Mixing
If $E={1/n:ninBbb{N} }$ and $f(x)= 1,xin E,;f(x)=0,xnotin E $, then show that $f$ is integrable and to...
$begingroup$
Let $E={1/n:ninBbb{N} }$ and define a function $f:[0,1]toBbb{R}$ by
begin{align} f(x)=begin{cases} 1,&xin E\0,&xnotin E .end{cases} end{align}
I want to show that $f$ is integrable and to compute $int^{1}_{0}f(x)dx$.
Proof of integrability.
Let $epsilon>0$ and $xin E$. Then, there exists $N$ such that $x=frac{1}{N}<frac{epsilon}{2}.$
Consider a partition $P={x_0,x_1,cdots,x_k}$ of $[0,1]$ such that
$$frac{1}{N}=x_0<x_1<cdots<x_k=1.$$
Take a refinement, $P_0={0,1/N}cup P:={t_0,t_1,cdots,t_m}$ such that $t_m-t_2<frac{1}{N^2}$.
begin{align} U(f,P_0)-L(f,P_0)&=sum^{m}_{j=1}(M_j-m_j)Delta t_j \&=sum^{m}_{j=1}M_jDelta t_j \&=M_1Delta t_1+sum^{m}_{j=2}M_jDelta t_j\&leq 1cdotfrac{1}{N}+Nsum^{m}_{j=2}Delta t_j\&< 1cdotfrac{1}{N}+Ncdotfrac{1}{N^2} \&< frac{epsilon}{2}+frac{epsilon}{2}\&=epsilon end{align}
Hence, $f$ is integrable. So, there a simple way I can compute $int^{1}_{0}f(x)dx$?
real-analysis
edited Jan 30 at 8:17
YuiTo Cheng
2,1863937
asked Jan 30 at 5:59
MichealMicheal
26511
$endgroup$
|
show 3 more comments
$begingroup$
Let $E={1/n:ninBbb{N} }$ and define a function $f:[0,1]toBbb{R}$ by
begin{align} f(x)=begin{cases} 1,&xin E\0,&xnotin E .end{cases} end{align}
I want to show that $f$ is integrable and to compute $int^{1}_{0}f(x)dx$.
Proof of integrability.
Let $epsilon>0$ and $xin E$. Then, there exists $N$ such that $x=frac{1}{N}<frac{epsilon}{2}.$
Consider a partition $P={x_0,x_1,cdots,x_k}$ of $[0,1]$ such that
$$frac{1}{N}=x_0<x_1<cdots<x_k=1.$$
Take a refinement, $P_0={0,1/N}cup P:={t_0,t_1,cdots,t_m}$ such that $t_m-t_2<frac{1}{N^2}$.
begin{align} U(f,P_0)-L(f,P_0)&=sum^{m}_{j=1}(M_j-m_j)Delta t_j \&=sum^{m}_{j=1}M_jDelta t_j \&=M_1Delta t_1+sum^{m}_{j=2}M_jDelta t_j\&leq 1cdotfrac{1}{N}+Nsum^{m}_{j=2}Delta t_j\&< 1cdotfrac{1}{N}+Ncdotfrac{1}{N^2} \&< frac{epsilon}{2}+frac{epsilon}{2}\&=epsilon end{align}
Hence, $f$ is integrable. So, there a simple way I can compute $int^{1}_{0}f(x)dx$?
real-analysis
edited Jan 30 at 8:17
YuiTo Cheng
2,1863937
asked Jan 30 at 5:59
MichealMicheal
26511
$endgroup$
$begingroup$
Your proof seems to reuse "$n$" 2-3 different ways.
$endgroup$
– Michael
Jan 30 at 6:16
$begingroup$
@Michael: Thanks! I've edited it.
$endgroup$
– Micheal
Jan 30 at 6:19
$begingroup$
In what sense did you edit? The first line still says $x = 1/n, n in mathbb{N}$, I assume that means "$x = 1/n$ for some $n in mathbb{N}$." But your next line starts talking about "for all $n in mathbb{N}$." You need to clarify if you are using a particular $n$ that is fixed throughout, or not. (equivalently, is $x$ fixed throughout, or not).
$endgroup$
– Michael
Jan 30 at 6:27
$begingroup$
@Michael: Okay, let me do that.
$endgroup$
– Micheal
Jan 30 at 6:29
$begingroup$
Your function is quite similar to the Dirichlet Function. In particular, though, your function is only $1$ at rational points of the form $1/n$ while Dirichlet's function is $1$ at all rational points. However, like with Dirichlet's function, and other similar functions like Thomae's function, the function is Riemann integrable, with the integral result being $0$ over any range, just like Greg Martin's answer states for your case.
$endgroup$
– John Omielan
Jan 30 at 6:33
|
show 3 more comments
$begingroup$
Let $E={1/n:ninBbb{N} }$ and define a function $f:[0,1]toBbb{R}$ by
begin{align} f(x)=begin{cases} 1,&xin E\0,&xnotin E .end{cases} end{align}
I want to show that $f$ is integrable and to compute $int^{1}_{0}f(x)dx$.
Proof of integrability.
Let $epsilon>0$ and $xin E$. Then, there exists $N$ such that $x=frac{1}{N}<frac{epsilon}{2}.$
Consider a partition $P={x_0,x_1,cdots,x_k}$ of $[0,1]$ such that
$$frac{1}{N}=x_0<x_1<cdots<x_k=1.$$
Take a refinement, $P_0={0,1/N}cup P:={t_0,t_1,cdots,t_m}$ such that $t_m-t_2<frac{1}{N^2}$.
begin{align} U(f,P_0)-L(f,P_0)&=sum^{m}_{j=1}(M_j-m_j)Delta t_j \&=sum^{m}_{j=1}M_jDelta t_j \&=M_1Delta t_1+sum^{m}_{j=2}M_jDelta t_j\&leq 1cdotfrac{1}{N}+Nsum^{m}_{j=2}Delta t_j\&< 1cdotfrac{1}{N}+Ncdotfrac{1}{N^2} \&< frac{epsilon}{2}+frac{epsilon}{2}\&=epsilon end{align}
Hence, $f$ is integrable. So, there a simple way I can compute $int^{1}_{0}f(x)dx$?
real-analysis
edited Jan 30 at 8:17
YuiTo Cheng
2,1863937
asked Jan 30 at 5:59
MichealMicheal
26511
$endgroup$
Let $E={1/n:ninBbb{N} }$ and define a function $f:[0,1]toBbb{R}$ by
begin{align} f(x)=begin{cases} 1,&xin E\0,&xnotin E .end{cases} end{align}
I want to show that $f$ is integrable and to compute $int^{1}_{0}f(x)dx$.
Proof of integrability.
Let $epsilon>0$ and $xin E$. Then, there exists $N$ such that $x=frac{1}{N}<frac{epsilon}{2}.$
Consider a partition $P={x_0,x_1,cdots,x_k}$ of $[0,1]$ such that
$$frac{1}{N}=x_0<x_1<cdots<x_k=1.$$
Take a refinement, $P_0={0,1/N}cup P:={t_0,t_1,cdots,t_m}$ such that $t_m-t_2<frac{1}{N^2}$.
begin{align} U(f,P_0)-L(f,P_0)&=sum^{m}_{j=1}(M_j-m_j)Delta t_j \&=sum^{m}_{j=1}M_jDelta t_j \&=M_1Delta t_1+sum^{m}_{j=2}M_jDelta t_j\&leq 1cdotfrac{1}{N}+Nsum^{m}_{j=2}Delta t_j\&< 1cdotfrac{1}{N}+Ncdotfrac{1}{N^2} \&< frac{epsilon}{2}+frac{epsilon}{2}\&=epsilon end{align}
Hence, $f$ is integrable. So, there a simple way I can compute $int^{1}_{0}f(x)dx$?
real-analysis
real-analysis
edited Jan 30 at 8:17
YuiTo Cheng
2,1863937
asked Jan 30 at 5:59
MichealMicheal
26511
edited Jan 30 at 8:17
YuiTo Cheng
2,1863937
asked Jan 30 at 5:59
MichealMicheal
26511
edited Jan 30 at 8:17
YuiTo Cheng
2,1863937
edited Jan 30 at 8:17
YuiTo Cheng
2,1863937
edited Jan 30 at 8:17
YuiTo Cheng
2,1863937
2,1863937
asked Jan 30 at 5:59
MichealMicheal
26511
asked Jan 30 at 5:59
MichealMicheal
26511
asked Jan 30 at 5:59
MichealMicheal
26511
26511
$begingroup$
Your proof seems to reuse "$n$" 2-3 different ways.
$endgroup$
– Michael
Jan 30 at 6:16
$begingroup$
@Michael: Thanks! I've edited it.
$endgroup$
– Micheal
Jan 30 at 6:19
$begingroup$
In what sense did you edit? The first line still says $x = 1/n, n in mathbb{N}$, I assume that means "$x = 1/n$ for some $n in mathbb{N}$." But your next line starts talking about "for all $n in mathbb{N}$." You need to clarify if you are using a particular $n$ that is fixed throughout, or not. (equivalently, is $x$ fixed throughout, or not).
$endgroup$
– Michael
Jan 30 at 6:27
$begingroup$
@Michael: Okay, let me do that.
$endgroup$
– Micheal
Jan 30 at 6:29
$begingroup$
Your function is quite similar to the Dirichlet Function. In particular, though, your function is only $1$ at rational points of the form $1/n$ while Dirichlet's function is $1$ at all rational points. However, like with Dirichlet's function, and other similar functions like Thomae's function, the function is Riemann integrable, with the integral result being $0$ over any range, just like Greg Martin's answer states for your case.
$endgroup$
– John Omielan
Jan 30 at 6:33
|
show 3 more comments
$begingroup$
Your proof seems to reuse "$n$" 2-3 different ways.
$endgroup$
– Michael
Jan 30 at 6:16
$begingroup$
@Michael: Thanks! I've edited it.
$endgroup$
– Micheal
Jan 30 at 6:19
$begingroup$
In what sense did you edit? The first line still says $x = 1/n, n in mathbb{N}$, I assume that means "$x = 1/n$ for some $n in mathbb{N}$." But your next line starts talking about "for all $n in mathbb{N}$." You need to clarify if you are using a particular $n$ that is fixed throughout, or not. (equivalently, is $x$ fixed throughout, or not).
$endgroup$
– Michael
Jan 30 at 6:27
$begingroup$
@Michael: Okay, let me do that.
$endgroup$
– Micheal
Jan 30 at 6:29
$begingroup$
Your function is quite similar to the Dirichlet Function. In particular, though, your function is only $1$ at rational points of the form $1/n$ while Dirichlet's function is $1$ at all rational points. However, like with Dirichlet's function, and other similar functions like Thomae's function, the function is Riemann integrable, with the integral result being $0$ over any range, just like Greg Martin's answer states for your case.
$endgroup$
– John Omielan
Jan 30 at 6:33
$begingroup$
Your proof seems to reuse "$n$" 2-3 different ways.
$endgroup$
– Michael
Jan 30 at 6:16
$begingroup$
Your proof seems to reuse "$n$" 2-3 different ways.
$endgroup$
– Michael
Jan 30 at 6:16
$begingroup$
@Michael: Thanks! I've edited it.
$endgroup$
– Micheal
Jan 30 at 6:19
$begingroup$
@Michael: Thanks! I've edited it.
$endgroup$
– Micheal
Jan 30 at 6:19
$begingroup$
In what sense did you edit? The first line still says $x = 1/n, n in mathbb{N}$, I assume that means "$x = 1/n$ for some $n in mathbb{N}$." But your next line starts talking about "for all $n in mathbb{N}$." You need to clarify if you are using a particular $n$ that is fixed throughout, or not. (equivalently, is $x$ fixed throughout, or not).
$endgroup$
– Michael
Jan 30 at 6:27
$begingroup$
In what sense did you edit? The first line still says $x = 1/n, n in mathbb{N}$, I assume that means "$x = 1/n$ for some $n in mathbb{N}$." But your next line starts talking about "for all $n in mathbb{N}$." You need to clarify if you are using a particular $n$ that is fixed throughout, or not. (equivalently, is $x$ fixed throughout, or not).
$endgroup$
– Michael
Jan 30 at 6:27
$begingroup$
@Michael: Okay, let me do that.
$endgroup$
– Micheal
Jan 30 at 6:29
$begingroup$
@Michael: Okay, let me do that.
$endgroup$
– Micheal
Jan 30 at 6:29
$begingroup$
Your function is quite similar to the Dirichlet Function. In particular, though, your function is only $1$ at rational points of the form $1/n$ while Dirichlet's function is $1$ at all rational points. However, like with Dirichlet's function, and other similar functions like Thomae's function, the function is Riemann integrable, with the integral result being $0$ over any range, just like Greg Martin's answer states for your case.
$endgroup$
– John Omielan
Jan 30 at 6:33
$begingroup$
Your function is quite similar to the Dirichlet Function. In particular, though, your function is only $1$ at rational points of the form $1/n$ while Dirichlet's function is $1$ at all rational points. However, like with Dirichlet's function, and other similar functions like Thomae's function, the function is Riemann integrable, with the integral result being $0$ over any range, just like Greg Martin's answer states for your case.
$endgroup$
– John Omielan
Jan 30 at 6:33
|
show 3 more comments
1 Answer
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You proved that $0 le U(f,P_0)-L(f,P_0)<varepsilon$ whenever $P_0$ is a partition that is sufficiently fine in terms of $varepsilon$. But it's easy to see that $L(f,P_0) = 0$ for this function. So you really have proved that $0 le U(f,P_0) < varepsilon$, which is enough to conclude that the integral equals $0$.
answered Jan 30 at 6:30
Greg MartinGreg Martin
36.5k23565
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add a comment |
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$begingroup$
You proved that $0 le U(f,P_0)-L(f,P_0)<varepsilon$ whenever $P_0$ is a partition that is sufficiently fine in terms of $varepsilon$. But it's easy to see that $L(f,P_0) = 0$ for this function. So you really have proved that $0 le U(f,P_0) < varepsilon$, which is enough to conclude that the integral equals $0$.
answered Jan 30 at 6:30
Greg MartinGreg Martin
36.5k23565
$endgroup$
add a comment |
$begingroup$
You proved that $0 le U(f,P_0)-L(f,P_0)<varepsilon$ whenever $P_0$ is a partition that is sufficiently fine in terms of $varepsilon$. But it's easy to see that $L(f,P_0) = 0$ for this function. So you really have proved that $0 le U(f,P_0) < varepsilon$, which is enough to conclude that the integral equals $0$.
answered Jan 30 at 6:30
Greg MartinGreg Martin
36.5k23565
$endgroup$
add a comment |
$begingroup$
You proved that $0 le U(f,P_0)-L(f,P_0)<varepsilon$ whenever $P_0$ is a partition that is sufficiently fine in terms of $varepsilon$. But it's easy to see that $L(f,P_0) = 0$ for this function. So you really have proved that $0 le U(f,P_0) < varepsilon$, which is enough to conclude that the integral equals $0$.
answered Jan 30 at 6:30
Greg MartinGreg Martin
36.5k23565
$endgroup$
You proved that $0 le U(f,P_0)-L(f,P_0)<varepsilon$ whenever $P_0$ is a partition that is sufficiently fine in terms of $varepsilon$. But it's easy to see that $L(f,P_0) = 0$ for this function. So you really have proved that $0 le U(f,P_0) < varepsilon$, which is enough to conclude that the integral equals $0$.
answered Jan 30 at 6:30
Greg MartinGreg Martin
36.5k23565
answered Jan 30 at 6:30
Greg MartinGreg Martin
36.5k23565
answered Jan 30 at 6:30
Greg MartinGreg Martin
36.5k23565
answered Jan 30 at 6:30
Greg MartinGreg Martin
36.5k23565
36.5k23565
add a comment |
add a comment |
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$begingroup$
Your proof seems to reuse "$n$" 2-3 different ways.
$endgroup$
– Michael
Jan 30 at 6:16
$begingroup$
@Michael: Thanks! I've edited it.
$endgroup$
– Micheal
Jan 30 at 6:19
$begingroup$
In what sense did you edit? The first line still says $x = 1/n, n in mathbb{N}$, I assume that means "$x = 1/n$ for some $n in mathbb{N}$." But your next line starts talking about "for all $n in mathbb{N}$." You need to clarify if you are using a particular $n$ that is fixed throughout, or not. (equivalently, is $x$ fixed throughout, or not).
$endgroup$
– Michael
Jan 30 at 6:27
$begingroup$
@Michael: Okay, let me do that.
$endgroup$
– Micheal
Jan 30 at 6:29
$begingroup$
Your function is quite similar to the Dirichlet Function. In particular, though, your function is only $1$ at rational points of the form $1/n$ while Dirichlet's function is $1$ at all rational points. However, like with Dirichlet's function, and other similar functions like Thomae's function, the function is Riemann integrable, with the integral result being $0$ over any range, just like Greg Martin's answer states for your case.
$endgroup$
– John Omielan
Jan 30 at 6:33