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Prove that this function is Riemann Integrable on $[0,1]$ [duplicate]
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This question already has an answer here:
Using the definition of Riemann integrability, prove that $f $is Riemann integrable on $[0,1]$
1 answer
Let $E:={frac{1}{n}: nin mathbb{N}}$. Define $f$ on $[0,1]$ by
$f(x)=begin{cases}1 ~~~~~~~~text{if $xin E$}\ 0~~~~~~~~text{if $xnotin E$ }end{cases}$.
Show that f is Riemann integrable.
Here is my attempt.
Let $epsilon >0$, define a partition $P_epsilon$ of $[0,1]$ where $$P_epsilon={0<x_1<x_2<...<1}$$
On each subinterval $[x_{i-1},x_i]$ define the following sets:
$$A= {i:[x_{i-1},x_i]cap E=emptyset}$$
$$B= {i:[x_{i-1},x_i]cap Eneq emptyset}$$
Then $Acap B=emptyset $, for $iin A$, $m_i=M_i=0$ and for $iin B$, $m_i=0$, $M_i=1$
Thus,
begin{split}
U(f,P_epsilon)-L(f,P_epsilon)&= sum_{iin A} (M_i-m_i)Delta x_i+sum_{iin B} (M_i-m_i)Delta x_i\
&=sum_{iin B} 1 times Delta x_i\
end{split}
Here is where I have a hard time in estimating the mesh of $P_epsilon$ to get $U(f,P_epsilon)-L(f,P_epsilon)<epsilon$.
I have seen different approaches to solving this problem but I think this approach seems more understanding.
Can someone please review my approach, modify or give me hints.
real-analysis riemann-integration
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Jan 23 at 21:18
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Using the definition of Riemann integrability, prove that $f $is Riemann integrable on $[0,1]$
1 answer
Let $E:={frac{1}{n}: nin mathbb{N}}$. Define $f$ on $[0,1]$ by
$f(x)=begin{cases}1 ~~~~~~~~text{if $xin E$}\ 0~~~~~~~~text{if $xnotin E$ }end{cases}$.
Show that f is Riemann integrable.
Here is my attempt.
Let $epsilon >0$, define a partition $P_epsilon$ of $[0,1]$ where $$P_epsilon={0<x_1<x_2<...<1}$$
On each subinterval $[x_{i-1},x_i]$ define the following sets:
$$A= {i:[x_{i-1},x_i]cap E=emptyset}$$
$$B= {i:[x_{i-1},x_i]cap Eneq emptyset}$$
Then $Acap B=emptyset $, for $iin A$, $m_i=M_i=0$ and for $iin B$, $m_i=0$, $M_i=1$
Thus,
begin{split}
U(f,P_epsilon)-L(f,P_epsilon)&= sum_{iin A} (M_i-m_i)Delta x_i+sum_{iin B} (M_i-m_i)Delta x_i\
&=sum_{iin B} 1 times Delta x_i\
end{split}
Here is where I have a hard time in estimating the mesh of $P_epsilon$ to get $U(f,P_epsilon)-L(f,P_epsilon)<epsilon$.
I have seen different approaches to solving this problem but I think this approach seems more understanding.
Can someone please review my approach, modify or give me hints.
real-analysis riemann-integration
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marked as duplicate by RRL real-analysis
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Jan 23 at 21:18
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
To evaluate the integral, there is a partition $P_{epsilon,N}$ where $[0,frac{1}{N+1}]$ is one of the subintervals and each point $1, frac{1}{2}, ldots, frac{1}{N}$ belongs to a subinterval of length $epsilon 2^{-(n+1)}$. We then have $$0 leqslant int_0^1f leqslant U(P_{epsilon,N},f) = frac{epsilon}{2} sum_{n=1}^N 2^{-n} + frac{1}{N+1}leqslant frac{epsilon}{2}+ frac{1}{N+1}$$ Now take $N$ so large that the RHS is less than $epsilon$. Since this is true for any $epsilon > 0$ we must have $int_0^1 f = 0$.
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– RRL
Jan 23 at 21:51
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Very well understood @RRL. But could this approach above be modified?
$endgroup$
– user397197
Jan 23 at 21:56
$begingroup$
Well I marked this as duplicate because it has been asked so many times here. But I understand you want some comments on your approach. To make your approach work you just need to add more detail about the partition. It needs to be constructed so that intervals in the set $A$ where $sup f = 1$ can be of arbitrarily small total length. I just showed you how to do that.
$endgroup$
– RRL
Jan 23 at 21:59
$begingroup$
I specified the lengths of intervals in $A$. The other intervals can be anything because there is no contribution to the upper sum.
$endgroup$
– RRL
Jan 23 at 22:01
$begingroup$
Thanks a lot @RRL
$endgroup$
– user397197
Jan 23 at 22:01
add a comment |
$begingroup$
This question already has an answer here:
Using the definition of Riemann integrability, prove that $f $is Riemann integrable on $[0,1]$
1 answer
Let $E:={frac{1}{n}: nin mathbb{N}}$. Define $f$ on $[0,1]$ by
$f(x)=begin{cases}1 ~~~~~~~~text{if $xin E$}\ 0~~~~~~~~text{if $xnotin E$ }end{cases}$.
Show that f is Riemann integrable.
Here is my attempt.
Let $epsilon >0$, define a partition $P_epsilon$ of $[0,1]$ where $$P_epsilon={0<x_1<x_2<...<1}$$
On each subinterval $[x_{i-1},x_i]$ define the following sets:
$$A= {i:[x_{i-1},x_i]cap E=emptyset}$$
$$B= {i:[x_{i-1},x_i]cap Eneq emptyset}$$
Then $Acap B=emptyset $, for $iin A$, $m_i=M_i=0$ and for $iin B$, $m_i=0$, $M_i=1$
Thus,
begin{split}
U(f,P_epsilon)-L(f,P_epsilon)&= sum_{iin A} (M_i-m_i)Delta x_i+sum_{iin B} (M_i-m_i)Delta x_i\
&=sum_{iin B} 1 times Delta x_i\
end{split}
Here is where I have a hard time in estimating the mesh of $P_epsilon$ to get $U(f,P_epsilon)-L(f,P_epsilon)<epsilon$.
I have seen different approaches to solving this problem but I think this approach seems more understanding.
Can someone please review my approach, modify or give me hints.
real-analysis riemann-integration
$endgroup$
This question already has an answer here:
Using the definition of Riemann integrability, prove that $f $is Riemann integrable on $[0,1]$
1 answer
Let $E:={frac{1}{n}: nin mathbb{N}}$. Define $f$ on $[0,1]$ by
$f(x)=begin{cases}1 ~~~~~~~~text{if $xin E$}\ 0~~~~~~~~text{if $xnotin E$ }end{cases}$.
Show that f is Riemann integrable.
Here is my attempt.
Let $epsilon >0$, define a partition $P_epsilon$ of $[0,1]$ where $$P_epsilon={0<x_1<x_2<...<1}$$
On each subinterval $[x_{i-1},x_i]$ define the following sets:
$$A= {i:[x_{i-1},x_i]cap E=emptyset}$$
$$B= {i:[x_{i-1},x_i]cap Eneq emptyset}$$
Then $Acap B=emptyset $, for $iin A$, $m_i=M_i=0$ and for $iin B$, $m_i=0$, $M_i=1$
Thus,
begin{split}
U(f,P_epsilon)-L(f,P_epsilon)&= sum_{iin A} (M_i-m_i)Delta x_i+sum_{iin B} (M_i-m_i)Delta x_i\
&=sum_{iin B} 1 times Delta x_i\
end{split}
Here is where I have a hard time in estimating the mesh of $P_epsilon$ to get $U(f,P_epsilon)-L(f,P_epsilon)<epsilon$.
I have seen different approaches to solving this problem but I think this approach seems more understanding.
Can someone please review my approach, modify or give me hints.
This question already has an answer here:
Using the definition of Riemann integrability, prove that $f $is Riemann integrable on $[0,1]$
1 answer
real-analysis riemann-integration
real-analysis riemann-integration
edited Jan 23 at 21:38
asked Jan 23 at 20:43
user397197
marked as duplicate by RRL real-analysis
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Jan 23 at 21:18
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by RRL real-analysis
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Jan 23 at 21:18
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
To evaluate the integral, there is a partition $P_{epsilon,N}$ where $[0,frac{1}{N+1}]$ is one of the subintervals and each point $1, frac{1}{2}, ldots, frac{1}{N}$ belongs to a subinterval of length $epsilon 2^{-(n+1)}$. We then have $$0 leqslant int_0^1f leqslant U(P_{epsilon,N},f) = frac{epsilon}{2} sum_{n=1}^N 2^{-n} + frac{1}{N+1}leqslant frac{epsilon}{2}+ frac{1}{N+1}$$ Now take $N$ so large that the RHS is less than $epsilon$. Since this is true for any $epsilon > 0$ we must have $int_0^1 f = 0$.
$endgroup$
– RRL
Jan 23 at 21:51
$begingroup$
Very well understood @RRL. But could this approach above be modified?
$endgroup$
– user397197
Jan 23 at 21:56
$begingroup$
Well I marked this as duplicate because it has been asked so many times here. But I understand you want some comments on your approach. To make your approach work you just need to add more detail about the partition. It needs to be constructed so that intervals in the set $A$ where $sup f = 1$ can be of arbitrarily small total length. I just showed you how to do that.
$endgroup$
– RRL
Jan 23 at 21:59
$begingroup$
I specified the lengths of intervals in $A$. The other intervals can be anything because there is no contribution to the upper sum.
$endgroup$
– RRL
Jan 23 at 22:01
$begingroup$
Thanks a lot @RRL
$endgroup$
– user397197
Jan 23 at 22:01
add a comment |
$begingroup$
To evaluate the integral, there is a partition $P_{epsilon,N}$ where $[0,frac{1}{N+1}]$ is one of the subintervals and each point $1, frac{1}{2}, ldots, frac{1}{N}$ belongs to a subinterval of length $epsilon 2^{-(n+1)}$. We then have $$0 leqslant int_0^1f leqslant U(P_{epsilon,N},f) = frac{epsilon}{2} sum_{n=1}^N 2^{-n} + frac{1}{N+1}leqslant frac{epsilon}{2}+ frac{1}{N+1}$$ Now take $N$ so large that the RHS is less than $epsilon$. Since this is true for any $epsilon > 0$ we must have $int_0^1 f = 0$.
$endgroup$
– RRL
Jan 23 at 21:51
$begingroup$
Very well understood @RRL. But could this approach above be modified?
$endgroup$
– user397197
Jan 23 at 21:56
$begingroup$
Well I marked this as duplicate because it has been asked so many times here. But I understand you want some comments on your approach. To make your approach work you just need to add more detail about the partition. It needs to be constructed so that intervals in the set $A$ where $sup f = 1$ can be of arbitrarily small total length. I just showed you how to do that.
$endgroup$
– RRL
Jan 23 at 21:59
$begingroup$
I specified the lengths of intervals in $A$. The other intervals can be anything because there is no contribution to the upper sum.
$endgroup$
– RRL
Jan 23 at 22:01
$begingroup$
Thanks a lot @RRL
$endgroup$
– user397197
Jan 23 at 22:01
$begingroup$
To evaluate the integral, there is a partition $P_{epsilon,N}$ where $[0,frac{1}{N+1}]$ is one of the subintervals and each point $1, frac{1}{2}, ldots, frac{1}{N}$ belongs to a subinterval of length $epsilon 2^{-(n+1)}$. We then have $$0 leqslant int_0^1f leqslant U(P_{epsilon,N},f) = frac{epsilon}{2} sum_{n=1}^N 2^{-n} + frac{1}{N+1}leqslant frac{epsilon}{2}+ frac{1}{N+1}$$ Now take $N$ so large that the RHS is less than $epsilon$. Since this is true for any $epsilon > 0$ we must have $int_0^1 f = 0$.
$endgroup$
– RRL
Jan 23 at 21:51
$begingroup$
To evaluate the integral, there is a partition $P_{epsilon,N}$ where $[0,frac{1}{N+1}]$ is one of the subintervals and each point $1, frac{1}{2}, ldots, frac{1}{N}$ belongs to a subinterval of length $epsilon 2^{-(n+1)}$. We then have $$0 leqslant int_0^1f leqslant U(P_{epsilon,N},f) = frac{epsilon}{2} sum_{n=1}^N 2^{-n} + frac{1}{N+1}leqslant frac{epsilon}{2}+ frac{1}{N+1}$$ Now take $N$ so large that the RHS is less than $epsilon$. Since this is true for any $epsilon > 0$ we must have $int_0^1 f = 0$.
$endgroup$
– RRL
Jan 23 at 21:51
$begingroup$
Very well understood @RRL. But could this approach above be modified?
$endgroup$
– user397197
Jan 23 at 21:56
$begingroup$
Very well understood @RRL. But could this approach above be modified?
$endgroup$
– user397197
Jan 23 at 21:56
$begingroup$
Well I marked this as duplicate because it has been asked so many times here. But I understand you want some comments on your approach. To make your approach work you just need to add more detail about the partition. It needs to be constructed so that intervals in the set $A$ where $sup f = 1$ can be of arbitrarily small total length. I just showed you how to do that.
$endgroup$
– RRL
Jan 23 at 21:59
$begingroup$
Well I marked this as duplicate because it has been asked so many times here. But I understand you want some comments on your approach. To make your approach work you just need to add more detail about the partition. It needs to be constructed so that intervals in the set $A$ where $sup f = 1$ can be of arbitrarily small total length. I just showed you how to do that.
$endgroup$
– RRL
Jan 23 at 21:59
$begingroup$
I specified the lengths of intervals in $A$. The other intervals can be anything because there is no contribution to the upper sum.
$endgroup$
– RRL
Jan 23 at 22:01
$begingroup$
I specified the lengths of intervals in $A$. The other intervals can be anything because there is no contribution to the upper sum.
$endgroup$
– RRL
Jan 23 at 22:01
$begingroup$
Thanks a lot @RRL
$endgroup$
– user397197
Jan 23 at 22:01
$begingroup$
Thanks a lot @RRL
$endgroup$
– user397197
Jan 23 at 22:01
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint:
Let $epsilon>0$ and consider the partition $P$, such that:
$1). epsilon$ is a partition point, so that if $n>1/ϵ, 1/nin [0,epsilon].$ Let $Nge 2$ be the smallest integer for which this happens
and
$2). $ there are $x_kin P$, such that $x_k<1/k<x_{k+1}$ and $x_{k+1}−x_k<epsilon/2k$ for $k<N$.
answered Jan 23 at 21:05
MatematletaMatematleta
11.5k2920
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint:
Let $epsilon>0$ and consider the partition $P$, such that:
$1). epsilon$ is a partition point, so that if $n>1/ϵ, 1/nin [0,epsilon].$ Let $Nge 2$ be the smallest integer for which this happens
and
$2). $ there are $x_kin P$, such that $x_k<1/k<x_{k+1}$ and $x_{k+1}−x_k<epsilon/2k$ for $k<N$.
answered Jan 23 at 21:05
MatematletaMatematleta
11.5k2920
$endgroup$
add a comment |
$begingroup$
Hint:
Let $epsilon>0$ and consider the partition $P$, such that:
$1). epsilon$ is a partition point, so that if $n>1/ϵ, 1/nin [0,epsilon].$ Let $Nge 2$ be the smallest integer for which this happens
and
$2). $ there are $x_kin P$, such that $x_k<1/k<x_{k+1}$ and $x_{k+1}−x_k<epsilon/2k$ for $k<N$.
answered Jan 23 at 21:05
MatematletaMatematleta
11.5k2920
$endgroup$
add a comment |
$begingroup$
Hint:
Let $epsilon>0$ and consider the partition $P$, such that:
$1). epsilon$ is a partition point, so that if $n>1/ϵ, 1/nin [0,epsilon].$ Let $Nge 2$ be the smallest integer for which this happens
and
$2). $ there are $x_kin P$, such that $x_k<1/k<x_{k+1}$ and $x_{k+1}−x_k<epsilon/2k$ for $k<N$.
answered Jan 23 at 21:05
MatematletaMatematleta
11.5k2920
$endgroup$
Hint:
Let $epsilon>0$ and consider the partition $P$, such that:
$1). epsilon$ is a partition point, so that if $n>1/ϵ, 1/nin [0,epsilon].$ Let $Nge 2$ be the smallest integer for which this happens
and
$2). $ there are $x_kin P$, such that $x_k<1/k<x_{k+1}$ and $x_{k+1}−x_k<epsilon/2k$ for $k<N$.
answered Jan 23 at 21:05
MatematletaMatematleta
11.5k2920
answered Jan 23 at 21:05
MatematletaMatematleta
11.5k2920
answered Jan 23 at 21:05
MatematletaMatematleta
11.5k2920
answered Jan 23 at 21:05
MatematletaMatematleta
11.5k2920
11.5k2920
add a comment |
add a comment |
$begingroup$
To evaluate the integral, there is a partition $P_{epsilon,N}$ where $[0,frac{1}{N+1}]$ is one of the subintervals and each point $1, frac{1}{2}, ldots, frac{1}{N}$ belongs to a subinterval of length $epsilon 2^{-(n+1)}$. We then have $$0 leqslant int_0^1f leqslant U(P_{epsilon,N},f) = frac{epsilon}{2} sum_{n=1}^N 2^{-n} + frac{1}{N+1}leqslant frac{epsilon}{2}+ frac{1}{N+1}$$ Now take $N$ so large that the RHS is less than $epsilon$. Since this is true for any $epsilon > 0$ we must have $int_0^1 f = 0$.
$endgroup$
– RRL
Jan 23 at 21:51
$begingroup$
Very well understood @RRL. But could this approach above be modified?
$endgroup$
– user397197
Jan 23 at 21:56
$begingroup$
Well I marked this as duplicate because it has been asked so many times here. But I understand you want some comments on your approach. To make your approach work you just need to add more detail about the partition. It needs to be constructed so that intervals in the set $A$ where $sup f = 1$ can be of arbitrarily small total length. I just showed you how to do that.
$endgroup$
– RRL
Jan 23 at 21:59
$begingroup$
I specified the lengths of intervals in $A$. The other intervals can be anything because there is no contribution to the upper sum.
$endgroup$
– RRL
Jan 23 at 22:01
$begingroup$
Thanks a lot @RRL
$endgroup$
– user397197
Jan 23 at 22:01