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Riemann integral on a single point
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Let $f$ be a continuous function over $mathbbR$. Let $cin(-infty,infty)$. Then, it is believed that $int_c^cf(xi)dxi=0$ in the sense of Riemann integration. Is this just a definition? But the definition of the Riemann integral on wikipedia or on text book always start with a closed interval [a,b] with $b>a$ and then partition the interval. So, the definition of Riemann integral does not cover the case $a=b$. So, $int_c^cf(xi)dxi=0$ seems to be a property of a Riemann integral inferred by its definition. Can anyone tell me if $int_c^cf(xi)dxi=0$ is a definition or not and show me some proof?
integration
asked Jul 30 at 5:24
Lin Xuelei
305
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up vote
2
down vote
favorite
Let $f$ be a continuous function over $mathbbR$. Let $cin(-infty,infty)$. Then, it is believed that $int_c^cf(xi)dxi=0$ in the sense of Riemann integration. Is this just a definition? But the definition of the Riemann integral on wikipedia or on text book always start with a closed interval [a,b] with $b>a$ and then partition the interval. So, the definition of Riemann integral does not cover the case $a=b$. So, $int_c^cf(xi)dxi=0$ seems to be a property of a Riemann integral inferred by its definition. Can anyone tell me if $int_c^cf(xi)dxi=0$ is a definition or not and show me some proof?
integration
asked Jul 30 at 5:24
Lin Xuelei
305
1
Let $I$ a real interval and $f$ a function being Riemann integrable on $I$, then for all $a le b le c in I$ We have : $int_a^c f(x) dx = int_a^b f(x) dx + int_b^c f(x) dx$.
– nicomezi
Jul 30 at 6:56
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $f$ be a continuous function over $mathbbR$. Let $cin(-infty,infty)$. Then, it is believed that $int_c^cf(xi)dxi=0$ in the sense of Riemann integration. Is this just a definition? But the definition of the Riemann integral on wikipedia or on text book always start with a closed interval [a,b] with $b>a$ and then partition the interval. So, the definition of Riemann integral does not cover the case $a=b$. So, $int_c^cf(xi)dxi=0$ seems to be a property of a Riemann integral inferred by its definition. Can anyone tell me if $int_c^cf(xi)dxi=0$ is a definition or not and show me some proof?
integration
asked Jul 30 at 5:24
Lin Xuelei
305
Let $f$ be a continuous function over $mathbbR$. Let $cin(-infty,infty)$. Then, it is believed that $int_c^cf(xi)dxi=0$ in the sense of Riemann integration. Is this just a definition? But the definition of the Riemann integral on wikipedia or on text book always start with a closed interval [a,b] with $b>a$ and then partition the interval. So, the definition of Riemann integral does not cover the case $a=b$. So, $int_c^cf(xi)dxi=0$ seems to be a property of a Riemann integral inferred by its definition. Can anyone tell me if $int_c^cf(xi)dxi=0$ is a definition or not and show me some proof?
integration
asked Jul 30 at 5:24
Lin Xuelei
305
asked Jul 30 at 5:24
Lin Xuelei
305
asked Jul 30 at 5:24
Lin Xuelei
305
asked Jul 30 at 5:24
Lin Xuelei
305
305
1
Let $I$ a real interval and $f$ a function being Riemann integrable on $I$, then for all $a le b le c in I$ We have : $int_a^c f(x) dx = int_a^b f(x) dx + int_b^c f(x) dx$.
– nicomezi
Jul 30 at 6:56
add a comment |Â
1
Let $I$ a real interval and $f$ a function being Riemann integrable on $I$, then for all $a le b le c in I$ We have : $int_a^c f(x) dx = int_a^b f(x) dx + int_b^c f(x) dx$.
– nicomezi
Jul 30 at 6:56
1
1
Let $I$ a real interval and $f$ a function being Riemann integrable on $I$, then for all $a le b le c in I$ We have : $int_a^c f(x) dx = int_a^b f(x) dx + int_b^c f(x) dx$.
– nicomezi
Jul 30 at 6:56
Let $I$ a real interval and $f$ a function being Riemann integrable on $I$, then for all $a le b le c in I$ We have : $int_a^c f(x) dx = int_a^b f(x) dx + int_b^c f(x) dx$.
– nicomezi
Jul 30 at 6:56
add a comment |Â
3 Answers
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1
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Yes, $int_c^cf(xi)dxi=0$ is a definition. As you pointed out, the standard definition of the Riemann integral does not cover this case. However, using the Riemann-Stieltjes integral we can define integration on a single point, in which case the equality $int_c^cf(xi)dxi=0$ becomes a theorem.
answered Jul 30 at 5:35
apanpapan3
1038
add a comment |Â
up vote
12
down vote
It does not have to be a definition. It can follow from the definition of the Riemann-Darboux integral.
If $f$ is bounded and by a partition of $c$ we mean a single degenerate interval $[c,c]$ with length $c-c = 0$, then for (vacuously any) lower and upper sum we have
$$L(P,f) = inf_x in cf(x) (c-c) = 0,$$ $$U(P,f)= sup_x in cf(x) (c-c) =0,$$
so
$$sup_P L(P,f) = inf_PU(P,f) =int_c^c f(x) ,dx = 0$$
Similarly, the Lebesgue integral $int_E f = 0$ for $E = c$ or any other zero-measure set. This is not simply an arbitrary definition -- it can be proved.
Alternate Proof
We can also "prove" this using the change-of variable theorem for Riemann integrals. Take $g:x in [a,b] mapsto c$. With $f$ continuous on $g([a,b]) = c$ and $g' = 0$ we have
$$int_c ^c f(x) ,dx = int_g(a)^g(b) f(x) , dx = int_a^b f(g(t))g'(t) , dt = int_a^b f(c) cdot 0 , dt = 0$$
This is less of a proof and more of a consistency check showing that if the integral over $[c,c]$ is purely by definition then it cannot be arbitrary.
answered Jul 30 at 5:54
RRL
43.4k42260
3
This is correct, however the point in the question is that whatever definition we want to use (and propositions therefrom) one must assume the the sums are calculated on partitions $[a,b]$ such that $a <b$. If we drop this assumption one might as well just defined it as zero and that's it.
– gented
Jul 30 at 8:59
Thanks for your detailed answer and for Gented's explain. Yes, now I think $int_c^cf(xi)dxi$ can be regarded as a definition. Since, in RRL's proof, the definition of Riemann integration is more general and robust than the one provided on wikipedia.
– Lin Xuelei
Jul 31 at 10:17
add a comment |Â
up vote
0
down vote
According to Apostol's Calculus Vol. 1 (second edition) page 74, it is a definition that $int_c^cf(x)dx=0$.
If the integral on a single point hasn't been defined, one cannot prove what its value is since it's meaningless.
Note:
It is reasonable to define $int_c^cf(x)dx=0$ because it is clear that the area of the line segment between the point $(c, 0)$ and $(c, f(c))$ is zero. Moreover, if we want to define it so that the theorem of additivity
$$ int_a^bf(x)dx+int_b^cf(x)dx=int_a^cf(x)dx$$
remains valid, then we must have
$$ int_a^cf(x)dx+int_c^cf(x)dx=int_a^cf(x)dx,$$
so $int_c^cf(x)dx=0$.
answered Jul 30 at 5:44
user529760
507216
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Yes, $int_c^cf(xi)dxi=0$ is a definition. As you pointed out, the standard definition of the Riemann integral does not cover this case. However, using the Riemann-Stieltjes integral we can define integration on a single point, in which case the equality $int_c^cf(xi)dxi=0$ becomes a theorem.
answered Jul 30 at 5:35
apanpapan3
1038
add a comment |Â
up vote
1
down vote
accepted
Yes, $int_c^cf(xi)dxi=0$ is a definition. As you pointed out, the standard definition of the Riemann integral does not cover this case. However, using the Riemann-Stieltjes integral we can define integration on a single point, in which case the equality $int_c^cf(xi)dxi=0$ becomes a theorem.
answered Jul 30 at 5:35
apanpapan3
1038
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Yes, $int_c^cf(xi)dxi=0$ is a definition. As you pointed out, the standard definition of the Riemann integral does not cover this case. However, using the Riemann-Stieltjes integral we can define integration on a single point, in which case the equality $int_c^cf(xi)dxi=0$ becomes a theorem.
answered Jul 30 at 5:35
apanpapan3
1038
Yes, $int_c^cf(xi)dxi=0$ is a definition. As you pointed out, the standard definition of the Riemann integral does not cover this case. However, using the Riemann-Stieltjes integral we can define integration on a single point, in which case the equality $int_c^cf(xi)dxi=0$ becomes a theorem.
answered Jul 30 at 5:35
apanpapan3
1038
answered Jul 30 at 5:35
apanpapan3
1038
answered Jul 30 at 5:35
apanpapan3
1038
answered Jul 30 at 5:35
apanpapan3
1038
1038
add a comment |Â
add a comment |Â
up vote
12
down vote
It does not have to be a definition. It can follow from the definition of the Riemann-Darboux integral.
If $f$ is bounded and by a partition of $c$ we mean a single degenerate interval $[c,c]$ with length $c-c = 0$, then for (vacuously any) lower and upper sum we have
$$L(P,f) = inf_x in cf(x) (c-c) = 0,$$ $$U(P,f)= sup_x in cf(x) (c-c) =0,$$
so
$$sup_P L(P,f) = inf_PU(P,f) =int_c^c f(x) ,dx = 0$$
Similarly, the Lebesgue integral $int_E f = 0$ for $E = c$ or any other zero-measure set. This is not simply an arbitrary definition -- it can be proved.
Alternate Proof
We can also "prove" this using the change-of variable theorem for Riemann integrals. Take $g:x in [a,b] mapsto c$. With $f$ continuous on $g([a,b]) = c$ and $g' = 0$ we have
$$int_c ^c f(x) ,dx = int_g(a)^g(b) f(x) , dx = int_a^b f(g(t))g'(t) , dt = int_a^b f(c) cdot 0 , dt = 0$$
This is less of a proof and more of a consistency check showing that if the integral over $[c,c]$ is purely by definition then it cannot be arbitrary.
answered Jul 30 at 5:54
RRL
43.4k42260
3
This is correct, however the point in the question is that whatever definition we want to use (and propositions therefrom) one must assume the the sums are calculated on partitions $[a,b]$ such that $a <b$. If we drop this assumption one might as well just defined it as zero and that's it.
– gented
Jul 30 at 8:59
Thanks for your detailed answer and for Gented's explain. Yes, now I think $int_c^cf(xi)dxi$ can be regarded as a definition. Since, in RRL's proof, the definition of Riemann integration is more general and robust than the one provided on wikipedia.
– Lin Xuelei
Jul 31 at 10:17
add a comment |Â
up vote
12
down vote
It does not have to be a definition. It can follow from the definition of the Riemann-Darboux integral.
If $f$ is bounded and by a partition of $c$ we mean a single degenerate interval $[c,c]$ with length $c-c = 0$, then for (vacuously any) lower and upper sum we have
$$L(P,f) = inf_x in cf(x) (c-c) = 0,$$ $$U(P,f)= sup_x in cf(x) (c-c) =0,$$
so
$$sup_P L(P,f) = inf_PU(P,f) =int_c^c f(x) ,dx = 0$$
Similarly, the Lebesgue integral $int_E f = 0$ for $E = c$ or any other zero-measure set. This is not simply an arbitrary definition -- it can be proved.
Alternate Proof
We can also "prove" this using the change-of variable theorem for Riemann integrals. Take $g:x in [a,b] mapsto c$. With $f$ continuous on $g([a,b]) = c$ and $g' = 0$ we have
$$int_c ^c f(x) ,dx = int_g(a)^g(b) f(x) , dx = int_a^b f(g(t))g'(t) , dt = int_a^b f(c) cdot 0 , dt = 0$$
This is less of a proof and more of a consistency check showing that if the integral over $[c,c]$ is purely by definition then it cannot be arbitrary.
answered Jul 30 at 5:54
RRL
43.4k42260
3
This is correct, however the point in the question is that whatever definition we want to use (and propositions therefrom) one must assume the the sums are calculated on partitions $[a,b]$ such that $a <b$. If we drop this assumption one might as well just defined it as zero and that's it.
– gented
Jul 30 at 8:59
Thanks for your detailed answer and for Gented's explain. Yes, now I think $int_c^cf(xi)dxi$ can be regarded as a definition. Since, in RRL's proof, the definition of Riemann integration is more general and robust than the one provided on wikipedia.
– Lin Xuelei
Jul 31 at 10:17
add a comment |Â
up vote
12
down vote
up vote
12
down vote
It does not have to be a definition. It can follow from the definition of the Riemann-Darboux integral.
If $f$ is bounded and by a partition of $c$ we mean a single degenerate interval $[c,c]$ with length $c-c = 0$, then for (vacuously any) lower and upper sum we have
$$L(P,f) = inf_x in cf(x) (c-c) = 0,$$ $$U(P,f)= sup_x in cf(x) (c-c) =0,$$
so
$$sup_P L(P,f) = inf_PU(P,f) =int_c^c f(x) ,dx = 0$$
Similarly, the Lebesgue integral $int_E f = 0$ for $E = c$ or any other zero-measure set. This is not simply an arbitrary definition -- it can be proved.
Alternate Proof
We can also "prove" this using the change-of variable theorem for Riemann integrals. Take $g:x in [a,b] mapsto c$. With $f$ continuous on $g([a,b]) = c$ and $g' = 0$ we have
$$int_c ^c f(x) ,dx = int_g(a)^g(b) f(x) , dx = int_a^b f(g(t))g'(t) , dt = int_a^b f(c) cdot 0 , dt = 0$$
This is less of a proof and more of a consistency check showing that if the integral over $[c,c]$ is purely by definition then it cannot be arbitrary.
answered Jul 30 at 5:54
RRL
43.4k42260
It does not have to be a definition. It can follow from the definition of the Riemann-Darboux integral.
If $f$ is bounded and by a partition of $c$ we mean a single degenerate interval $[c,c]$ with length $c-c = 0$, then for (vacuously any) lower and upper sum we have
$$L(P,f) = inf_x in cf(x) (c-c) = 0,$$ $$U(P,f)= sup_x in cf(x) (c-c) =0,$$
so
$$sup_P L(P,f) = inf_PU(P,f) =int_c^c f(x) ,dx = 0$$
Similarly, the Lebesgue integral $int_E f = 0$ for $E = c$ or any other zero-measure set. This is not simply an arbitrary definition -- it can be proved.
Alternate Proof
We can also "prove" this using the change-of variable theorem for Riemann integrals. Take $g:x in [a,b] mapsto c$. With $f$ continuous on $g([a,b]) = c$ and $g' = 0$ we have
$$int_c ^c f(x) ,dx = int_g(a)^g(b) f(x) , dx = int_a^b f(g(t))g'(t) , dt = int_a^b f(c) cdot 0 , dt = 0$$
This is less of a proof and more of a consistency check showing that if the integral over $[c,c]$ is purely by definition then it cannot be arbitrary.
answered Jul 30 at 5:54
RRL
43.4k42260
edited Jul 30 at 7:49
answered Jul 30 at 5:54
RRL
43.4k42260
answered Jul 30 at 5:54
RRL
43.4k42260
answered Jul 30 at 5:54
RRL
43.4k42260
43.4k42260
3
This is correct, however the point in the question is that whatever definition we want to use (and propositions therefrom) one must assume the the sums are calculated on partitions $[a,b]$ such that $a <b$. If we drop this assumption one might as well just defined it as zero and that's it.
– gented
Jul 30 at 8:59
Thanks for your detailed answer and for Gented's explain. Yes, now I think $int_c^cf(xi)dxi$ can be regarded as a definition. Since, in RRL's proof, the definition of Riemann integration is more general and robust than the one provided on wikipedia.
– Lin Xuelei
Jul 31 at 10:17
add a comment |Â
3
This is correct, however the point in the question is that whatever definition we want to use (and propositions therefrom) one must assume the the sums are calculated on partitions $[a,b]$ such that $a <b$. If we drop this assumption one might as well just defined it as zero and that's it.
– gented
Jul 30 at 8:59
Thanks for your detailed answer and for Gented's explain. Yes, now I think $int_c^cf(xi)dxi$ can be regarded as a definition. Since, in RRL's proof, the definition of Riemann integration is more general and robust than the one provided on wikipedia.
– Lin Xuelei
Jul 31 at 10:17
3
3
This is correct, however the point in the question is that whatever definition we want to use (and propositions therefrom) one must assume the the sums are calculated on partitions $[a,b]$ such that $a <b$. If we drop this assumption one might as well just defined it as zero and that's it.
– gented
Jul 30 at 8:59
This is correct, however the point in the question is that whatever definition we want to use (and propositions therefrom) one must assume the the sums are calculated on partitions $[a,b]$ such that $a <b$. If we drop this assumption one might as well just defined it as zero and that's it.
– gented
Jul 30 at 8:59
Thanks for your detailed answer and for Gented's explain. Yes, now I think $int_c^cf(xi)dxi$ can be regarded as a definition. Since, in RRL's proof, the definition of Riemann integration is more general and robust than the one provided on wikipedia.
– Lin Xuelei
Jul 31 at 10:17
Thanks for your detailed answer and for Gented's explain. Yes, now I think $int_c^cf(xi)dxi$ can be regarded as a definition. Since, in RRL's proof, the definition of Riemann integration is more general and robust than the one provided on wikipedia.
– Lin Xuelei
Jul 31 at 10:17
add a comment |Â
up vote
0
down vote
According to Apostol's Calculus Vol. 1 (second edition) page 74, it is a definition that $int_c^cf(x)dx=0$.
If the integral on a single point hasn't been defined, one cannot prove what its value is since it's meaningless.
Note:
It is reasonable to define $int_c^cf(x)dx=0$ because it is clear that the area of the line segment between the point $(c, 0)$ and $(c, f(c))$ is zero. Moreover, if we want to define it so that the theorem of additivity
$$ int_a^bf(x)dx+int_b^cf(x)dx=int_a^cf(x)dx$$
remains valid, then we must have
$$ int_a^cf(x)dx+int_c^cf(x)dx=int_a^cf(x)dx,$$
so $int_c^cf(x)dx=0$.
answered Jul 30 at 5:44
user529760
507216
add a comment |Â
up vote
0
down vote
According to Apostol's Calculus Vol. 1 (second edition) page 74, it is a definition that $int_c^cf(x)dx=0$.
If the integral on a single point hasn't been defined, one cannot prove what its value is since it's meaningless.
Note:
It is reasonable to define $int_c^cf(x)dx=0$ because it is clear that the area of the line segment between the point $(c, 0)$ and $(c, f(c))$ is zero. Moreover, if we want to define it so that the theorem of additivity
$$ int_a^bf(x)dx+int_b^cf(x)dx=int_a^cf(x)dx$$
remains valid, then we must have
$$ int_a^cf(x)dx+int_c^cf(x)dx=int_a^cf(x)dx,$$
so $int_c^cf(x)dx=0$.
answered Jul 30 at 5:44
user529760
507216
add a comment |Â
up vote
0
down vote
up vote
0
down vote
According to Apostol's Calculus Vol. 1 (second edition) page 74, it is a definition that $int_c^cf(x)dx=0$.
If the integral on a single point hasn't been defined, one cannot prove what its value is since it's meaningless.
Note:
It is reasonable to define $int_c^cf(x)dx=0$ because it is clear that the area of the line segment between the point $(c, 0)$ and $(c, f(c))$ is zero. Moreover, if we want to define it so that the theorem of additivity
$$ int_a^bf(x)dx+int_b^cf(x)dx=int_a^cf(x)dx$$
remains valid, then we must have
$$ int_a^cf(x)dx+int_c^cf(x)dx=int_a^cf(x)dx,$$
so $int_c^cf(x)dx=0$.
answered Jul 30 at 5:44
user529760
507216
According to Apostol's Calculus Vol. 1 (second edition) page 74, it is a definition that $int_c^cf(x)dx=0$.
If the integral on a single point hasn't been defined, one cannot prove what its value is since it's meaningless.
Note:
It is reasonable to define $int_c^cf(x)dx=0$ because it is clear that the area of the line segment between the point $(c, 0)$ and $(c, f(c))$ is zero. Moreover, if we want to define it so that the theorem of additivity
$$ int_a^bf(x)dx+int_b^cf(x)dx=int_a^cf(x)dx$$
remains valid, then we must have
$$ int_a^cf(x)dx+int_c^cf(x)dx=int_a^cf(x)dx,$$
so $int_c^cf(x)dx=0$.
answered Jul 30 at 5:44
user529760
507216
edited Jul 30 at 12:58
answered Jul 30 at 5:44
user529760
507216
answered Jul 30 at 5:44
user529760
507216
answered Jul 30 at 5:44
user529760
507216
507216
add a comment |Â
add a comment |Â
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1
Let $I$ a real interval and $f$ a function being Riemann integrable on $I$, then for all $a le b le c in I$ We have : $int_a^c f(x) dx = int_a^b f(x) dx + int_b^c f(x) dx$.
– nicomezi
Jul 30 at 6:56