Lets get started.
Unlike the Riemann integral, interestingly, we will see that no notion of any limit, formally speaking, will be needed in defining the Darboux integral. Indeed, many people, myself included, find the definition of the Darboux integral simpler than the definition of the Riemann integral, although both notions of integrability (as we will see) are equivalent.
Before we define the Darboux integral, two preliminary concepts must be introduced.
Lets start with one of the properties of the Real Numbers, the so called 'least upper bound' or 'completeness' property.
Definition 1.2 (Least Upper Bound / Completeness property of the Real numbers) -
For any non-empty subset $A$ of $\mathbb {R}$, if $A$ is bounded above, that is, if there exists a real number $b \in \mathbb {R}$ such that $b \geq x$ for all $ x \in A$, then there exists a real number $c$ such that $c \geq x$ for all $x \in A$ with the additional property that for any $d \in \mathbb {R}$ such that $d \geq x$ for all $x \in A$ , $c \leq d$ . We call $c$ the 'least upper bound' of $A$ or we say $c = \sup A$ or $c$ is the 'supremum' of $A$.
The property stated above is actually a consequence of how we construct the real numbers, in fact, the goal of both of the most popular approaches to constructing the real numbers (Dedekind cuts, Cauchy sequences) is precisely to be able to equip the set of all real numbers with this very property. The completeness property of the Real numbers is absolutely fundamental to all of calculus and Real analysis, as it enables us to start working with sequences and their limits, which in turn enables us to start looking at the limits of real valued functions.
There are several different ways you can try to visualise what the completeness property is, one of these ways is to think about the completeness property as 'the real number line contains no holes'. The same cannot be said about the rationals. As an example, take the set $ S = \{x \in \mathbb {Q} | \ x^2 \leq 2\}$ . Try as you might, you will never be able to find a number $ c \in \mathbb {Q} $ such that $c$ is the least upper bound of $S$. Working in the real numbers, however, the answer is $\sqrt{2}$.
Now we will also define the 'infimum' or the 'greatest lower bound' of a subset of the real numbers that is bounded below.
Definition 1.3 (Greatest Lower Bound / Infimum) - If a nonempty subset of the real numbers, $A$ is bounded below, that is, there exists a real number $b$ such that for all $x \in A$, $ b \leq x$, then there exists a real number $c$ such that $c \leq x$ for all $x \in A$ which possesses the additional property that for any real number $d$ such that $d \leq x$ for all $x \in A$, $c \geq d$. We call $c$ the 'infimum' of $A$, or the 'greatest lower bound' of $A$. We may also write $ c = \inf A$.
The truth of Definition 1.3 follows from the least upper bound property, and a sketch of this proof has been presented below:
Proof (Definition 1.3) (Sketch) : Suppose $A \subset \mathbb {R}$ is bounded below, consider $ S = \{ x \in \mathbb {R} | x \leq y , \forall y \in A \}$ . Prove that $S$ is bounded above, now apply the least upper bound property. $\sup S$ must exist. Prove that $\sup S$ is equal to the greatest lower bound of $A$ (Proof by contradiction may be useful here)
$\square$
We may have also started with the 'infimum' property and proceeded to prove the 'supremum' property of the Real numbers, that is, the 'infimum' and 'supremum' properties of the Real numbers are logically equivalent (one implies the other).
Two more definitions and we'll be able to define the Darboux integral ( I promise these definitions won't be lengthy)
Definition 1.4 (Boundedness) - A function $f$ is said to be 'bounded' on $[a,b]$ if there exists a real number $M > 0 $ such that for all $x$ in $[a,b]$ , $|f(x)| \leq M$.
Definition 1.5 (Darboux Upper and Lower Sums) - The Darboux upper sum of a function $f$ on $[a,b]$ (provided the function is defined over $[a,b]$) with respect to a partition $P=\{x_0=a,x_1,x_2,...,x_{n-1},x_n=b\}$ of $[a,b]$ is $ U(f,P) = \sum_{i=1}^n (\sup{f}_{x \in [x_{i-1}, x_i]})(x_i - x_{i-1})$
The Darboux lower sum of a function $f$ on $[a,b]$ with respect to a partition $P=\{x_0=a,x_1,x_2,...,x_{n-1},x_n=b\}$ of $[a,b]$ is $ L(f,P) = \sum_{i=1}^n (\inf{f}_{x \in [x_{i-1}, x_i]})(x_i - x_{i-1})$
It may seem a bit strange as to why we use $\inf$ and $\sup$ rather than $\min$ and $\max$. This is because it is possible that $f$ is not continuous, or not continuous everywhere, and therefore $f$ may not itself attain extreme values on some interval.
The diagram above depicts a Darboux upper sum for a function that is continuous over $[a,b]$ (therefore $\sup {f}_{x \in [x_{i-1},x_i]} = \max{f}_{x \in [x_{i-1},x_i]}$)
Rather than using the busy $\sup {f}_{x \in [x_{i-1},x_i]}$ and $\inf {f}_{x \in [x_{i-1},x_i]}$ we will denote $M_i = \sup {f}_{x \in [x_{i-1},x_i]}$ and $m_i = \inf {f}_{x \in [x_{i-1},x_i]}$ later in this post.
Another way of thinking about these upper sums are as 'over-approximations'. It follows that Darboux lower sums can be thought of as 'under-approximations' to the area under $f(x)$ on $[a,b]$.
We will often be interchanging between the notation $U(f,P)$ for a Darboux upper sum of $f$ over $[a,b]$ with respect to the partition $P$ and $\sum_{i=1}^n (M_i)(x_i - x_{i-1}) = U(f,P)$. Likewise, we will also frequently use the notation $L(f,P)$ to denote the Darboux lower sum of $f$ over $[a,b]$ with respect to the partition $P$.
Without further ado, let us define the Darboux integral.
Definition 1.4 (Darboux Integral) - A bounded function $f$ is said to be Darboux integrable if $\inf\{U(f,P) | P \ is \ a \ partition \ of \ [a,b]\} = \sup\{L(f,P) | P \ is \ a \ partition \ of \ [a,b]\}$. Furthermore, if $f$ is Darboux integrable on $[a,b]$ we write $\inf\{U(f,P) | P \ is \ a \ partition \ of \ [a,b]\} = \sup\{L(f,P) | P \ is \ a \ partition \ of \ [a,b]\} = \int_{a}^b f$
Before we go any further, how can we be so sure $\inf\{U(f,P) | P \ is \ a \ partition \ of \ [a,b]\}$ and
$\sup\{L(f,P) | P \ is \ a \ partition \ of \ [a,b]\}$ exist?
It turns out since we assumed $f$ was bounded, we can indeed assume this, the explicit proof of this fact is presented below:
Proof ( $\inf{U(f,P)}$ and $\sup{L(f,P)}$ exist) - Recall that $f$ is bounded, therefore for all $x \in [a,b]$ we have $|f| \leq M$. Let $P = \{x_0 = a, x_1, ..., x_{n-1}, x_n\}$ .
We have
$L(f,P) = \sum_{i=1}^n (m_i)(x_i - x_{i-1}) \leq M(\sum_{i=1}^n (x_i - x_{i-1})) = M(b-a)$.
Therefore $L(f,P)$ is bounded above for any arbitrary partition $P$ of $[a,b]$.
It thus follows by the least upper bound property of the Real numbers that $\sup\{L(f,P) | P \ is \ a \ partition \ of \ [a,b]\}$ exists. Similarly, we can prove that $\inf\{U(f,P) | P \ is \ a \ partition \ of \ [a,b]\}$ exists.
$\square$
Now back to Definition 1.4, although this definition does not technically involve limits, it does seem very limit-esque. One can imagine the 'area' under $f(x)$ over $[a,b]$ being 'squeezed' by both the lower and upper Riemann sums and thus forced to converge to a certain value.
Before we entertain this concept of an area being 'squeezed' by upper and lower sums any further, let us prove several lemmas regarding the supremum and infimum that will be useful to us later in the next blog post (Introduction to real analysis 1.3).
Lemma 1.0 (Property of Supremum/Infimum) - Suppose $c$ is an upper or lower bound of a nonempty set $A \subset \mathbb{R}$. Then $c = \sup A$ or $c = \inf A$ if and only if for all $\epsilon > 0$ , there exists $x \in A$ such that $| x - c | < \epsilon$.
Proof (Lemma 1.0) - We will prove one direction of Lemma 1.0 by contradiction. Suppose $c = \sup A$ for some nonempty set $A \subset \mathbb {R}$ that is bounded above. Now suppose $\forall x \in A$ we have $|x - c| = c - x \geq \epsilon$ then we have $ x \leq c - \epsilon $ for all $x$ in $A$. So we have $c - \epsilon$ is an upper bound of $A$. But we have already defined $c = \sup A$. Contradiction. Similarly, we can prove Lemma 1.0 for $c = \inf A$. (The proof is essentially the same).
To prove the other side, we must prove that $c$ is the least upper bound of $A$. First of all, we already know that $c$ is an upper bound of $A$ (as per the hypothesis of Lemma 1.0). Therefore $x\leq c$ for all $x \in A$. Now consider any upper bound of $A$, call it $d$. We will show that $c \leq d$. Suppose $c > d$, then $c-d > 0$. As per hypothesis, there exists a $x \in A$ such that $|x - c| < c - d$. This implies that $ c - x < c - d$ and that $-x< -d$ or $x > d$. But $d$ is an upper bound of $A$. This yields a contradiction.
$\square$
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