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My textbook, An Introduction to Laplace Transforms and Fourier Series, Second Edition, by Phil Dyke, says the following:

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

$$S = \sum_{r = 1}^n M_r \delta_r$$

$$s = \sum_{r = 1}^n m_r \delta_r$$

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 $m(b - a)$.

Shouldn’t the sentence

Let $M$ be the minimum of all possible $M_r$ and $m$ be the maximum of all possible $m_r$.

be “$M$ is the maximum of all possible $M_r$ and $m$ is the minimum of all possible $m_r$.”?

And shouldn’t the sentence

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 $m(b - a)$.

be “An upper bound for the set $S$ is therefore $M(b - a)$ and a lower bound for the set $S$ is $m(b - a)$.”?

Using the definitions of of supremum and infimum from Mathematical Analysis by Rudin (see below), the supremum is the least upper bound and the infimum is the greatest lower bound. So not only are supremum and "lower bound" actually different concepts — it seems that they would be contradictory concepts? After all, something cannot be both a supremum and a lower bound. And analogously for infimum and "upper bound”?

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I would greatly appreciate it if people would please take the time to clarify this.

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    $\begingroup$ Wow, that's some sloppy writing ... $\endgroup$ Commented Dec 26, 2019 at 21:17
  • $\begingroup$ @NoahSchweber Apparently this deserved a second edition! $\endgroup$ Commented Dec 26, 2019 at 21:36

2 Answers 2

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I don't have a copy of the textbook, so I can't be completely sure what they're trying to state in that section. However, a key thing about using the upper & lower Riemann sums is dealing with their convergence, if any, to a specific value. In that regard, you want to consider how the smallest values of the upper sums approaches the largest values of the lower sums. As such, the sentence

Let $M$ be the minimum of all possible $M_r$ and $m$ be the maximum of all possible $m_r$.

has an appropriate approach, but as stated in various places, including in the comment to this answer, it's poorly written, but what you're suggesting is not appropriate.

However, you're correct there's a mistake in the next sentence of

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 terms "supremum" and "infimum" should be switched around in that sentence.

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  • $\begingroup$ I would rather say that “Let $M$ be the minimum of all possible $M_r$” makes no sense. All possible $M_r$, where we’ve fixed a partition or no? If no, then there probably is no minimum; if so, then I guess taking the least of all $m_r$ would make $m(b-a)$ a lower bound for the sum (not set?!) $s$...This passage is frankly just incompetently written. $\endgroup$ Commented Dec 26, 2019 at 21:35
  • $\begingroup$ @KevinCarlson Thanks for the feedback. You're correct that sentence is not well written. I just meant it had the right idea compared to what the OP was suggesting instead. I've updated my answer to make this more clear. $\endgroup$ Commented Dec 26, 2019 at 21:41
  • $\begingroup$ “The minimum of all possible $M_r$ seems to have precious little to do with the minimum of the upper sums. After all, each $M_r$ is the height of just one rectangle. The minimum will not be achievable in general, and if it could, $M(b-a)$ defined in this way would usually be less than the integral. At least that’s how I read it. $\endgroup$
    – David K
    Commented Dec 27, 2019 at 0:04
  • $\begingroup$ Thanks for the answer! You're right: "Let $M$ be the minimum of all possible $M_r$ and $m$ be the maximum of all possible $m_r$." sounds reasonable when we have "A lower bound or infimum for the set $S$ is therefore $M(b - a)$ and an upper bound or supremum for the set $s$ is $m(b - a)$." The entire section was so poorly written that I confused myself even more trying to fix it. $\endgroup$ Commented Dec 27, 2019 at 5:12
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    $\begingroup$ @ThePointer You're welcome. I agree that section of text was very poorly worded, so I believe quite a few people would get confused by what they were trying to state, as well as when trying to fix it. $\endgroup$ Commented Dec 27, 2019 at 5:14
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I agree that the explanation, as you have transcribed it, is confusing.

Let $T\subset S$ be a bounded subset. Let the set $$U = \{ M\in S|\ \forall x \in T: x\leq M \}$$ be the set of all the upper bounds $M$ of $T$. We then define the supremum $$\sup T := M'\in U:\forall M\in U,M\geq M' $$

Analogously we take the set of all lower bounds $$L = \{ m\in S|\ \forall x \in T: x\geq m \}$$ and define accordingly the infimum $$\inf T := m'\in L:\forall m\in L,m\leq m' $$

Given this, I'd recommend reading this to understand correctly the construction in your book: https://en.wikipedia.org/wiki/Riemann_integral#Definition

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