1, x_2,ldots,x_{4N-5}}$ where $x_1=frac{1}{2N}$, $x_{4N-5}=1$, $x_{2k+1}-x_{2k}=frac{1}{N^3}$, and $x_{2k}<frac{1}{2N-k}<x_{2k+1}$.
Then, $$S_u(f,V)= 1cdot frac{1}{2N} + frac{1}{N^3} cdot (2N-1)<frac{1}{2N}+frac{2N}{N^3}=frac{1}{2N}+frac{2}{N^2}$$.
Since $Nge 5$, $frac{2}{N^2}<frac{1}{2N}$. Therefore, $S_u(f,V)<frac{1}{N}<epsilon$.
answered Jun 20 '14 at 14:13
BadoeBadoe
385111
answered Jun 20 '14 at 14:13
BadoeBadoe
385111
answered Jun 20 '14 at 14:13
BadoeBadoe
385111
answered Jun 20 '14 at 14:13
BadoeBadoe
385111
385111
add a comment |
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