$$$$

25  Axioms

Highlights of this Chapter: we give an axiomatic definition of integration, and use these axioms to propose a formula for computing integrals using sums.

The integral is meant to measure the (net) area. When $f$ is positive, for instance, we learn in calculus that ${\int }_a^{b}fdx$ should be the area under $f$ between $a$ and $b$. That is, it should be the area of the region $\mathcal{R}=\{(x,y)\in {\mathbb{R}}^2\mid a\le x\le b0\le y\le f(x)\}$. But how does one measure area? Perhaps surprisingly, this turns out to be much more difficult than it sounds, and a full resolution took until the beginnings of the 20th century, with the advent of measure theory.

n b Happily, we do not need the full generalities of this theory to introduce the single variable integration theories one first meets in analysis, and we can do something conceptually simpler. Following the example we set with our introduction to the elementary functions, we seek an axiomatic description of what integration is ‘about’, before we demand a procedure to calculate it. In this chapter we carry this out: proposing a simple set of axioms that anything worthy of being called an integral must satisfy, and then use these to produce a formula which can calculate integrals in every case we will need.

25.1 Characterizing Integration

If we are to propose axioms for integration, we must first think carefully about what we expect an integral to be. Looking back to calculus, we recall integration to be a procedure taking a function $f$ and a closed interval $[a,b]$ and producing a number, which we denote ${\int }_a^{b}f(x)dx$. That is, at its most basic level, an integral is a function taking an interval and a function defined on that interval to a number. To remind ourselves of the calculus notation (while still maintaining something distinct, to help us stay formal), we write this function as

$$\{f,I\}\mapsto {\int }_{I}f\in \mathbb{R}$$

Fixing the interval $I$, we can think of ${\int }_I\{\}$ as a map assigning real valued functions on $I$ to numbers, and fixing $f:\mathbb{R}\to \mathbb{R}$, we may think of ${\int }_{\{\}}f$ as a map assigning intervals to numbers (the integral of $f$ over that interval). From each of these perspectives there are some obvious properties that anything worthy of being called an integral must satisfy:

• It gets rectangles right: If $f(x)=k\in \mathbb{R}$ is a constant function, its graph is a horizontal line, which encloses a rectangle over an interval $I=[a,b]$. But we know the area of a rectangle should be its base times its height! So ${\int }_{[a,b]}k$ should be $k(b-a)$.

• It gets inequalities right If $f(x)\le g(x)$ on an interval $I$, the graph of $f$ lies fully underneath the graph of $g$, so the net area under $f$ cannot be greater than the area under $g$. That is, $f\le g$ on $I$ should imply ${\int }_{I}f\le {\int }_{I}g$.

• Area is additive Given an interval $I=[a,b]$ one should be able to find the net area under $f$ by breaking the interval into pieces, finding the area under $f$ on each piece, and adding up the results. This is simplest when we consider just two pieces: for $c\in [a,b]$ we should have ${\int }_{[a,b]}f={\int }_{[a,c]}f+{\int }_{[c,b]}f$.

Since these are properties we clearly want an integral to have, we might wish to take them directly as axioms: and basically that is what we will do! But there is a slight subtlety we need to contend with: the collection of all real valued functions contains some wild beasts, and we shouldn’t be so hasty as to assume that it makes sense to measure the area under the curve of every function (indeed, for our theory, it will turn out that the *area under the function which is 1 on the rationals and zero on the irrationals, is undefined). So, instead of insisting these hold for all functions, we will formalize an integral as declaring a subset of functions to be integrable, and only imposing these axioms on that subset.

Definition 25.1 (Axiomatic Integration) For any closed interval $J=[a,b]$ we denote by $\mathcal{I}(J)$ the set of integrable functions on $J$. Then a collection of functions $\mathcal{I}(J)\to \mathbb{R}$ is an integral, and denoted $$f\mapsto {\int }_{J}f$$ if it satisfies the following axioms:

• If $k\in \mathbb{R}$ then $f(x)=k$ is an element of $\mathcal{I}([a,b])$ for any interval $[a,b]$ and $${\int }_{[a,b]}k=k(b-a).$$

• If $f,g\in \mathcal{I}([a,b])$ and $f(x)\le g(x)$ for all $x\in [a,b]$ then $${\int }_{[a,b]}f\le {\int }_{[a,b]}g$$

• If $[a,b]$ is an interval and $c\in (a,b)$, then $f\in \mathcal{I}([a,b])$ if and only if $f\in \mathcal{I}([a,c])$ and $f\in \mathcal{I}([c,b])$. Furthermore, in this case their values are related by $${\int }_{[a,b]}f={\int }_{[a,c]}f+{\int }_{[c,b]}f$$

Note these axioms do not aim to uniquely specify an integral, but rather to delineate properties that anything worthy of being called an integral must have. Over the past two centuries, there has been quite a lot of work done studying the possible different integrals - all the different functions that satisfy this definition. However through all this work a beautiful story has emerged: for all their differences, all the various constructions give exactly the same answers for the continuous functions, and those answers can be calculated directly from the axioms themselves! Its this streamlined, abstract thread that we will pursue in this course.

25.1.1 $⧫$ Improper Integrals

We have axiomatized the integral for bounded functions on closed intervals, but the definition can be naturally extended to unbounded intervals and (certain) unbounded functions via limits.

Definition 25.2 (Improper Integrals: Unbounded Intervals) The integral of a bounded function $f$ on a ray $[a,\mathrm{\infty })$ is defined as a limit of its integrals over growing closed intervals $${\int }_{[a,\mathrm{\infty })}f:=\underset{b\to \mathrm{\infty }}{lim}{\int }_{[a,b]}f$$ with the analogous definition for rays $(-\mathrm{\infty },b]$. The integral over the entire real line is defined by taking each endpoint to $\pm \mathrm{\infty }$ separately

$${\int }_{\mathbb{R}}f:=\underset{\begin{array}{c}a\to -\mathrm{\infty }\\ b\to \mathrm{\infty }\end{array}}{lim}{\int }_{[a,b]}f$$

That is, both orders $\underset{a\to -\mathrm{\infty }}{lim}\underset{b\to \mathrm{\infty }}{lim}$ and $\underset{b\to \mathrm{\infty }}{lim}\underset{a\to -\mathrm{\infty }}{lim}$ exist and are equal.

Definition 25.3 (Improper Integrals: Unbounded Functions) If $f$ is defined on $(a,b]$ and integrable on each subinterval $[t,b]$ for $t>a$, we define the improper integral on $(a,b]$ as a limit $${\int }_{(a,b]}f=\underset{t\to a^{+}}{lim}{\int }_{[t,a]}f$$ Similarly for functions unbounded on $[a,b)$ but bounded on each $[a,t]\subset [a,b)$, we define ${\int }_{[a,b)}f=\underset{t\to b^{-}}{lim}{\int }_{[a,t]}f$.

If a function defined on $[a,b]$ is unbounded in a neighborhood of some $c\in (a,b)$ but is integrable on every subinterval missing $c$, we say the integral on $[a,b]$ exists if and only if the integral on $[a,c)$ and $(c,b]$ both exist, in which case we define it as equal to their sum.

25.2 The Integral as a Function

WHile our definition strictly only gives the value of integrals on one interval at a time, the axioms let us do a bit of work and define the Integral Function $F:[a,b]$ of an integrable function $f:[a,b]$ and study its properties.

Proposition 25.1 (The Integral as a Function) If $f\in \mathcal{I}([a,b])$ is an integrable function, then there exists a function $F:[a,b]\to \mathbb{R}$ defined by $$F(x)={\int }_{[a,x]}f$$

Proof. This is just subdivision at work: for any $x\in [a,b]$ we may write $$[a,b]=[a,x]\cup [x,b]$$. Then Axiom III implies that $f$ is integrable on $[a,x]$, and so the number ${\int }_{[a,x]}f$ is defined. This assignment describes a real valued function

$$x\mapsto {\int }_{[a,x]}f$$

We can learn much from the axioms about this integral function: they imply that when a function is integrable, its integral is continuous!

Theorem 25.1 If $f\in \mathcal{I}([a,b])$ is a bounded integrable function, then its integral $F(x)={\int }_{[a,x]}f$ is continuous.

Proof. Let $f$ be integrable and bounded by $M$ on $[a,b]$, and set $F(x)={\int }_{[a.x]}f$. Begin by choosing an $\epsilon >0$. We will prove something even strong than asked - that $f$ is uniformly continuous by finding a $\delta >0$ where if $|y-x|<\delta$ we have $|F(y)-F(x)|<\epsilon$. Let’s unpack this a bit: if $x<y$ are two points of $[a,b]$,

$$F(y)-F(x)={\int }_{[a,y]}f-{\int }_{[a,x]}f$$

But subdivision (Axiom III) implies $$\begin{array}{rl}F(y)& ={\int }_{[a,y]}f\\ & ={\int }_{[a,x]}f+{\int }_{[x,y]}f\\ & =F(x)+{\int }_{[x,y]}f\end{array}$$

Thus $F(y)-F(x)$ is just the integral of $f$ on the subinterval $[x,y]\subset [a,b]$. Because $f$ is bounded by $M$ we know $-M\le f(x)\le M$. By subdivsion, $f$ is then integrable on every sub-interval $I\subset [a,b]$, and by comparison (Axiom II) this implies $$-M|I|\le {\int }_{I}f\le M|I|$$

So, we choose $\delta =\epsilon /M$. This immediately yields what we want, as if $|y-x|<\delta$,

$$-\epsilon =-M\delta <-M|y-x|\le {\int }_{[x,y]}f\le M|y-x|<M\delta =\epsilon$$

Thus $|F(y)-F(x)|=|{\int }_{[x,y]}f,dx|<\epsilon$.

Remark 25.1. Of course, the proven result is not really stronger than what was asked, since we began on a closed interval, and we know that continuous on a closed interval implies uniformly continuous.

However, if you look carefully at the proof you see we nowhere used that the original domain was a closed interval! So what we have really proven is that the area function $F(x)={\int }_{[a,b]}f$ is uniformly continuous anytime $f$ is bounded!

As defined, the integral is only a function of $x$ for $x$ greater than the chosen starting endpoint. While this is what is desired in many applications, its also useful to be able to extend the definition to make sense for $x$ below the starting point as well. This is the integral often met in calculus, which we call the oriented integral as it changes sign when the interval from $a$ to $x$ is traced backwards.

Definition 25.4 (The Oriented Integral) Given a function $f$ which is integrable on the interval between $a$ and $b$, the oriented integral of $f$ is denoted ${\int }_a^{b}f$ and equals $${\int }_a^{b}f=\{\begin{array}{ll}{\int }_{[a,b]}f& a\le b\\ -{\int }_{[b,a]}f& b<a\end{array}$$

Corollary 25.1 Given any function $f$ which is integrable on an interval $[a,b]$, for any $c\in [a,b]$ the oriented integral defines a function $[a,b]\to \mathbb{R}$ by $$x\mapsto {\int }_c^{x}f$$

25.3 How to Compute?

Just as we can use these axioms to prove theorems about any possible integral, we can also use it to compute values. We content ourselves with a simple example here, to illustrate, and follow with a more general discussion.

25.3.1 Integrating f(x)=x

Given the function $f(x)=x$, lets temporarily assume that we have some integral $\int$ satisfying the axioms above, and also that this integral considers $x$ to be integrable on the interval $[0,1]$. From here, it turns out the axioms unambiguously determine its value, allowing us to prove

Theorem 25.2 If $\int$ is any integral for which $f(x)=x$ is integrable, then ${\int }_{[0,1]}x=\frac{1}{2}$.

We proceed in steps. First, note that on the interval $[0,1]$ we know (by definition) that $f(x)=x$ is between $0$ and $1$. So, by axiom 2 $$0\le x\le 1⟹{\int }_{[0,1]}0\le {\int }_{[0,1]}x\le {\int }_{[0,1]}1$$

The upper and lower bounds here are constants, and so we can evaluate their integrals by axiom 1, giving

$$0\le {\int }_{[0,1]}x\le 1$$

This of course is a pretty terrible estimate; but we can easily use the same ideas to do better! Indeed, since we’ve assumed that $x$ is integrable on $[0,1]$ axiom 3 ensures us that its also integrable on $I_1=[0,1/2]$ and $I_2=[1/2,1]$ (or any other subintervals, for that matter). On these smaller intervals, we have better understanding of the behavior of $f(x)$, and thus better estimates:

$$x\in I_1⟹0\le f(x)=x\le 1/2x\in I_2⟹1/2\le f(x)=x\le 1$$

To each (constant!) bound we can apply axiom 1 to integrate, it and then apply axiom 2 to ensure the inequality is preserved. Thus

$$0={\int }_{I_1}0\le {\int }_{I_1}x\le {\int }_{I_1}\frac{1}{2}=\frac{1}{4}$$ $$\frac{1}{4}={\int }_{I_2}\frac{1}{2}\le {\int }_{I_2}x\le {\int }_{I_2}1=\frac{1}{2}$$

Adding these two intequalities gives $\frac{1}{4}\le {\int }_{I_1}x+{\int }_{I_2}x\le \frac{3}{4}$, but then using axiom 2 we recognize that the sum in the middle is just the subdivision of ${\int }_{[0,1]}x$ at $1/2$. Thus we have

$$\frac{1}{4}\le {\int }_{[0,1]}x\le \frac{3}{4}$$

Of course we can do better. If we divide $[0,1]$ into $N$ intervals of equal length $1/N$ (say $I_i=[(i-1)/N,i/N]$) then by using the subdivision axiom inductively, we find $f(x)=x$ is integrable on each, and

$${\int }_{[0,1]}x={\int }_{I_1}x+{\int }_{[1/n,1]}x={\int }_{I_1}x+{\int }_{I_2}x+{\int }_{[2/n,1]}x=\cdots =\sum _{i=1}^N{\int }_{I_i}x$$

On each of these intervals $I_i$ we can easily bound $f(x)=x$:since its monotone increasing, its smallest value is its left endpoint and its largest value is its right endpoint: $(i-1)/N\le x\le i/N$. Thus, by axioms 1 and 2

$$\frac{i-1}{N^2}={\int }_{I_i}\frac{i-1}{N}\le {\int }_{I_i}x\le {\int }_{I_i}\frac{i}{N}=\frac{i}{N^2}$$

Summing up these inequalities, and recalling ${\int }_{[0,1]}x=\sum _{i=0}^N{\int }_{I_i}x$, we find $$\sum _{i=1}^N\frac{i-1}{N^2}\le {\int }_{[0,1]}x\le \sum _{i=1}^N\frac{i}{N^2}$$

This inequality must hold for all values of $N$, which turns out to be enough to completely fix the value of ${\int }_{[0,1]}x$

Exercise 25.1 (Integrating $x$)  

• Call the lower estimate $L_N$ and the upper estimate $U_N$. Prove that as $N\to \mathrm{\infty }$, so long as one of these sums converges so does the other, and their values are equal. Thus, the constant sequence ${\int }_{[0,1]}x$ is squeezed between $L_N$ and $U_N$, so in the limit must also take their common value!

• Next, prove that $U_N\to \frac{1}{2}$ as $N\to \mathrm{\infty }$ *Hint: use previous homework, where we did summation by parts to find a formula for $\sum _{i=1}^{N}i=1+2+3\cdots +N$.

Throughout this entire calculation we’ve only used the axioms, and the assumption that $x$ is integrable. Thus, we’ve proven a pretty strong result: no matter how you try to precisely define integrals, there is an unambiguous choice for the value of ${\int }_{[0,1]}x$. If its defined at all, it must equal exactly $1/2$.

Exercise 25.2 (Integrating $x$, Part II) Extend the above result to show that if $x$ is integrable on $[a,b]$, then $${\int }_{[a,b]}x=\frac{b^2}{2}-\frac{a^2}{2}$$ (Its OK if you assume in your proof that $0<a<b$ to cut down on worrying about negative numbers, the proof of the general case is not much more work)

Hint: first, generalize the work we did together in the book above, from the interval $[0,1]$ to a general interval $[0,c]$ and prove that ${\int }_{[0,c]}x=\frac{c^2}{2}$. Then use the fact that you know this for all $c>0$ and the subdivision axiom to get what you want.*

Exercise 25.3 (Integrating $x^2$) Prove that if $x^2$ is integrable on $[0,c]$ that its value must be $${\int }_{[0,c]}{x}^2=\frac{c^3}{3}$$

Use this to deduce that for any interval $[a,b]$, (feel free to just do the case $a,b\ge 0$) $${\int }_{[a,b]}{x}^2=\frac{b^3}{3}-\frac{a^3}{3}$$

Hint: follow the similar process to what we did above: using the axioms to bound by sums, and then using the summation by parts formula from earlier in the course to calculate the limit

25.3.2 The Darboux Integral

We can take this kind of reasoning even farther, and propose a means of calculating integrals of arbitrary functions, whenever they are forced to exist by the axioms. The main idea is the same: to estimate an integral we use subdivision to break the domain into smaller and smaller pieces, and then use inequalities to get better and better estimates of $f$ on each piece. It will be useful to give a name to such a subdivision of the interval: a partition.

Definition 25.5 (Partitions) A partition of the interval $I=[a,b]$ is a finite ordered set $P=\{t_0,t_1,\dots ,t_n\}$ with $a=t_0<t_1<\dots <t_{n-1}<t_N=b$.

• $N$ is called the length of the partition
• We write $P_i=[t_i,t_{i+1}]$ for the $i^{th}$ interval of $P$, and $|P_i|=(t_{i+1}-t_i)$ for its width.
  
• The maxwidth of $P$ is the maximal width of the $P$’s intervals, $\mathrm{maxwidth}(P)=\underset{0\le i<N}{max}\{|P_i|\}$.
• The set of all partitions on a fixed interval $I$ is denoted ${\mathcal{P}}_I$.$${\mathcal{P}}_I=\{P:P\text{ is a partition of }I\}$$

On each partition, we can try to find bounds on the value of our function $f$. We no longer know the lower bound will occur at the left side and the upper bound at the right side, or even that the lower and upper bounds are even achieved by some points in the domain (we do know this when $f$ is continuous, by the extreme value theorem, but we can’t say much for a general $f$). But, even when the max and min do not exist the infimum and supremum always do (for any boounded nonempty set,by completeness).

Definition 25.6 (Upper and Lower Sums) Let $f$ be a function, and $P$ a partition of the closed interval $I$. For each segment $P_i=[t_i,t_{i+1}]$, we define $$m_i=\underset{x\in P_i}{inf}\{f(x)\}{M}_i=\underset{x\in P_i}{sup}\{f(x)\}$$

We then define the upper sum $U_I(f,P)$ and the lower sum $L_I(f,P)$ as

$$L_I(f,P)=\sum _{0\le i<N}{m}_i|P_i|$$ $$U_I(f,P)=\sum _{0\le i<N}{M}_i|P_i|$$s

Using the subdivision axiom on our partition, we see that if $f$ is integrable, we must have

$$\sum _{0\le i<N}{m}_i|P_i|\le \sum _{0\le i<N}{\int }_{I_i}f\le \sum _{0\le i<N}{M}_i|P_i|$$

This is great in that given any partition we can get some nice bounds on the possible values of our integral ${\int }_{I}f$, but we can’t focus on a single partition and need to think more generally. Let’s look at each inequality separately. What does it mean that $L_I(f,P)$ is less than or equal to the integral, for any partition? Turning this around, we are saying that ${\int }_{I}f$ is an upper bound for the lower sums. So this upper bound can’t be less than the least upper bound, meaning

$$\underset{P\in {\mathcal{P}}_I}{sup}{L}_I(f,P)\le {\int }_{I}f$$

Similarly, the second inequality tells us that ${\int }_{I}f$ is a lower bound for the set of all possible upper sums. It must then be less than or equal to the greatest lower bound, so

$${\int }_{I}f\le \underset{P\in {\mathcal{P}}_I}{inf}{U}_I(f,P)$$

These quantities prove extremely useful estimates, so we will give them a name:

Definition 25.7 (Upper and Lower Integrals) Let $f$ be a function on the closed interval $I$. Then we define the upper integral $U_I(f)$ and the lower integral $L_I(f)$ as $$U(f)=\underset{P\in {\mathcal{P}}_I}{inf}\{U_I(f,P)\}$$ $$L(f)=\underset{P\in {\mathcal{P}}_I}{sup}\{L_I(f,P)\}$$

It may happen that for a given function $f$, the upper integral and lower integral are not equal to one another. In this case, the best bounds we could think to construct from the axioms (looking over all possible partitions) aren’t enough to nail down a value for the integral of the function.

Exercise 25.4 Prove the characteristic function of the rationals has $U(\chi )=1$ and $L(\chi )=0$ on the interval $[0,1]$.

But sometimes we will find that $U(f)=L(f)$. Since we know $L(f)\le {\int }_{I}f$ and ${\int }_{I}f\le U(f)$ whenver the integral exists, this then uniquely specifies the value: if $f$ is integrable under any possible definition of the integral, its value must be this common quantity $U(f)=L(f)$ we computed here. Taking this one step further actually provides a really reasonable potential definition of an integral: given a function $f$ we compute the upper and lower sums: if they are not equal, we declare the function is not integrable. But if they are equal, we define the integral to be their common value (as we must, as this is the only option!). This definition is due to Darboux, and is called the Darboux integral.

Definition 25.8 (Darboux Integral) Let $f$ be a function on the closed interval $I$. Then $f$ is Darboux-Integrable on $I$ if $U(f)=L(f)$, and we define the integral to be this common value: $${\int }_{[a,b]}f=U(f)=L(f)$$

Because we used the axioms (and only the axioms) to come up with this construction, its perhaps not surprising that the resulting thing actually does satisfy the axioms, so is an example of an integral. But since we are working with limit-like quantities (infima and suprema) we should to be careful and actually check nothing goes wrong. This is the content of the next (optional) chapter.