Textbook Definition of Riemann Integral: Supremum, Infimum, Lower/Upper Bounds
Clash Royale CLAN TAG#URR8PPP
up vote
4
down vote
favorite
The following excerpt is from my Laplace transform textbook, where the author is trying to define the Riemann integral:
Let $F(x)$ be a function which is defined and is bounded in the interval $a le x le b$ and suppose that $m$ and $M$ are respectively the lower and upper bounds of $F(x)$ in this interval (written $[a, b]$ see Appendix C). Take a set of points $x_0 = a, x_1, x_2, dots, x_r − 1, x_r , dots, x_n = b$ and write $delta_r = x_r − x_r − 1$. Let $M_r$, $m_r$ be the bounds of $F(x)$ in the subinterval $(x_r − 1, x_r)$ and form the sums
$S = sum limits_r = 1^n M_r delta_r$
$s = sum limits_r = 1^n m_r delta_r$
These are called respectively the upper and lower Riemann sums corresponding to the mode of subdivision. It is certainly clear that $S ge s$. There are a variety of ways that can be used to partition the interval $(a, b)$ and each way will have (in general) different $M_r$ and $m_r$ leading to different $S$ and $s$. Let $M$ be the minimum of all possible $M_r$ and $m$ be the maximum of all possible $m_r$. A lower bound or supremum for the set $S$ is therefore $M(b − a)$ and an upper bound or infimum for the set $s$ is $m(b − a)$.
The supremum is defined to be the least upper bound; the infimum is defined to be the greatest lower bound.
So not only are supremum and "lower bound" actually different concepts -- they're contradictory concepts. Something cannot be both a supremum and a lower bound. This is analogously true for infimum and "upper bound".
Am I misunderstanding something, or has the author made an error?
For reference, I post the following from Rudin's Mathematical Analysis:
Thank you for any help.
UPDATE: I received a response from the author on this matter:
A more important point is the analysis of the Riemann integral. If you have access to the classic textbook "A Course of Analysis" by E G Phillips, 1962 Cambridge University Press the full story is in section 7.2 on page 166. In short, in order to have a uniquely defined integral, the supremum (I) or upper bound of all the lower sums has to be the same as the infimum (J) or lower bound of all the upper sums of all possible choices of subdivisions. These have to be the same (I=J) otherwise any integral will have many different values and would not be well defined. It is this that I have summarised on page 4, it is classical mathematical analysis.
Can people please review this further and comment on whether I (and others in the comments) misunderstood this, or whether we are correct?
real-analysis proof-explanation supremum-and-infimum riemann-integration riemann-sum
asked Jul 12 at 20:22
Wyuw
1388
 |Â
show 4 more comments
up vote
4
down vote
favorite
The following excerpt is from my Laplace transform textbook, where the author is trying to define the Riemann integral:
Let $F(x)$ be a function which is defined and is bounded in the interval $a le x le b$ and suppose that $m$ and $M$ are respectively the lower and upper bounds of $F(x)$ in this interval (written $[a, b]$ see Appendix C). Take a set of points $x_0 = a, x_1, x_2, dots, x_r − 1, x_r , dots, x_n = b$ and write $delta_r = x_r − x_r − 1$. Let $M_r$, $m_r$ be the bounds of $F(x)$ in the subinterval $(x_r − 1, x_r)$ and form the sums
$S = sum limits_r = 1^n M_r delta_r$
$s = sum limits_r = 1^n m_r delta_r$
These are called respectively the upper and lower Riemann sums corresponding to the mode of subdivision. It is certainly clear that $S ge s$. There are a variety of ways that can be used to partition the interval $(a, b)$ and each way will have (in general) different $M_r$ and $m_r$ leading to different $S$ and $s$. Let $M$ be the minimum of all possible $M_r$ and $m$ be the maximum of all possible $m_r$. A lower bound or supremum for the set $S$ is therefore $M(b − a)$ and an upper bound or infimum for the set $s$ is $m(b − a)$.
The supremum is defined to be the least upper bound; the infimum is defined to be the greatest lower bound.
So not only are supremum and "lower bound" actually different concepts -- they're contradictory concepts. Something cannot be both a supremum and a lower bound. This is analogously true for infimum and "upper bound".
Am I misunderstanding something, or has the author made an error?
For reference, I post the following from Rudin's Mathematical Analysis:
Thank you for any help.
UPDATE: I received a response from the author on this matter:
A more important point is the analysis of the Riemann integral. If you have access to the classic textbook "A Course of Analysis" by E G Phillips, 1962 Cambridge University Press the full story is in section 7.2 on page 166. In short, in order to have a uniquely defined integral, the supremum (I) or upper bound of all the lower sums has to be the same as the infimum (J) or lower bound of all the upper sums of all possible choices of subdivisions. These have to be the same (I=J) otherwise any integral will have many different values and would not be well defined. It is this that I have summarised on page 4, it is classical mathematical analysis.
Can people please review this further and comment on whether I (and others in the comments) misunderstood this, or whether we are correct?
real-analysis proof-explanation supremum-and-infimum riemann-integration riemann-sum
asked Jul 12 at 20:22
Wyuw
1388
1
Yes, you're right, he's got them switched. Poor proofreading.
– saulspatz
Jul 12 at 20:28
@saulspatz Ok, thank you for your help.
– Wyuw
Jul 12 at 20:30
3
Both of the last two sentences are screwed up. The second-to-last sentence should probably be "$M$ is the maximum of all possible $M_r$ and $m$ is the minimum of all possible $m_r$". The last sentence, as you have discerned, should probably be "An upper bound for the number $S$ is therefore $M(b-a)$ and a lower bound for the number $s$ is $m(b-a)$."
– Lee Mosher
Jul 12 at 20:44
@LeeMosher Yes, I think you are correct. I will edit the main post with the formal definitions of supremum and infimum for ease of reference. Thank you for the help.
– Wyuw
Jul 12 at 21:06
The last two sentences can be replaced with the following: "All sums of type $S$ are bounded below by $m(b-a) $ because $M_rgeq m$ and all sums of type $s$ are bounded above by $M(b-a) $ as $m_rleq M$.
– Paramanand Singh
Jul 13 at 1:03
 |Â
show 4 more comments
up vote
4
down vote
favorite
up vote
4
down vote
favorite
The following excerpt is from my Laplace transform textbook, where the author is trying to define the Riemann integral:
Let $F(x)$ be a function which is defined and is bounded in the interval $a le x le b$ and suppose that $m$ and $M$ are respectively the lower and upper bounds of $F(x)$ in this interval (written $[a, b]$ see Appendix C). Take a set of points $x_0 = a, x_1, x_2, dots, x_r − 1, x_r , dots, x_n = b$ and write $delta_r = x_r − x_r − 1$. Let $M_r$, $m_r$ be the bounds of $F(x)$ in the subinterval $(x_r − 1, x_r)$ and form the sums
$S = sum limits_r = 1^n M_r delta_r$
$s = sum limits_r = 1^n m_r delta_r$
These are called respectively the upper and lower Riemann sums corresponding to the mode of subdivision. It is certainly clear that $S ge s$. There are a variety of ways that can be used to partition the interval $(a, b)$ and each way will have (in general) different $M_r$ and $m_r$ leading to different $S$ and $s$. Let $M$ be the minimum of all possible $M_r$ and $m$ be the maximum of all possible $m_r$. A lower bound or supremum for the set $S$ is therefore $M(b − a)$ and an upper bound or infimum for the set $s$ is $m(b − a)$.
The supremum is defined to be the least upper bound; the infimum is defined to be the greatest lower bound.
So not only are supremum and "lower bound" actually different concepts -- they're contradictory concepts. Something cannot be both a supremum and a lower bound. This is analogously true for infimum and "upper bound".
Am I misunderstanding something, or has the author made an error?
For reference, I post the following from Rudin's Mathematical Analysis:
Thank you for any help.
UPDATE: I received a response from the author on this matter:
A more important point is the analysis of the Riemann integral. If you have access to the classic textbook "A Course of Analysis" by E G Phillips, 1962 Cambridge University Press the full story is in section 7.2 on page 166. In short, in order to have a uniquely defined integral, the supremum (I) or upper bound of all the lower sums has to be the same as the infimum (J) or lower bound of all the upper sums of all possible choices of subdivisions. These have to be the same (I=J) otherwise any integral will have many different values and would not be well defined. It is this that I have summarised on page 4, it is classical mathematical analysis.
Can people please review this further and comment on whether I (and others in the comments) misunderstood this, or whether we are correct?
real-analysis proof-explanation supremum-and-infimum riemann-integration riemann-sum
asked Jul 12 at 20:22
Wyuw
1388
The following excerpt is from my Laplace transform textbook, where the author is trying to define the Riemann integral:
Let $F(x)$ be a function which is defined and is bounded in the interval $a le x le b$ and suppose that $m$ and $M$ are respectively the lower and upper bounds of $F(x)$ in this interval (written $[a, b]$ see Appendix C). Take a set of points $x_0 = a, x_1, x_2, dots, x_r − 1, x_r , dots, x_n = b$ and write $delta_r = x_r − x_r − 1$. Let $M_r$, $m_r$ be the bounds of $F(x)$ in the subinterval $(x_r − 1, x_r)$ and form the sums
$S = sum limits_r = 1^n M_r delta_r$
$s = sum limits_r = 1^n m_r delta_r$
These are called respectively the upper and lower Riemann sums corresponding to the mode of subdivision. It is certainly clear that $S ge s$. There are a variety of ways that can be used to partition the interval $(a, b)$ and each way will have (in general) different $M_r$ and $m_r$ leading to different $S$ and $s$. Let $M$ be the minimum of all possible $M_r$ and $m$ be the maximum of all possible $m_r$. A lower bound or supremum for the set $S$ is therefore $M(b − a)$ and an upper bound or infimum for the set $s$ is $m(b − a)$.
The supremum is defined to be the least upper bound; the infimum is defined to be the greatest lower bound.
So not only are supremum and "lower bound" actually different concepts -- they're contradictory concepts. Something cannot be both a supremum and a lower bound. This is analogously true for infimum and "upper bound".
Am I misunderstanding something, or has the author made an error?
For reference, I post the following from Rudin's Mathematical Analysis:
Thank you for any help.
UPDATE: I received a response from the author on this matter:
A more important point is the analysis of the Riemann integral. If you have access to the classic textbook "A Course of Analysis" by E G Phillips, 1962 Cambridge University Press the full story is in section 7.2 on page 166. In short, in order to have a uniquely defined integral, the supremum (I) or upper bound of all the lower sums has to be the same as the infimum (J) or lower bound of all the upper sums of all possible choices of subdivisions. These have to be the same (I=J) otherwise any integral will have many different values and would not be well defined. It is this that I have summarised on page 4, it is classical mathematical analysis.
Can people please review this further and comment on whether I (and others in the comments) misunderstood this, or whether we are correct?
real-analysis proof-explanation supremum-and-infimum riemann-integration riemann-sum
asked Jul 12 at 20:22
Wyuw
1388
edited Aug 28 at 12:52
asked Jul 12 at 20:22
Wyuw
1388
asked Jul 12 at 20:22
Wyuw
1388
asked Jul 12 at 20:22
Wyuw
1388
1388
1
Yes, you're right, he's got them switched. Poor proofreading.
– saulspatz
Jul 12 at 20:28
@saulspatz Ok, thank you for your help.
– Wyuw
Jul 12 at 20:30
3
Both of the last two sentences are screwed up. The second-to-last sentence should probably be "$M$ is the maximum of all possible $M_r$ and $m$ is the minimum of all possible $m_r$". The last sentence, as you have discerned, should probably be "An upper bound for the number $S$ is therefore $M(b-a)$ and a lower bound for the number $s$ is $m(b-a)$."
– Lee Mosher
Jul 12 at 20:44
@LeeMosher Yes, I think you are correct. I will edit the main post with the formal definitions of supremum and infimum for ease of reference. Thank you for the help.
– Wyuw
Jul 12 at 21:06
The last two sentences can be replaced with the following: "All sums of type $S$ are bounded below by $m(b-a) $ because $M_rgeq m$ and all sums of type $s$ are bounded above by $M(b-a) $ as $m_rleq M$.
– Paramanand Singh
Jul 13 at 1:03
 |Â
show 4 more comments
1
Yes, you're right, he's got them switched. Poor proofreading.
– saulspatz
Jul 12 at 20:28
@saulspatz Ok, thank you for your help.
– Wyuw
Jul 12 at 20:30
3
Both of the last two sentences are screwed up. The second-to-last sentence should probably be "$M$ is the maximum of all possible $M_r$ and $m$ is the minimum of all possible $m_r$". The last sentence, as you have discerned, should probably be "An upper bound for the number $S$ is therefore $M(b-a)$ and a lower bound for the number $s$ is $m(b-a)$."
– Lee Mosher
Jul 12 at 20:44
@LeeMosher Yes, I think you are correct. I will edit the main post with the formal definitions of supremum and infimum for ease of reference. Thank you for the help.
– Wyuw
Jul 12 at 21:06
The last two sentences can be replaced with the following: "All sums of type $S$ are bounded below by $m(b-a) $ because $M_rgeq m$ and all sums of type $s$ are bounded above by $M(b-a) $ as $m_rleq M$.
– Paramanand Singh
Jul 13 at 1:03
1
1
Yes, you're right, he's got them switched. Poor proofreading.
– saulspatz
Jul 12 at 20:28
Yes, you're right, he's got them switched. Poor proofreading.
– saulspatz
Jul 12 at 20:28
@saulspatz Ok, thank you for your help.
– Wyuw
Jul 12 at 20:30
@saulspatz Ok, thank you for your help.
– Wyuw
Jul 12 at 20:30
3
3
Both of the last two sentences are screwed up. The second-to-last sentence should probably be "$M$ is the maximum of all possible $M_r$ and $m$ is the minimum of all possible $m_r$". The last sentence, as you have discerned, should probably be "An upper bound for the number $S$ is therefore $M(b-a)$ and a lower bound for the number $s$ is $m(b-a)$."
– Lee Mosher
Jul 12 at 20:44
Both of the last two sentences are screwed up. The second-to-last sentence should probably be "$M$ is the maximum of all possible $M_r$ and $m$ is the minimum of all possible $m_r$". The last sentence, as you have discerned, should probably be "An upper bound for the number $S$ is therefore $M(b-a)$ and a lower bound for the number $s$ is $m(b-a)$."
– Lee Mosher
Jul 12 at 20:44
@LeeMosher Yes, I think you are correct. I will edit the main post with the formal definitions of supremum and infimum for ease of reference. Thank you for the help.
– Wyuw
Jul 12 at 21:06
@LeeMosher Yes, I think you are correct. I will edit the main post with the formal definitions of supremum and infimum for ease of reference. Thank you for the help.
– Wyuw
Jul 12 at 21:06
The last two sentences can be replaced with the following: "All sums of type $S$ are bounded below by $m(b-a) $ because $M_rgeq m$ and all sums of type $s$ are bounded above by $M(b-a) $ as $m_rleq M$.
– Paramanand Singh
Jul 13 at 1:03
The last two sentences can be replaced with the following: "All sums of type $S$ are bounded below by $m(b-a) $ because $M_rgeq m$ and all sums of type $s$ are bounded above by $M(b-a) $ as $m_rleq M$.
– Paramanand Singh
Jul 13 at 1:03
 |Â
show 4 more comments
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2849042%2ftextbook-definition-of-riemann-integral-supremum-infimum-lower-upper-bounds%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
1
Yes, you're right, he's got them switched. Poor proofreading.
– saulspatz
Jul 12 at 20:28
@saulspatz Ok, thank you for your help.
– Wyuw
Jul 12 at 20:30
3
Both of the last two sentences are screwed up. The second-to-last sentence should probably be "$M$ is the maximum of all possible $M_r$ and $m$ is the minimum of all possible $m_r$". The last sentence, as you have discerned, should probably be "An upper bound for the number $S$ is therefore $M(b-a)$ and a lower bound for the number $s$ is $m(b-a)$."
– Lee Mosher
Jul 12 at 20:44
@LeeMosher Yes, I think you are correct. I will edit the main post with the formal definitions of supremum and infimum for ease of reference. Thank you for the help.
– Wyuw
Jul 12 at 21:06
The last two sentences can be replaced with the following: "All sums of type $S$ are bounded below by $m(b-a) $ because $M_rgeq m$ and all sums of type $s$ are bounded above by $M(b-a) $ as $m_rleq M$.
– Paramanand Singh
Jul 13 at 1:03