Real Analysis: Integrals
Darboux Integrals
Motivation: Finding the Area Under the Curve
Given a real function $f,$ we would like to find the area between $f$ and the $x$-axis over a given interval $[a, b].$ We can use basic geometry to calculate the areas of rectangles or triangles in the cases when $f$ is constant or linear, for example, but what about all of the other cases?
The basic approach of all methods to answering this question is to split the interval into a number of smaller subintervals. Within each interval, $f$ is evaluated at some point in the subinterval, and the product of that value of $f$ and the length of the subinterval is taken as an estimate of the area for that subinterval. The sum of these estimates forms the estimate for the entire interval. Intuitively, the more subintervals there are, the closer the estimate will be. The true area of the total interval is thereby determined by taking the limit of the sum of these areas as the widths of the subintervals goes to $0.$ This if this limit exists, its value is called the integral of $f$ from $a$ to $b.$
Measuring Intervals
The length $\mu$ of a closed interval $I = [a, b]$ is given by $\mu(I) = b - a.$
The notation $\mu$ is chosen to align with the more general concept of measure. Roughly speaking, measure can be thought of as a generalization of length in the same way that a topological space can be thought of as a generalization of a metric space.
Finite Interval Subdivisions
A finite interval subdivision, or simply subdivision, of a non-singleton closed interval $I = [a, b]$ is an ordered set $P = \{p_1, \ldots, p_n\}$ with the following properties:
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Each element of $P$ is called a subinterval and is itself a non-singleton closed interval.
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$\bigcup P = I.$
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$\inf(p_1) = a.$
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$\sup(p_n) = b.$
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$\sup(p_i) = \inf(p_{i+1})$ for $i \in \{1, \ldots, n-1\}.$
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The length of the longest subinterval, $\max\limits_{p_i \in P}(\mu(p_i)),$ is called the norm of $P.$
For example $P = \{ [0, 1], [1, 3], [3, 4], [4, 5] \}$ is a subdivision of $[0, 5]$ whose norm is $2.$ A subdivision with $n$ elements has $n-1$ points of overlap between its members; in the aforementioned example, these points are $1,$ $3,$ and $4.$
Refinements of Finite Interval Subdivisions
A subdivision $P_2$ is said to be finer than a subdivision $P_1$ if every element of $P_1$ can be expressed as the union of elements from $P_2.$ In this case, $P_2$ is called a refinement of $P_1,$ and $P_1$ is said to be coarser than $P_2.$
If $P_1$ and $P_2$ are subdivisions, then their common refinement is the set of all intersections of each of their subintervals, and is denoted $P_1 \# P_2:$
$$P_1 \# P_2 = \{ p_1 \cap p_2 : p_1 \in P_1 \text{ and } p_2 \in P_2 \}.$$
The Darboux Integral
Let $f$ be a real function defined along an interval $[a, b],$ and let $P$ be a finite interval subdivision of $[a, b].$ The lower Darboux sum of $f$ with respect to $P$ is defined as
$$L(f, P) = \sum\limits_{p \in P} \inf\limits_{x \in p}(f(x)) \mu(p).$$
Likewise, the upper Darboux sum of $f$ with respect to $P$ is defined as
$$U(f, P) = \sum\limits_{p \in P} \sup\limits_{x \in p}(f(x)) \mu(p).$$
The lower Darboux integral of $f$ is defined as the least upper bound of all lower Darboux sums,
$$L^*(f, [a, b]) = \sup L(f, P),$$
where the supremum is taken over all possible finite interval subdivision of $[a, b].$ Intuitively, you could consider each lower Darboux sum as an underestimate of the area under the curve, and the lower Darboux integral is the best possible underestimate of all.
Likewise, the upper Darboux integral of $f$ is
$$U^*(f, [a, b]) = \inf U(f, P)$$
where the infimum is likewise taken over all possible finite interval subdivisions of $[a, b].$ As with the lower Darboux integral, each upper Darboux sums can be considered an overestimate of the area under the curve, and the upper Darboux integral is the best possible overestimate of all.
If both the upper and lower Darboux integrals of $f$ exists and are equal to one another, then the value is simply called the Darboux integral of $f,$ whose value is denoted
$$\int\limits_{a}^{b} f(x) \, dx = L^*(f, [a, b]) = U^*(f, [a, b]).$$
The $\int$ symbol is called the integral sign, and $f,$ the function being integrated, is called the integrand. The expression is read aloud as "the integral from $a$ to $b$ of $f(x)$ with respect to $x.$" If the Darboux integral of $f$ exists, then $f$ is said to be Darboux integrable, or if the context is clear, simply integrable.
Darboux integrable functions on are also Riemann integrable, and we say that $f \in \mathscr{R}([a, b]).$ Riemann integrals are defined differently from Darboux integrals, but are in fact equivalent; for reasons explained in the next section on Riemann integrals, the Riemann terminology is much more common.
Is the Definition of the Darboux Integral Useful?
The foregoing definition of the Darboux integral is coherent enough - subdivide the interval $[a, b]$ into a subdivisions, take the two most extreme calculations for the areas under the curve determined by those subdivisions, and declare victory if the infimum and supremum of these calculations match in a sort of limit-like consideration of all possible subdivisions.
But how is one supposed to actually compute such a value? All the logical rigor in the world is meaningless if it does not allow us to actually use it. From the definition alone, it seems we could at best approximate an integral by using a computer to attempt to exhaustively hone in on a solution by calculating $U$ and $L$ for many subdivisions and call it a day if they seem to converge. But even that process requires that the computer to find the infimum and supremum of $f$ on arbitrary subintervals, which is a task all on its own.
Exhaustive computation is in fact the only known method for calculating certain integrals, and the next section on Riemann integrals will present more efficient approaches for this process. However, for many classes of integrals, these calculations can be avoided by applying the fundamental theorem of calculus. This storied theorem relates the processes of integration and differentiation and to produce closed form algebraic solutions. We start the journey to this theorem by proving several key properties of Darboux integrals.
The General Condition for Integrability
The general condition for integrability is an extremely useful theorem, and is used to prove other conditions for integrability as well as many algebraic properties of integrability. It states that a bounded function $f$ defined over $[a, b]$ is integrable on $[a, b]$ if and only if for every $\varepsilon > 0$ there exists a finite interval subdivision $P$ such that
$$U(f, P) - L(f, P) < \varepsilon.$$
The "if" part here is critical - if for any $\varepsilon > 0$ we can find a subdivision $P$ such that the inequality holds, then we can conclude that $f$ is integrable. This allows us to specifically construct a particular kind of subdivision according to the constraints of the particular result we are trying to prove, rather than consider all possible subdivisions.
Further Conditions for Integrability
For a bounded real function $f$ and an interval $[a, b],$ the following properties hold:
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Continuity implies integrability: If $f$ is continuous along $[a, b],$ then it is integrable on $[a, b].$
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Monotonicity implies integrability: If $f$ is monotonic along $[a, b],$ then it is integrable on $[a, b].$
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Finite discontinuities implies integrability: If $f$ only has finitely many discontinuities on $[a, b],$ then it is integrable on $[a, b].$
Much in the way that the simple algebraic laws of differentiability allow us to prove increasingly complicated classes of functions are differentiable, so, too, do these conditions allow us to ascertain the integrability of many classes of functions.
Problems
Finite Additivity: For a finite interval subdivision $P$ of $n$ elements of an interval $[a, b],$ show that
$$\sum\limits_{i=1}^{n} \mu(p_i) = \mu([a, b]).$$
Let $P$ be a finite interval subdivision of $[a, b]$ with $n$ elements. Furthermore, denote the lower and upper bounds of an element of $p \in P$ as $p_l$ and $p_u,$ respectively. Without loss of generality, order the elements of $P$ in such a way that $p_{i, u} = p_{i+1, l}.$ Now expand the $\mu$ term of the sum:
$\sum\limits_{i = 1}^n \mu(p_i) = \sum\limits_{i=1}^n p_{i,u} - p_{i, l}.$
Because $p_{i, u} = p_{i+1, l}$ for $i \in \{1, \ldots, n-1\},$ the righthand side forms a telescoping sum. Therefore
$\sum\limits_{i = 1}^n \mu(p_i) = p_{n, u} - p_{1, l}\\\sum\limits_{i = 1}^n \mu(p_i) = b - a\\\sum\limits_{i = 1}^n \mu(p_i) = \mu([a, b]).$
Given two finite interval subdivisions $P_1$ and $P_2$ of $[a, b],$ show that $P_1 \# P_2$ is finer than both $P_1$ and $P_2.$
Let $p \in P_1$ and consider each $p_i \in P_1 \# P_2.$ Let $A = \{ p_i : p \cap p_i \neq \varnothing\}.$ If $p_i \in A,$ then by definition of common refinement there is some $p_2 \in P_2$ such that $p_i = p_1 \cap p_2,$ and so it follows that $p_i \subseteq p_1.$ It follows that $\bigcup A \subseteq p.$
Consider $x \in p_1.$ Because $P_2$ is a finite interval subdivision of $[a, b],$ there is some $p_j \in P_2$ such that $x \in p_j.$ It follows that $p_1 \cap p_j \neq \varnothing,$ and so $p_j \in A.$ Therefore $p \subseteq \bigcup A.$ It follows that $p = \bigcup A,$ and so $P_1 \# P_2$ is finer than $P_1.$ The identical logic shows that $P_1 \# P_2$ is finer than $P_2.$
Show that if $P_2$ is finer than $P_1,$ then $L(f, P_1) \leq L(f, P_2).$
Let $f$ be a real function, let $[a, b]$ be an interval, and $P_1$ and $P_2$ be finite interval subdivisions of $[a, b]$ such that $P_2$ is finer than $P_1.$
Because $P_2$ is finer than $P_1,$ each $p \in P_1$ can be expressed as a union of elements in $P_2.$ Call this set $Q(p),$ where $Q(p) = \{ p_i \in P_2 : p_i \subseteq p\}.$ For each $p_i \in Q(p),$ we note that $\inf(p_i) \geq \inf(p).$ Likewise, note that $\mu(p) = \sum\limits_{p_i \in Q(p)} \mu(p_i).$ Therefore
$\inf(p)\mu(p) = \sum\limits_{p_i \in Q(p)}\inf(p)\mu(p_i)\\ \inf(p)\mu(p) \leq \sum\limits_{p_i \in Q(p)}\inf(p_i)\mu(p_i).$
We apply this fact to all elements of $P_1:$
$ L(f, P_1) = \sum\limits_{p_j \in P_1} \inf(p_j)\mu(p_j) \\ L(f, P_1) = \sum\limits_{p_j \in P_1}\sum\limits_{p_i \in Q(p_j)} \inf(p_j)\mu(p_i) \\ L(f, P_1) \leq \sum\limits_{p_j \in P_1}\sum\limits_{p_i \in Q(p_j)} \inf(p_i)\mu(p_i) $
Because $P_2$ is a refinement of $P_1,$ we can see that $\bigcup\limits_{p_j \in P_1} Q(p_j) = P_2.$ Therefore we can rearrange the sum on the righthand side:
$ \sum\limits_{p_j \in P_1}\sum\limits_{p_i \in Q(p_j)} \inf(p_i)\mu(p_i) = \sum\limits_{p_i \in P_2} \inf(p_i)\mu(p_i) $
The result follows:
$ L(f, P_1) \leq \sum\limits_{p_i \in P_2} \inf(p_i)\mu(p_i) \\ L(f, P_1) \leq L(f, P_2) $
Let $f$ be a real function defined on the interval $[a, b],$ and let $P$ be a finite interval subdivision of $[a, b].$ Show that
$$L(f, P) \leq U(f, P).$$
For a given $p_i \in P,$ it follows by definition of infimum and supremum that
$$\inf(f(p_i)) \leq \sup(f(p_i)).$$
Because $\mu(p_i) \geq 0,$ it follows from the ordering properties on $\mathbb{R}$ that
$$\inf(f(p_i))\mu(p_i) \leq \sup(f(p_i))\mu(p_i).$$
By the ordering properties on $\mathbb{R}$ again, we see that
$$\sum\limits_{p_i \in P} \inf(f(p_i)) \mu(p_i) \leq \sum\limits_{p_i \in P} \sup(f(p_i)) \mu(p_i).$$
It follows by definition of lower and upper Darboux sums that
$$L(f, P) \leq U(f, P).$$
Show that $L^*(f, [a, b]) \leq U^*(f, [a, b])$ for any given interval $[a, b],$ assuming the two are both defined.
Let $P_1$ and $P_2$ be two finite interval subdivisions of $(a, b),$ and let $P' = P_1 \# P_2.$ Since $P'$ is finer than both $P_1$ and $P_2,$ we see that
$$L(f, P_1) \leq L(f, P') \text{ and } U(f, P') \leq U(f, P_2).$$
Because $L(f, P') \leq U(f, P'),$ it follows that $L(f, P_1) \leq U(f, P_2).$ Since this is true regarding any finite interval subdivision $P_1,$ it follows that $L^*(f, [a, b]) \leq U(f, P_2).$ And again, since this is true regarding any finite interval subdivision $P_2,$ we conclude that $L^*(f, [a, b]) \leq U^*(f, [a, b]).$
Show that if $f \in \mathscr{R}([a, b]),$ and $P$ is any finite interval subdivision of $[a, b],$ then
$$L(f, P) \leq \int\limits_a^b f(x) \, dx \leq U(f, P).$$
Let $P$ be any finite interval subdivision of $[a, b].$ As shown in the prior proof,
$$L(f, P) \leq L^*(f, [a, b]) \leq U^*(f, [a, b]) \leq U(f, P).$$
Since $f \in \mathscr{R}([a, b]),$ it follows by the definition of Darboux integrability that
$$\int\limits_a^b f(x) \, dx = L^*(f, [a, b]) = U^*(f, [a, b]).$$
Therefore
$$L(f, P) \leq \int\limits_a^b f(x) \, dx \leq U(f, P).$$
Consider the irrational indicator function
$$f(x) = \left\{ \begin{array}{ll} 1 & x \in \mathbb{R} - \mathbb{Q}\\ 0 & x \in \mathbb{Q} \end{array} \right.$$
Plainly, $f(x)=1$ when $x$ is irrational and $f(x)=0$ when $x$ is rational.
Show that $f$ is not Riemann integrable on any interval.
Let $P$ be a nonempty finite interval subdivision of $[a, b].$ Consider some interval $p_i \in P.$ Then $p_i$ contains infinitely many rational and irrational numbers. It follows that $\inf(f(p_i)) = 0$ and $\sup(f(p_i)) = 1.$ Therefore $L(f, P) = 0$ and $U(f, P) = b - a.$ Since this is true for any $P,$ it follows that $L^*(f, [a, b]) = 0$ and $U^*(f, [a, b]) = b - a.$ Therefore $f$ is not Riemann-integrable on any interval.
General condition for integrability: Show that a function $f \in \mathscr{R}([a, b])$ if and only if for every $\varepsilon > 0$ there exists a finite interval subdivision $P$ of $[a, b]$ such that $U(f, P) - L(f, P) < \varepsilon.$
First, assume $f \in \mathscr{R}([a, b]).$ Then $L^*(f, [a, b]) = U^*(f, [a, b]).$ Pick some $\frac{\varepsilon}{2} > 0.$ Because $L^*(f, [a, b])$ is the least upper bound of all lower Riemann sums, there is some subdivision $P_1$ such that
$$L^*(f, [a, b]) - L(f, P_1) < \frac{\varepsilon}{2}.$$
Likewise, because $U^*(f, [a, b])$ is the greatest lower bound of all upper Riemann sums, there is some subdivision $P_2$ such that
$$U(f, P_2) - U^*(f, [a, b]) < \frac{\varepsilon}{2}.$$
Take $P' = P_1 \# P_2.$ Then
$$L(f, P_1) \leq L(f, P') \quad \text{ and } \quad U(f, P') \leq U(f, P_2).$$
Therefore
$$L^*(f, [a, b]) - L(f, P') < \frac{\varepsilon}{2} \quad \text{ and } \quad U(f, P') - U^*(f, [a, b]) < \frac{\varepsilon}{2}.$$
Rearranging this gives us
$$L^*(f, [a, b]) < L(f, P') + \frac{\varepsilon}{2} \quad \text{ and } \quad U(f, P') < U^*(f, [a, b]) + \frac{\varepsilon}{2}.$$
Now add $\frac{\varepsilon}{2}$ to the lower sum's inequality:
$$L^*(f, [a, b]) + \frac{\varepsilon}{2} < L(f, P') + \varepsilon \quad \text{ and } \quad U(f, P') < U^*(f, [a, b]) + \frac{\varepsilon}{2}.$$
Since $L^*(f, [a, b]) = U^*(f, [a, b]),$ it follows by transitivity that
$$U(f, P') < L(f, P') + \varepsilon,$$
which simplifies to the desired result:
$$U(f, P') - L(f, P') < \varepsilon.$$
Conversely, assume for every $\varepsilon > 0$ there exists a finite interval subdivision $P$ such that $U(f, P) - L(f, P) < \varepsilon.$ Since $U^*(f, [a, b]) - L^*(f, [a, b]) \leq U(f, P) - L(f, P),$ it follows that $U^*(f, [a, b]) - L^*(f, [a, b]) < \varepsilon.$ Therefore $U^*(f, [a, b]) = L^*(f, [a, b]).$
Continuity implies integrability: Show that if $f$ is continuous on $[a, b],$ then $f \in \mathscr{R}([a, b]).$
Let $f$ be a continuous real function defined on $[a, b].$ Since $[a, b]$ is compact, it follows that $f$ is uniformly continuous. Thus, for every $\frac{\varepsilon}{b-a} > 0$ there exists a $\delta > 0$ such that $|f(x) - f(y)| < \frac{\varepsilon}{b-a}$ whenever $|x - y| < \delta$ for all $x, y \in [a, b].$
Let $P$ be a finite interval subdivision of $[a, b]$ with $n$ elements such that $\mu(p_i) < \delta$ for each $p_i \in P.$ It follows that $\sup(f(p_i)) - \inf(f(p_i)) < \frac{\varepsilon}{b-a}.$ Therefore
$U(f, P) - L(f, P) = \sum\limits_{i=1}^{n} (\sup(f(p_i)) - \inf(f(p_i))) \mu(p_i)\\ U(f, P) - L(f, P) < \sum\limits_{i=1}^{n} \frac{\varepsilon}{b-a} \mu(p_i)\\ U(f, P) - L(f, P) < \varepsilon.$
It follows that $f \in \mathscr{R}([a, b]).$
Monotonicity implies integrability: Show that if $f$ is monotonic on $[a, b],$ then $f \in \mathscr{R}([a, b]).$
Let $f$ be a real function defined on $[a, b].$ Assume $f$ is monotonically increasing. Then $f(x) \leq f(y)$ whenever $x \leq y.$ Pick a finite subdivision $P$ of $n$ elements such that $\mu(p_i) = \frac{b - a}{n}.$ It follows that
$$U(f, P) - L(f, P) = \sum\limits_{i=1}^{n} (\sup(f(p_i)) - \inf(f(p_i))) \mu(p_i).$$
Because $f$ is monotonically increasing, it follows that $\sup(f(p_i)) = \inf(f(p_{i+1}))$ for all $i \in \{1, \ldots, n-1\}.$ Therefore $U(f, P) - L(f, P)$ forms a telescoping sum. Therefore
$ U(f, P) - L(f, P) = (f(b) - f(a)) \sum\limits_{i=1}^{n} \mu(p_i) \\ U(f, P) - L(f, P) = (f(b) - f(a)) \dfrac{\mu(P)}{n} \\ U(f, P) - L(f, P) = \dfrac{(f(b) - f(a))(b-a)}{n} $
Pick $\varepsilon > 0.$ By the Archimedean property, there exists an $n \in \mathbb{N}$ such that $\dfrac{(f(b) - f(a))(b-a)}{n} < \varepsilon.$ It follows that $f \in \mathscr{R}([a, b]).$