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Riemann integrability on open subset of $mathbb{R}$
Let $f:mathbb{R}to mathbb{R}$ be a continuous function such that $f$ is Riemann integrable over $mathbb{R}$.
Is it true that $f$ is Riemann integrable over any open subset $Asubset mathbb{R}$??
real-analysis improper-integrals riemann-integration
asked Nov 20 '18 at 19:12
eleguitar
122114
|
show 1 more comment
Let $f:mathbb{R}to mathbb{R}$ be a continuous function such that $f$ is Riemann integrable over $mathbb{R}$.
Is it true that $f$ is Riemann integrable over any open subset $Asubset mathbb{R}$??
real-analysis improper-integrals riemann-integration
asked Nov 20 '18 at 19:12
eleguitar
122114
What do you mean by $f$ is Riemann integrable over $mathbb{R}$? You might find your answer there.
– Jacky Chong
Nov 20 '18 at 19:15
What is your definition of integrable?
– Federico
Nov 20 '18 at 19:15
I mean improper integral as defined here en.m.wikipedia.org/wiki/Improper_integral
– eleguitar
Nov 20 '18 at 19:29
A function which has an improper integral restricts to an integrable function on any closed interval, and thus restricts to an integrable function on a closed interval that contains $A$, but as RRL points out this may not work for open intervals depending on how you define the integral.
– Monstrous Moonshiner
Nov 20 '18 at 19:51
I understand. But It works on open subsets whose boundary have measure zero like $(0,+infty)$, right?
– eleguitar
Nov 20 '18 at 20:02
|
show 1 more comment
Let $f:mathbb{R}to mathbb{R}$ be a continuous function such that $f$ is Riemann integrable over $mathbb{R}$.
Is it true that $f$ is Riemann integrable over any open subset $Asubset mathbb{R}$??
real-analysis improper-integrals riemann-integration
asked Nov 20 '18 at 19:12
eleguitar
122114
Let $f:mathbb{R}to mathbb{R}$ be a continuous function such that $f$ is Riemann integrable over $mathbb{R}$.
Is it true that $f$ is Riemann integrable over any open subset $Asubset mathbb{R}$??
real-analysis improper-integrals riemann-integration
real-analysis improper-integrals riemann-integration
asked Nov 20 '18 at 19:12
eleguitar
122114
asked Nov 20 '18 at 19:12
eleguitar
122114
asked Nov 20 '18 at 19:12
eleguitar
122114
asked Nov 20 '18 at 19:12
eleguitar
122114
asked Nov 20 '18 at 19:12
eleguitar
122114
122114
What do you mean by $f$ is Riemann integrable over $mathbb{R}$? You might find your answer there.
– Jacky Chong
Nov 20 '18 at 19:15
What is your definition of integrable?
– Federico
Nov 20 '18 at 19:15
I mean improper integral as defined here en.m.wikipedia.org/wiki/Improper_integral
– eleguitar
Nov 20 '18 at 19:29
A function which has an improper integral restricts to an integrable function on any closed interval, and thus restricts to an integrable function on a closed interval that contains $A$, but as RRL points out this may not work for open intervals depending on how you define the integral.
– Monstrous Moonshiner
Nov 20 '18 at 19:51
I understand. But It works on open subsets whose boundary have measure zero like $(0,+infty)$, right?
– eleguitar
Nov 20 '18 at 20:02
|
show 1 more comment
What do you mean by $f$ is Riemann integrable over $mathbb{R}$? You might find your answer there.
– Jacky Chong
Nov 20 '18 at 19:15
What is your definition of integrable?
– Federico
Nov 20 '18 at 19:15
I mean improper integral as defined here en.m.wikipedia.org/wiki/Improper_integral
– eleguitar
Nov 20 '18 at 19:29
A function which has an improper integral restricts to an integrable function on any closed interval, and thus restricts to an integrable function on a closed interval that contains $A$, but as RRL points out this may not work for open intervals depending on how you define the integral.
– Monstrous Moonshiner
Nov 20 '18 at 19:51
I understand. But It works on open subsets whose boundary have measure zero like $(0,+infty)$, right?
– eleguitar
Nov 20 '18 at 20:02
What do you mean by $f$ is Riemann integrable over $mathbb{R}$? You might find your answer there.
– Jacky Chong
Nov 20 '18 at 19:15
What do you mean by $f$ is Riemann integrable over $mathbb{R}$? You might find your answer there.
– Jacky Chong
Nov 20 '18 at 19:15
What is your definition of integrable?
– Federico
Nov 20 '18 at 19:15
What is your definition of integrable?
– Federico
Nov 20 '18 at 19:15
I mean improper integral as defined here en.m.wikipedia.org/wiki/Improper_integral
– eleguitar
Nov 20 '18 at 19:29
I mean improper integral as defined here en.m.wikipedia.org/wiki/Improper_integral
– eleguitar
Nov 20 '18 at 19:29
A function which has an improper integral restricts to an integrable function on any closed interval, and thus restricts to an integrable function on a closed interval that contains $A$, but as RRL points out this may not work for open intervals depending on how you define the integral.
– Monstrous Moonshiner
Nov 20 '18 at 19:51
A function which has an improper integral restricts to an integrable function on any closed interval, and thus restricts to an integrable function on a closed interval that contains $A$, but as RRL points out this may not work for open intervals depending on how you define the integral.
– Monstrous Moonshiner
Nov 20 '18 at 19:51
I understand. But It works on open subsets whose boundary have measure zero like $(0,+infty)$, right?
– eleguitar
Nov 20 '18 at 20:02
I understand. But It works on open subsets whose boundary have measure zero like $(0,+infty)$, right?
– eleguitar
Nov 20 '18 at 20:02
|
show 1 more comment
1 Answer
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Assuming you mean Riemann integrable on any closed, bounded interval and you have defined the Riemann integral over an arbitrary open set using the characteristic function, the answer is no.
We can construct an open set $A subset [0,1]$ whose boundary does not have measure zero. Take $f(x) = 1$.
Then the Riemann integral
$$int_A f := int_0^1 f(x) chi_A(x) , dx $$
does not exist
Construction of $A$
Let ${r_i}$ be an enumeration of the rational numbers in $(0,1)$. This is a countable set. We can take intervals such that $r_i in(a_i,b_i) subset (0,1)$ and $sum_{i geqslant1} (b_i - a_i) < 1$. The union
$$A = bigcup_{i geqslant 1}(a_i,b_i)$$
is an open set with boundary $partial A= [0,1] setminus A$ since the rationals are dense in $[0,1]$. However, $m(partial A) = m([0,1]) - m(A) > 1 - sum_{i geqslant 1}(b_i-a_i)> 0$.
answered Nov 20 '18 at 19:25
RRL
49.1k42573
This is interesting, care to elaborate? Namely, how do you construct such an open set, and how do you show that the Riemann integral does not exist?
– Monstrous Moonshiner
Nov 20 '18 at 19:28
1
See Spivak, Calculus on Manifolds, problem 3-11 where a non-rectifiable set in $mathbb{R}$ is constructed. Then the characteristic function is discontinuous on a set of non-zero measure so it is not Riemann integrable.
– RRL
Nov 20 '18 at 19:30
1
In $mathbb{R}^n$ where $n$ may be $1$, Riemann integrals are defined on rectangles (intervals) first and then exptended to arbitrary sets using the characteristic function. If the boundary is not of measure $0$, then the integral will not exist unless the function approaches $0$ at every boundary point -- and this can't happen for the function $f(x) = 1$.
– RRL
Nov 20 '18 at 19:34
2
@MonstrousMoonshiner: That is how an improper Riemann integral over an open set in $A subset mathbb{R}^n$ can be defined. It requires that the function be Riemann integrable over those closed subintervals. But existence of an improper integral does not mean the function-- in this case $chi_A$ -- is Riemann integrable over $A$. I will elaborate in the anser about the construction of this pathological set $A$.
– RRL
Nov 20 '18 at 20:12
2
See also "fat Cantor set".
– RRL
Nov 20 '18 at 20:22
|
show 2 more comments
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Assuming you mean Riemann integrable on any closed, bounded interval and you have defined the Riemann integral over an arbitrary open set using the characteristic function, the answer is no.
We can construct an open set $A subset [0,1]$ whose boundary does not have measure zero. Take $f(x) = 1$.
Then the Riemann integral
$$int_A f := int_0^1 f(x) chi_A(x) , dx $$
does not exist
Construction of $A$
Let ${r_i}$ be an enumeration of the rational numbers in $(0,1)$. This is a countable set. We can take intervals such that $r_i in(a_i,b_i) subset (0,1)$ and $sum_{i geqslant1} (b_i - a_i) < 1$. The union
$$A = bigcup_{i geqslant 1}(a_i,b_i)$$
is an open set with boundary $partial A= [0,1] setminus A$ since the rationals are dense in $[0,1]$. However, $m(partial A) = m([0,1]) - m(A) > 1 - sum_{i geqslant 1}(b_i-a_i)> 0$.
answered Nov 20 '18 at 19:25
RRL
49.1k42573
This is interesting, care to elaborate? Namely, how do you construct such an open set, and how do you show that the Riemann integral does not exist?
– Monstrous Moonshiner
Nov 20 '18 at 19:28
1
See Spivak, Calculus on Manifolds, problem 3-11 where a non-rectifiable set in $mathbb{R}$ is constructed. Then the characteristic function is discontinuous on a set of non-zero measure so it is not Riemann integrable.
– RRL
Nov 20 '18 at 19:30
1
In $mathbb{R}^n$ where $n$ may be $1$, Riemann integrals are defined on rectangles (intervals) first and then exptended to arbitrary sets using the characteristic function. If the boundary is not of measure $0$, then the integral will not exist unless the function approaches $0$ at every boundary point -- and this can't happen for the function $f(x) = 1$.
– RRL
Nov 20 '18 at 19:34
2
@MonstrousMoonshiner: That is how an improper Riemann integral over an open set in $A subset mathbb{R}^n$ can be defined. It requires that the function be Riemann integrable over those closed subintervals. But existence of an improper integral does not mean the function-- in this case $chi_A$ -- is Riemann integrable over $A$. I will elaborate in the anser about the construction of this pathological set $A$.
– RRL
Nov 20 '18 at 20:12
2
See also "fat Cantor set".
– RRL
Nov 20 '18 at 20:22
|
show 2 more comments
Assuming you mean Riemann integrable on any closed, bounded interval and you have defined the Riemann integral over an arbitrary open set using the characteristic function, the answer is no.
We can construct an open set $A subset [0,1]$ whose boundary does not have measure zero. Take $f(x) = 1$.
Then the Riemann integral
$$int_A f := int_0^1 f(x) chi_A(x) , dx $$
does not exist
Construction of $A$
Let ${r_i}$ be an enumeration of the rational numbers in $(0,1)$. This is a countable set. We can take intervals such that $r_i in(a_i,b_i) subset (0,1)$ and $sum_{i geqslant1} (b_i - a_i) < 1$. The union
$$A = bigcup_{i geqslant 1}(a_i,b_i)$$
is an open set with boundary $partial A= [0,1] setminus A$ since the rationals are dense in $[0,1]$. However, $m(partial A) = m([0,1]) - m(A) > 1 - sum_{i geqslant 1}(b_i-a_i)> 0$.
answered Nov 20 '18 at 19:25
RRL
49.1k42573
This is interesting, care to elaborate? Namely, how do you construct such an open set, and how do you show that the Riemann integral does not exist?
– Monstrous Moonshiner
Nov 20 '18 at 19:28
1
See Spivak, Calculus on Manifolds, problem 3-11 where a non-rectifiable set in $mathbb{R}$ is constructed. Then the characteristic function is discontinuous on a set of non-zero measure so it is not Riemann integrable.
– RRL
Nov 20 '18 at 19:30
1
In $mathbb{R}^n$ where $n$ may be $1$, Riemann integrals are defined on rectangles (intervals) first and then exptended to arbitrary sets using the characteristic function. If the boundary is not of measure $0$, then the integral will not exist unless the function approaches $0$ at every boundary point -- and this can't happen for the function $f(x) = 1$.
– RRL
Nov 20 '18 at 19:34
2
@MonstrousMoonshiner: That is how an improper Riemann integral over an open set in $A subset mathbb{R}^n$ can be defined. It requires that the function be Riemann integrable over those closed subintervals. But existence of an improper integral does not mean the function-- in this case $chi_A$ -- is Riemann integrable over $A$. I will elaborate in the anser about the construction of this pathological set $A$.
– RRL
Nov 20 '18 at 20:12
2
See also "fat Cantor set".
– RRL
Nov 20 '18 at 20:22
|
show 2 more comments
Assuming you mean Riemann integrable on any closed, bounded interval and you have defined the Riemann integral over an arbitrary open set using the characteristic function, the answer is no.
We can construct an open set $A subset [0,1]$ whose boundary does not have measure zero. Take $f(x) = 1$.
Then the Riemann integral
$$int_A f := int_0^1 f(x) chi_A(x) , dx $$
does not exist
Construction of $A$
Let ${r_i}$ be an enumeration of the rational numbers in $(0,1)$. This is a countable set. We can take intervals such that $r_i in(a_i,b_i) subset (0,1)$ and $sum_{i geqslant1} (b_i - a_i) < 1$. The union
$$A = bigcup_{i geqslant 1}(a_i,b_i)$$
is an open set with boundary $partial A= [0,1] setminus A$ since the rationals are dense in $[0,1]$. However, $m(partial A) = m([0,1]) - m(A) > 1 - sum_{i geqslant 1}(b_i-a_i)> 0$.
answered Nov 20 '18 at 19:25
RRL
49.1k42573
Assuming you mean Riemann integrable on any closed, bounded interval and you have defined the Riemann integral over an arbitrary open set using the characteristic function, the answer is no.
We can construct an open set $A subset [0,1]$ whose boundary does not have measure zero. Take $f(x) = 1$.
Then the Riemann integral
$$int_A f := int_0^1 f(x) chi_A(x) , dx $$
does not exist
Construction of $A$
Let ${r_i}$ be an enumeration of the rational numbers in $(0,1)$. This is a countable set. We can take intervals such that $r_i in(a_i,b_i) subset (0,1)$ and $sum_{i geqslant1} (b_i - a_i) < 1$. The union
$$A = bigcup_{i geqslant 1}(a_i,b_i)$$
is an open set with boundary $partial A= [0,1] setminus A$ since the rationals are dense in $[0,1]$. However, $m(partial A) = m([0,1]) - m(A) > 1 - sum_{i geqslant 1}(b_i-a_i)> 0$.
answered Nov 20 '18 at 19:25
RRL
49.1k42573
edited Nov 20 '18 at 20:29
answered Nov 20 '18 at 19:25
RRL
49.1k42573
answered Nov 20 '18 at 19:25
RRL
49.1k42573
answered Nov 20 '18 at 19:25
RRL
49.1k42573
49.1k42573
This is interesting, care to elaborate? Namely, how do you construct such an open set, and how do you show that the Riemann integral does not exist?
– Monstrous Moonshiner
Nov 20 '18 at 19:28
1
See Spivak, Calculus on Manifolds, problem 3-11 where a non-rectifiable set in $mathbb{R}$ is constructed. Then the characteristic function is discontinuous on a set of non-zero measure so it is not Riemann integrable.
– RRL
Nov 20 '18 at 19:30
1
In $mathbb{R}^n$ where $n$ may be $1$, Riemann integrals are defined on rectangles (intervals) first and then exptended to arbitrary sets using the characteristic function. If the boundary is not of measure $0$, then the integral will not exist unless the function approaches $0$ at every boundary point -- and this can't happen for the function $f(x) = 1$.
– RRL
Nov 20 '18 at 19:34
2
@MonstrousMoonshiner: That is how an improper Riemann integral over an open set in $A subset mathbb{R}^n$ can be defined. It requires that the function be Riemann integrable over those closed subintervals. But existence of an improper integral does not mean the function-- in this case $chi_A$ -- is Riemann integrable over $A$. I will elaborate in the anser about the construction of this pathological set $A$.
– RRL
Nov 20 '18 at 20:12
2
See also "fat Cantor set".
– RRL
Nov 20 '18 at 20:22
|
show 2 more comments
This is interesting, care to elaborate? Namely, how do you construct such an open set, and how do you show that the Riemann integral does not exist?
– Monstrous Moonshiner
Nov 20 '18 at 19:28
1
See Spivak, Calculus on Manifolds, problem 3-11 where a non-rectifiable set in $mathbb{R}$ is constructed. Then the characteristic function is discontinuous on a set of non-zero measure so it is not Riemann integrable.
– RRL
Nov 20 '18 at 19:30
1
In $mathbb{R}^n$ where $n$ may be $1$, Riemann integrals are defined on rectangles (intervals) first and then exptended to arbitrary sets using the characteristic function. If the boundary is not of measure $0$, then the integral will not exist unless the function approaches $0$ at every boundary point -- and this can't happen for the function $f(x) = 1$.
– RRL
Nov 20 '18 at 19:34
2
@MonstrousMoonshiner: That is how an improper Riemann integral over an open set in $A subset mathbb{R}^n$ can be defined. It requires that the function be Riemann integrable over those closed subintervals. But existence of an improper integral does not mean the function-- in this case $chi_A$ -- is Riemann integrable over $A$. I will elaborate in the anser about the construction of this pathological set $A$.
– RRL
Nov 20 '18 at 20:12
2
See also "fat Cantor set".
– RRL
Nov 20 '18 at 20:22
This is interesting, care to elaborate? Namely, how do you construct such an open set, and how do you show that the Riemann integral does not exist?
– Monstrous Moonshiner
Nov 20 '18 at 19:28
This is interesting, care to elaborate? Namely, how do you construct such an open set, and how do you show that the Riemann integral does not exist?
– Monstrous Moonshiner
Nov 20 '18 at 19:28
1
1
See Spivak, Calculus on Manifolds, problem 3-11 where a non-rectifiable set in $mathbb{R}$ is constructed. Then the characteristic function is discontinuous on a set of non-zero measure so it is not Riemann integrable.
– RRL
Nov 20 '18 at 19:30
See Spivak, Calculus on Manifolds, problem 3-11 where a non-rectifiable set in $mathbb{R}$ is constructed. Then the characteristic function is discontinuous on a set of non-zero measure so it is not Riemann integrable.
– RRL
Nov 20 '18 at 19:30
1
1
In $mathbb{R}^n$ where $n$ may be $1$, Riemann integrals are defined on rectangles (intervals) first and then exptended to arbitrary sets using the characteristic function. If the boundary is not of measure $0$, then the integral will not exist unless the function approaches $0$ at every boundary point -- and this can't happen for the function $f(x) = 1$.
– RRL
Nov 20 '18 at 19:34
In $mathbb{R}^n$ where $n$ may be $1$, Riemann integrals are defined on rectangles (intervals) first and then exptended to arbitrary sets using the characteristic function. If the boundary is not of measure $0$, then the integral will not exist unless the function approaches $0$ at every boundary point -- and this can't happen for the function $f(x) = 1$.
– RRL
Nov 20 '18 at 19:34
2
2
@MonstrousMoonshiner: That is how an improper Riemann integral over an open set in $A subset mathbb{R}^n$ can be defined. It requires that the function be Riemann integrable over those closed subintervals. But existence of an improper integral does not mean the function-- in this case $chi_A$ -- is Riemann integrable over $A$. I will elaborate in the anser about the construction of this pathological set $A$.
– RRL
Nov 20 '18 at 20:12
@MonstrousMoonshiner: That is how an improper Riemann integral over an open set in $A subset mathbb{R}^n$ can be defined. It requires that the function be Riemann integrable over those closed subintervals. But existence of an improper integral does not mean the function-- in this case $chi_A$ -- is Riemann integrable over $A$. I will elaborate in the anser about the construction of this pathological set $A$.
– RRL
Nov 20 '18 at 20:12
2
2
See also "fat Cantor set".
– RRL
Nov 20 '18 at 20:22
See also "fat Cantor set".
– RRL
Nov 20 '18 at 20:22
|
show 2 more comments
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What do you mean by $f$ is Riemann integrable over $mathbb{R}$? You might find your answer there.
– Jacky Chong
Nov 20 '18 at 19:15
What is your definition of integrable?
– Federico
Nov 20 '18 at 19:15
I mean improper integral as defined here en.m.wikipedia.org/wiki/Improper_integral
– eleguitar
Nov 20 '18 at 19:29
A function which has an improper integral restricts to an integrable function on any closed interval, and thus restricts to an integrable function on a closed interval that contains $A$, but as RRL points out this may not work for open intervals depending on how you define the integral.
– Monstrous Moonshiner
Nov 20 '18 at 19:51
I understand. But It works on open subsets whose boundary have measure zero like $(0,+infty)$, right?
– eleguitar
Nov 20 '18 at 20:02