Mixing
Sufficient condition for Riemann integrability
$begingroup$
There are many ways to define the Riemann Integral. I am using this one, where I denote $sigma(f,P^{*})$ the Riemann Sum relative to a tagged partition $P^{*}$:
$textbf{Definition}$
We say that a function $f:[a,b] to mathbb{R}$ is Riemann-Integrable if exist the limit:
$I=lim_{||P|| to 0} sigma(f,P^{*})$
and then we write $I=int_{a}^{b}f(x)dx$.
The limit exist in the sense that given $epsilon > 0$, there is $delta > 0$ such that for any partition $P$ of $[a,b]$ with $||P|| < delta$ and for any tagged partition $P^{*}$, we have:
$$|sigma(f,P^{*}) - I| < epsilon$$
By definition, if $f$ is integrable in $[a.b]$, then given $epsilon < 0 $ exists two tagged partition $P^{*}$ and $P^{**}$ such that:
$$|sigma(f,P^{*}) - I| < epsilon / 2$$
$$|sigma(f,P^{**}) - I| < epsilon / 2$$
hence, $|sigma(f,P^{*}) - sigma(f,P^{**})| < epsilon$. Therefore, is a necessary condition for integrability that one can find a partition $P$ such that two Riemann Sums relative to $P$ are very close together, no matter what the scalars that we pick in $P^{*}$ and $P^{**}$.
I am pretty sure that this is a sufficient condition as well. But I have any ideas how to prove it. Can anyone help with this?
riemann-integration
asked Oct 20 '18 at 22:29
fortranfortran
162
$endgroup$
add a comment |
$begingroup$
There are many ways to define the Riemann Integral. I am using this one, where I denote $sigma(f,P^{*})$ the Riemann Sum relative to a tagged partition $P^{*}$:
$textbf{Definition}$
We say that a function $f:[a,b] to mathbb{R}$ is Riemann-Integrable if exist the limit:
$I=lim_{||P|| to 0} sigma(f,P^{*})$
and then we write $I=int_{a}^{b}f(x)dx$.
The limit exist in the sense that given $epsilon > 0$, there is $delta > 0$ such that for any partition $P$ of $[a,b]$ with $||P|| < delta$ and for any tagged partition $P^{*}$, we have:
$$|sigma(f,P^{*}) - I| < epsilon$$
By definition, if $f$ is integrable in $[a.b]$, then given $epsilon < 0 $ exists two tagged partition $P^{*}$ and $P^{**}$ such that:
$$|sigma(f,P^{*}) - I| < epsilon / 2$$
$$|sigma(f,P^{**}) - I| < epsilon / 2$$
hence, $|sigma(f,P^{*}) - sigma(f,P^{**})| < epsilon$. Therefore, is a necessary condition for integrability that one can find a partition $P$ such that two Riemann Sums relative to $P$ are very close together, no matter what the scalars that we pick in $P^{*}$ and $P^{**}$.
I am pretty sure that this is a sufficient condition as well. But I have any ideas how to prove it. Can anyone help with this?
riemann-integration
asked Oct 20 '18 at 22:29
fortranfortran
162
$endgroup$
$begingroup$
It seems valid since one can choose a Riemann sum arbitrarily close to a Darboux sum.
$endgroup$
– Tony Piccolo
Nov 24 '18 at 6:58
add a comment |
$begingroup$
There are many ways to define the Riemann Integral. I am using this one, where I denote $sigma(f,P^{*})$ the Riemann Sum relative to a tagged partition $P^{*}$:
$textbf{Definition}$
We say that a function $f:[a,b] to mathbb{R}$ is Riemann-Integrable if exist the limit:
$I=lim_{||P|| to 0} sigma(f,P^{*})$
and then we write $I=int_{a}^{b}f(x)dx$.
The limit exist in the sense that given $epsilon > 0$, there is $delta > 0$ such that for any partition $P$ of $[a,b]$ with $||P|| < delta$ and for any tagged partition $P^{*}$, we have:
$$|sigma(f,P^{*}) - I| < epsilon$$
By definition, if $f$ is integrable in $[a.b]$, then given $epsilon < 0 $ exists two tagged partition $P^{*}$ and $P^{**}$ such that:
$$|sigma(f,P^{*}) - I| < epsilon / 2$$
$$|sigma(f,P^{**}) - I| < epsilon / 2$$
hence, $|sigma(f,P^{*}) - sigma(f,P^{**})| < epsilon$. Therefore, is a necessary condition for integrability that one can find a partition $P$ such that two Riemann Sums relative to $P$ are very close together, no matter what the scalars that we pick in $P^{*}$ and $P^{**}$.
I am pretty sure that this is a sufficient condition as well. But I have any ideas how to prove it. Can anyone help with this?
riemann-integration
asked Oct 20 '18 at 22:29
fortranfortran
162
$endgroup$
There are many ways to define the Riemann Integral. I am using this one, where I denote $sigma(f,P^{*})$ the Riemann Sum relative to a tagged partition $P^{*}$:
$textbf{Definition}$
We say that a function $f:[a,b] to mathbb{R}$ is Riemann-Integrable if exist the limit:
$I=lim_{||P|| to 0} sigma(f,P^{*})$
and then we write $I=int_{a}^{b}f(x)dx$.
The limit exist in the sense that given $epsilon > 0$, there is $delta > 0$ such that for any partition $P$ of $[a,b]$ with $||P|| < delta$ and for any tagged partition $P^{*}$, we have:
$$|sigma(f,P^{*}) - I| < epsilon$$
By definition, if $f$ is integrable in $[a.b]$, then given $epsilon < 0 $ exists two tagged partition $P^{*}$ and $P^{**}$ such that:
$$|sigma(f,P^{*}) - I| < epsilon / 2$$
$$|sigma(f,P^{**}) - I| < epsilon / 2$$
hence, $|sigma(f,P^{*}) - sigma(f,P^{**})| < epsilon$. Therefore, is a necessary condition for integrability that one can find a partition $P$ such that two Riemann Sums relative to $P$ are very close together, no matter what the scalars that we pick in $P^{*}$ and $P^{**}$.
I am pretty sure that this is a sufficient condition as well. But I have any ideas how to prove it. Can anyone help with this?
riemann-integration
riemann-integration
asked Oct 20 '18 at 22:29
fortranfortran
162
asked Oct 20 '18 at 22:29
fortranfortran
162
asked Oct 20 '18 at 22:29
fortranfortran
162
asked Oct 20 '18 at 22:29
fortranfortran
162
asked Oct 20 '18 at 22:29
fortranfortran
162
162
$begingroup$
It seems valid since one can choose a Riemann sum arbitrarily close to a Darboux sum.
$endgroup$
– Tony Piccolo
Nov 24 '18 at 6:58
add a comment |
$begingroup$
It seems valid since one can choose a Riemann sum arbitrarily close to a Darboux sum.
$endgroup$
– Tony Piccolo
Nov 24 '18 at 6:58
$begingroup$
It seems valid since one can choose a Riemann sum arbitrarily close to a Darboux sum.
$endgroup$
– Tony Piccolo
Nov 24 '18 at 6:58
$begingroup$
It seems valid since one can choose a Riemann sum arbitrarily close to a Darboux sum.
$endgroup$
– Tony Piccolo
Nov 24 '18 at 6:58
add a comment |
2 Answers
2
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oldest
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$begingroup$
There is such a sufficient (Cauchy) condition. The correct statement is:
Suppose for any $epsilon > 0$ there exists $delta > 0$ such that
$|sigma(f,P) - sigma(f,P')| < epsilon$ for all partitions $P$ and
$P'$ with $|P|, , |P'| < delta$ and for any choice of tags.
Then $f$ is Riemann integrable.
In proving this one must first show the existence of a viable candidate for the value of the integral and then show that it satisfies the required definition.
First -- and I will leave this to you for now -- construct a decreasing sequence of positive numbers $delta_n$ and partitions $P_n$ with $|P_n| < delta_n$ such that for any partition $P$ with $|P| < delta_n$ we have (for any choice of tags)
$$|sigma(f,P) - sigma(f,P_n)| < 1/n$$
Hence, if $m geqslant n$ then $|sigma(f,P_m) - sigma(f,P_n)| < 1/n$. The sequence $sigma(f,P_n)$ is a Cauchy sequence and must converge to a real number $I$.
To show that $I$ satisfies the definition of the integral $int_a^b f(x) , dx$, for any $epsilon >0$ take $n$ such that $1/n < epsilon/2$ and sufficiently large such that $|sigma(f,P_n) - I| < epsilon/2$. If $P$ is a partition with $|P| < delta_n$, then it follows that
$$|sigma(f,P) - I| leqslant |sigma(f,P) - sigma(f,P_n)| + |sigma(f,P_n) - I| < 1/n + epsilon/2 < epsilon $$
answered Oct 20 '18 at 22:44
RRLRRL
51.2k42573
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add a comment |
$begingroup$
Your claim is founded.
For convenence I rewrite the statement.
Let $f:[a,b] to mathbb{R}$ be a bounded function.
If for every $varepsilon>0$ there exists a partition $P$ of $[a,b]$ such that $|sigma (f,P^*)- sigma (f,P^{**})|<varepsilon$ for any Riemann sums $sigma (f,P^*)$ and $sigma (f,P^{**})$ for $f$ corresponding to $P$, then there exists $I=lim_{|P| to 0} sigma (f,P^*)$, the limit being independent of the tags of $P$.
If we choose the tags conveniently using the definitions of sup and inf, we get $$S(f,P)-s(f,P)<varepsilon$$ for some $P$.
So the arbitrariness of $,varepsilon,$ gives $$S(f)=s(f)$$ That is the number $I$ we are searching for since $$S(f)=lim_{|P| to 0} S(f,P)$$ $$s(f)=lim_{|P| to 0} s(f,P)$$ and $$s(f,P) le sigma (f,P^*) le S(f,P)$$ for every $P$ however tagged.
answered Jan 13 at 18:42
Tony PiccoloTony Piccolo
3,2152719
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2 Answers
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2 Answers
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$begingroup$
There is such a sufficient (Cauchy) condition. The correct statement is:
Suppose for any $epsilon > 0$ there exists $delta > 0$ such that
$|sigma(f,P) - sigma(f,P')| < epsilon$ for all partitions $P$ and
$P'$ with $|P|, , |P'| < delta$ and for any choice of tags.
Then $f$ is Riemann integrable.
In proving this one must first show the existence of a viable candidate for the value of the integral and then show that it satisfies the required definition.
First -- and I will leave this to you for now -- construct a decreasing sequence of positive numbers $delta_n$ and partitions $P_n$ with $|P_n| < delta_n$ such that for any partition $P$ with $|P| < delta_n$ we have (for any choice of tags)
$$|sigma(f,P) - sigma(f,P_n)| < 1/n$$
Hence, if $m geqslant n$ then $|sigma(f,P_m) - sigma(f,P_n)| < 1/n$. The sequence $sigma(f,P_n)$ is a Cauchy sequence and must converge to a real number $I$.
To show that $I$ satisfies the definition of the integral $int_a^b f(x) , dx$, for any $epsilon >0$ take $n$ such that $1/n < epsilon/2$ and sufficiently large such that $|sigma(f,P_n) - I| < epsilon/2$. If $P$ is a partition with $|P| < delta_n$, then it follows that
$$|sigma(f,P) - I| leqslant |sigma(f,P) - sigma(f,P_n)| + |sigma(f,P_n) - I| < 1/n + epsilon/2 < epsilon $$
answered Oct 20 '18 at 22:44
RRLRRL
51.2k42573
$endgroup$
add a comment |
$begingroup$
There is such a sufficient (Cauchy) condition. The correct statement is:
Suppose for any $epsilon > 0$ there exists $delta > 0$ such that
$|sigma(f,P) - sigma(f,P')| < epsilon$ for all partitions $P$ and
$P'$ with $|P|, , |P'| < delta$ and for any choice of tags.
Then $f$ is Riemann integrable.
In proving this one must first show the existence of a viable candidate for the value of the integral and then show that it satisfies the required definition.
First -- and I will leave this to you for now -- construct a decreasing sequence of positive numbers $delta_n$ and partitions $P_n$ with $|P_n| < delta_n$ such that for any partition $P$ with $|P| < delta_n$ we have (for any choice of tags)
$$|sigma(f,P) - sigma(f,P_n)| < 1/n$$
Hence, if $m geqslant n$ then $|sigma(f,P_m) - sigma(f,P_n)| < 1/n$. The sequence $sigma(f,P_n)$ is a Cauchy sequence and must converge to a real number $I$.
To show that $I$ satisfies the definition of the integral $int_a^b f(x) , dx$, for any $epsilon >0$ take $n$ such that $1/n < epsilon/2$ and sufficiently large such that $|sigma(f,P_n) - I| < epsilon/2$. If $P$ is a partition with $|P| < delta_n$, then it follows that
$$|sigma(f,P) - I| leqslant |sigma(f,P) - sigma(f,P_n)| + |sigma(f,P_n) - I| < 1/n + epsilon/2 < epsilon $$
answered Oct 20 '18 at 22:44
RRLRRL
51.2k42573
$endgroup$
add a comment |
$begingroup$
There is such a sufficient (Cauchy) condition. The correct statement is:
Suppose for any $epsilon > 0$ there exists $delta > 0$ such that
$|sigma(f,P) - sigma(f,P')| < epsilon$ for all partitions $P$ and
$P'$ with $|P|, , |P'| < delta$ and for any choice of tags.
Then $f$ is Riemann integrable.
In proving this one must first show the existence of a viable candidate for the value of the integral and then show that it satisfies the required definition.
First -- and I will leave this to you for now -- construct a decreasing sequence of positive numbers $delta_n$ and partitions $P_n$ with $|P_n| < delta_n$ such that for any partition $P$ with $|P| < delta_n$ we have (for any choice of tags)
$$|sigma(f,P) - sigma(f,P_n)| < 1/n$$
Hence, if $m geqslant n$ then $|sigma(f,P_m) - sigma(f,P_n)| < 1/n$. The sequence $sigma(f,P_n)$ is a Cauchy sequence and must converge to a real number $I$.
To show that $I$ satisfies the definition of the integral $int_a^b f(x) , dx$, for any $epsilon >0$ take $n$ such that $1/n < epsilon/2$ and sufficiently large such that $|sigma(f,P_n) - I| < epsilon/2$. If $P$ is a partition with $|P| < delta_n$, then it follows that
$$|sigma(f,P) - I| leqslant |sigma(f,P) - sigma(f,P_n)| + |sigma(f,P_n) - I| < 1/n + epsilon/2 < epsilon $$
answered Oct 20 '18 at 22:44
RRLRRL
51.2k42573
$endgroup$
There is such a sufficient (Cauchy) condition. The correct statement is:
Suppose for any $epsilon > 0$ there exists $delta > 0$ such that
$|sigma(f,P) - sigma(f,P')| < epsilon$ for all partitions $P$ and
$P'$ with $|P|, , |P'| < delta$ and for any choice of tags.
Then $f$ is Riemann integrable.
In proving this one must first show the existence of a viable candidate for the value of the integral and then show that it satisfies the required definition.
First -- and I will leave this to you for now -- construct a decreasing sequence of positive numbers $delta_n$ and partitions $P_n$ with $|P_n| < delta_n$ such that for any partition $P$ with $|P| < delta_n$ we have (for any choice of tags)
$$|sigma(f,P) - sigma(f,P_n)| < 1/n$$
Hence, if $m geqslant n$ then $|sigma(f,P_m) - sigma(f,P_n)| < 1/n$. The sequence $sigma(f,P_n)$ is a Cauchy sequence and must converge to a real number $I$.
To show that $I$ satisfies the definition of the integral $int_a^b f(x) , dx$, for any $epsilon >0$ take $n$ such that $1/n < epsilon/2$ and sufficiently large such that $|sigma(f,P_n) - I| < epsilon/2$. If $P$ is a partition with $|P| < delta_n$, then it follows that
$$|sigma(f,P) - I| leqslant |sigma(f,P) - sigma(f,P_n)| + |sigma(f,P_n) - I| < 1/n + epsilon/2 < epsilon $$
answered Oct 20 '18 at 22:44
RRLRRL
51.2k42573
edited Oct 20 '18 at 23:52
answered Oct 20 '18 at 22:44
RRLRRL
51.2k42573
answered Oct 20 '18 at 22:44
RRLRRL
51.2k42573
answered Oct 20 '18 at 22:44
RRLRRL
51.2k42573
51.2k42573
add a comment |
add a comment |
$begingroup$
Your claim is founded.
For convenence I rewrite the statement.
Let $f:[a,b] to mathbb{R}$ be a bounded function.
If for every $varepsilon>0$ there exists a partition $P$ of $[a,b]$ such that $|sigma (f,P^*)- sigma (f,P^{**})|<varepsilon$ for any Riemann sums $sigma (f,P^*)$ and $sigma (f,P^{**})$ for $f$ corresponding to $P$, then there exists $I=lim_{|P| to 0} sigma (f,P^*)$, the limit being independent of the tags of $P$.
If we choose the tags conveniently using the definitions of sup and inf, we get $$S(f,P)-s(f,P)<varepsilon$$ for some $P$.
So the arbitrariness of $,varepsilon,$ gives $$S(f)=s(f)$$ That is the number $I$ we are searching for since $$S(f)=lim_{|P| to 0} S(f,P)$$ $$s(f)=lim_{|P| to 0} s(f,P)$$ and $$s(f,P) le sigma (f,P^*) le S(f,P)$$ for every $P$ however tagged.
answered Jan 13 at 18:42
Tony PiccoloTony Piccolo
3,2152719
$endgroup$
add a comment |
$begingroup$
Your claim is founded.
For convenence I rewrite the statement.
Let $f:[a,b] to mathbb{R}$ be a bounded function.
If for every $varepsilon>0$ there exists a partition $P$ of $[a,b]$ such that $|sigma (f,P^*)- sigma (f,P^{**})|<varepsilon$ for any Riemann sums $sigma (f,P^*)$ and $sigma (f,P^{**})$ for $f$ corresponding to $P$, then there exists $I=lim_{|P| to 0} sigma (f,P^*)$, the limit being independent of the tags of $P$.
If we choose the tags conveniently using the definitions of sup and inf, we get $$S(f,P)-s(f,P)<varepsilon$$ for some $P$.
So the arbitrariness of $,varepsilon,$ gives $$S(f)=s(f)$$ That is the number $I$ we are searching for since $$S(f)=lim_{|P| to 0} S(f,P)$$ $$s(f)=lim_{|P| to 0} s(f,P)$$ and $$s(f,P) le sigma (f,P^*) le S(f,P)$$ for every $P$ however tagged.
answered Jan 13 at 18:42
Tony PiccoloTony Piccolo
3,2152719
$endgroup$
add a comment |
$begingroup$
Your claim is founded.
For convenence I rewrite the statement.
Let $f:[a,b] to mathbb{R}$ be a bounded function.
If for every $varepsilon>0$ there exists a partition $P$ of $[a,b]$ such that $|sigma (f,P^*)- sigma (f,P^{**})|<varepsilon$ for any Riemann sums $sigma (f,P^*)$ and $sigma (f,P^{**})$ for $f$ corresponding to $P$, then there exists $I=lim_{|P| to 0} sigma (f,P^*)$, the limit being independent of the tags of $P$.
If we choose the tags conveniently using the definitions of sup and inf, we get $$S(f,P)-s(f,P)<varepsilon$$ for some $P$.
So the arbitrariness of $,varepsilon,$ gives $$S(f)=s(f)$$ That is the number $I$ we are searching for since $$S(f)=lim_{|P| to 0} S(f,P)$$ $$s(f)=lim_{|P| to 0} s(f,P)$$ and $$s(f,P) le sigma (f,P^*) le S(f,P)$$ for every $P$ however tagged.
answered Jan 13 at 18:42
Tony PiccoloTony Piccolo
3,2152719
$endgroup$
Your claim is founded.
For convenence I rewrite the statement.
Let $f:[a,b] to mathbb{R}$ be a bounded function.
If for every $varepsilon>0$ there exists a partition $P$ of $[a,b]$ such that $|sigma (f,P^*)- sigma (f,P^{**})|<varepsilon$ for any Riemann sums $sigma (f,P^*)$ and $sigma (f,P^{**})$ for $f$ corresponding to $P$, then there exists $I=lim_{|P| to 0} sigma (f,P^*)$, the limit being independent of the tags of $P$.
If we choose the tags conveniently using the definitions of sup and inf, we get $$S(f,P)-s(f,P)<varepsilon$$ for some $P$.
So the arbitrariness of $,varepsilon,$ gives $$S(f)=s(f)$$ That is the number $I$ we are searching for since $$S(f)=lim_{|P| to 0} S(f,P)$$ $$s(f)=lim_{|P| to 0} s(f,P)$$ and $$s(f,P) le sigma (f,P^*) le S(f,P)$$ for every $P$ however tagged.
answered Jan 13 at 18:42
Tony PiccoloTony Piccolo
3,2152719
edited Jan 14 at 17:25
answered Jan 13 at 18:42
Tony PiccoloTony Piccolo
3,2152719
answered Jan 13 at 18:42
Tony PiccoloTony Piccolo
3,2152719
answered Jan 13 at 18:42
Tony PiccoloTony Piccolo
3,2152719
3,2152719
add a comment |
add a comment |
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$begingroup$
It seems valid since one can choose a Riemann sum arbitrarily close to a Darboux sum.
$endgroup$
– Tony Piccolo
Nov 24 '18 at 6:58