$$$$

28  Definition

Highlights of this Chapter: we give an axiomatic definition of the integral, and use these axioms to prove the fundamental theorem of calculus, as well as several corollaries such as $u$-substitution and integration by parts.

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 $\mathcal{R}\subset {\mathbb{R}}^2$ below:

$${\int }_a^{b}fdx=\mathrm{Area}\left(\mathcal{R}\right)$$ $$\mathcal{R}=\{(x,y)\in {\mathbb{R}}^2\mid a\le x\le b0\le y\le f(x)\}$$

Thus a good theory of area would immediately lead to a good theory of integration. But how does one measure area? Perhaps surprisingly, this turns out to be much more difficult than it sounds - and all the difficulties weren’t worked out until the beginning of the 20th century with the advent of measure theory.

We will not need the full power of this theory here - areas under the graphs of functions are a special enough case that we can develop a theory of integration independently. But our beginnings will be the same: area is a concept we struggle to define explicitly even though we know many rules it should behave. Thus, area is a prime target to try and characterize axiomatically, and then seek an explicit definition that realizes our axioms.

What are some natural axioms for area? Perhaps the most fundamental is that area is additive: if $U,V$ are two disjoint subsets of the plane, then $$\mathrm{Area}(U\cup V)=\mathrm{Area}(U)+\mathrm{Area}(V)$$

It turns out this simple rule alone is enough to provide some axioms, which completely determine the theory of integration (for continuous functions, at least).

Remark 28.1. In fact, the definition of a measure is just a slight generalization of this: a measure $\mu$ is a function from a collection $M$ of subsets (called measurable subsets) of a space $X$ to $\mathbb{R}$ such that

• For all $A\in M$, $\mu (A)\ge 0$
• If $A_n$ is a countable set of disjoint sets in $M$ then $$\mu \left(\bigcup _i{A}_i\right)=\sum _i\mu (A_i)$$

A measure on ${\mathbb{R}}^n$ is typically also asked to be translation invariant or more generally isometry invariant: if $A,B$ are congruent subsets then $\mu (A)=\mu (B)$.

28.1 Properties of Area

We will try to produce some axioms for integration as a net area. Consider a function $f$ on an interval $[a,b]$ (picture a positive function, if you want the analogy with area to be exact). We will write

$${\int }_{[a,b]}fdx$$

For the integral of $f$. A first axiom: given a constant function $f(x)=k$, the region under its graph is a rectangle, and the area of a rectangle is base times height. Thus,

$${\int }_{[a,b]}kdx=k(b-a)$$

Next, if $f$ and $g$ are two functions on $[a,b]$ where $f(x)\le g(x)$ for all $x$, then the graph of $f$ lies underneath the graph of $g$ so since area is additive, we should require our integral to satisfy

$${\int }_{[a,b]}fdx\le {\int }_{[a,b]}gdx$$

And, finally if $c\in [a,b]$ we can divide the interval into two intervals $[a,c]$ and $[c,b]$ disjoint except for their boundary point $c$. Since area is additive, we should impose

$${\int }_{[a,b]}fdx={\int }_{[a,c]}fdx+{\int }_{[c,b]}fdx$$

These rules do not seem like much, but we will see that they are quite powerful: they completely determine the behavior of the integral for continuous functions.

28.2 Axioms

To take an axiomatic treatment seriously, we need to first make these rules more precise. The integral we have described is a function: it takes an input (a function on an interval) and gives a unique output (a real number). But what is the domain? A first hopeful thought might be “all functions” - but we might want to be wary of imposing this from the outset. After all, we have seen that real analysis allows many monstrous functions (like the function which is 0 on the irrationals and 1 on the rationals) that we might not want to - or even be able to - assign an area to!

In fact, this worry is quite real: the Mathematician Guiseppe Vitali showed in 1905 that there are subsets of $\mathbb{R}$ which cannot coherently be given a length with the example below.

Example 28.1 ($★$ Vitali Sets) Two nubmers $x$ and $y$ are said to be rationally related if $x-y\in \mathbb{Q}$. This defines an equivalence relation on the interval $[0,1]$, and we choose $V$ to be a set of equivalence class representatives (that is, $V$ contains exactly one element from each equivalence class).

We now show that $V$ has the following property: a countable number of disjoint copies of $V$ cover all of $[0,1]$. Let $R$ be the set of all rational numbers in $[-1,1]$, and note that $R$ is countable. We can then define for each $r$ the set $V_r=\{v+r\mid v\in V\}$, and note that if $r\ne s$ then $V_r$ and $V_s$ are disjoint. But, the union of all the $V_r$ contains the interval $[0,1]$ since every point $x\in [0,1]$ is equivalent to some $v\in V$ by definition, meaning there is some $r\in \mathbb{Q}$ with $v+r=x$, and for both $v$ and $x$ to lie in $[0,1]$ requires $|v-x|=|r|\le 1$ so $r\in [-1,1]$. Indeed, the union itself must be a subset of $[-1,2]$.

Now, if $V$ can be assigned a length by our measure $\mu$, there are two options either $\mu (V)=0$ or $\mu (V)\ne 0$. We will show both lead to contradiction, so in fact $V$ cannot be assigned such a length.

First, if $\mu (V)=0$ then $\mu (V_r)=0$ for all $r$ as these are just translated copies of $V$. And, as the $V_r$ are all disjoint, we can compute the total measure by adding the measure of each individually: $$\mu ({\cup }_r{V}_r)=0+0+0+\cdots =0$$ But this union *contains the unit interval $[0,1]$, which has length 1! And $1>0$ so this is a contradiction.

A similar argument prevents $V$ from having any finite measure. If $\mu (V)=ϵ$ for any positive $ϵ$, then the area of the union diverges $$\mu ({\cup }_r{V}_r)=ϵ+ϵ+\cdots =\mathrm{\infty }$$ But the union s contained in $[-1,2]$ which has length $3$, and as $3<\mathrm{\infty }$ this is a contradiction as well.

It follows from this that certain subsets of the plane cannot be given an area: in particular, if $V$ is one of Vitali’s non-measurable sets, then the set of points under the graph of

$${\chi }_V(x)=\{\begin{array}{ll}1& x\in V\\ 0& x\notin V\end{array}$$

cannot be coherently assigned an area, and thus we cannot assign a value to the integral of ${\chi }_V$ on the interval $[0,1]$.

Remark 28.2. This construction of non-measurable sets is a fancier version of the following argument that there can be no uniform probability distribution on that natural numbers: (for instance, this is the kind of thing you implicitly assume exists when you ask someone to pick a random number)

Say you want to assign each integer the same probability $ϵ$. Recall the total probability needs to be $100\mathrm{\%}$: this leads to a problem because we need to solve $$1=ϵ+ϵ+ϵ+ϵ+\cdots$$

And if $ϵ\ne 0$ this sum diverges, but if $ϵ=0$ this sum is zero, and in neither case is it $1$.

Thus, because we’ve realized that trying to assign an area to all subsets of the plane (or even all regions under the graph of a function) is too much to ask, we need to specify as part of our theory a class of ‘integrable functions’, and impose our axioms only on those.

Definition 28.1 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 $$f\mapsto {\int }_{J}fdx$$ 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 $${\int }_{[a,b]}kdx=k(b-a).$$

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

• 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 $${\int }_{[a,b]}fdx={\int }_{[a,c]}fdx+{\int }_{[c,b]}fdx$$

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.

Example 28.2 (The “Constant Integral”) The first axiom tells us that if a constant is integrable, then we must have ${\int }_{[a,b]}kdx=k(b-a)$.

So, let $\mathcal{C}$ be the set of constant functions, and define an integral on $\mathcal{C}$ exactly by this formula. Then our integral satsifeies

• Axiom I, by definition
• Axiom II: if $k<K$ and $b-a>0$ then $k(b-a)<K(b-a)$ so $$k\le K⟹\int [a,b]kdx\le {\int }_{[a,b]}Kdx$$
• Axiom III: Since for any $c$ whatsoever $b-a=b-c+c-a$ we have $k(b-a)=k(b-c)+k(c-a)$. Using this for an arbitrary $c\in (a,b)$ yields $${\int }_{[a,b]}kdx={\int }_{[a,c]}kdx+{\int }_{[c,b]}kdx$$

Thus, the assignment $\{k,[a,b]\}\mapsto k(b-a)$ defines an integral on the space of constant functions.

This integral is not particularly useful, as it is undefined for any non-constant function. One can make it slightly better by extending to an integral for linear functions $f(x)=mx+b$.

Exercise 28.1 (The “Linear Integral”) Let $\mathcal{L}(J)$ denote the set of linear functions $y=mx+b$ on the interval $J$. Show that the following rule defines an integral on $\mathcal{L}(J)$, satisfying the axioms.

$${\int }_{[u,v]}^{\mathrm{Lin}}mx+bdx=m\frac{v^2-u^2}{2}+b(v-u)$$

Here we have spelled out the domain (of functions) clearly for the proposed integral, and given a formula by fiat. This is not the usual means of constructing an integral of course, as it requires we sort of already know the answer! The usual way we will let $\mathcal{I}(J)$ be determined is to write down a particular definition for the integral (as a limiting process of some kind) and then take $\mathcal{I}(J)$ to be the set of all functions on $J$ for which that process converges.

28.2.1 Properties from the Axioms

In all of the following we assume that $\int$ is some integral satisfying the axioms above axioms.

Proposition 28.1 If $\{c\}$ is the degenerate closed interval containing a single point, and $f$ is a function which is integrable on any interval containing $a$, then $${\int }_{\{a\}}fdx=0$$

Proof. Let $f$ be integrable on the interval $[u,v]$ and $a\in [u,v]$ be a point. Without loss of generality we can in fact take $a$ to be one of the endpoints of the interval, by subdivision: if $a\in (u,v)$ then Axiom III implies that $f$ is integrable on $[u,a]$ and on $[a,v]$ as well.

Thus, we assume $f$ is integrable on $[a,v]$, and further subdivide this interval as $$[a,v]=[a,a]\cup [a,v]=\{a\}\cup [a,v]$$

By subdivision, we see that $f$ is integrable on $\{a\}$ and that

$${\int }_{[a,v]}fdx={\int }_{\{a\}}fdx+{\int }_{[a,v]}fdx$$

Subtracting the common integral over $[a,v]$ from both sides yields the result,

$${\int }_{\{a\}}fdx=0$$

Proposition 28.2 If $f\in \mathcal{I}([a,b])$ is an integrable function, then there exists a function $F:[a,b]\to \mathbb{R}$ defined by $$F(x)={\int }_{[a,x]}fdx$$

Proof. Again 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 ${\int }_{[a,x]}fdx$ is defined. This assignment describes a real valued function

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

The above proposition has a short proof because it did not claim much: we learned nothing about the nature of the area function $F$. Drawing Calculus I style pictures of $F$ makes one readily believe more should be possible - and indeed it is. With one additional assumption (that $f$ is bounded on the interval in question) we can prove a pretty strong claim - its area function is continuous!

Theorem 28.1 If $f\in \mathcal{I}([a,b])$ is a bounded integrable function, then its integral $F(x)={\int }_{[a,x]}fdx$ is continuous.

Proof. Let $f$ be integrable and bounded by $M$ on $[a,b]$, and set $F(x)={\int }_{[a.x]}fdx$. Begin by choosing an $ϵ>0$. 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 $|F(y)-F(x)|<ϵ$. Let’s unpack this a bit: if $x<y$ are two points of $[a,b]$,

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

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

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 $$-M|I|\le {\int }_{I}fdx\le M|I|$$

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

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

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

Remark 28.3. 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 $F(x)={\int }_{[a,b]}fdx$ is uniformly continuous anytime $f$ is bounded!

28.3 The Fundamental Theorem

We’ve already seen that these meager axioms hide great power: we could prove that the integral of a bounded function was continuous directly without anything else! But this is only the start of an incredible story. Here, we jump straight to the main event - and prove that these axioms characterize the fundamental theorem of calculus!

Theorem 28.2 (The Fundamental Theorem of Calculus) Let $f$ be a continuous function and assume that $f$ is integrable on $[a,b]$. Denote its area function by $$F(x)={\int }_{[a,x]}fdx$$ Then $F$ is differentiable, and for all points $x\in (a,b)$, $$F^{\prime }=f$$

Proof. Because $f$ is continuous on a closed interval, it is bounded (by the Extreme Value theorem), and so the area function $F$ is continuous (Theorem 28.1).

Choose an arbitrary $c\in (a,b)$. We wish to show that $F^{\prime }(c)=f(c)$: that is, we need $$\underset{x\to c}{lim}\frac{F(x)-F(c)}{x-c}=f(c)$$ In terms of $ϵ$s and $\delta$s, this means for arbitrary $ϵ$ we need to find a $\delta$ such that if $x$ is within $\delta$ of $c$, this difference quotient is within $ϵ$ of $f(c)$.

It will be convenient to separate this argument into two cases, depending on if $x<c$ or $c<x$ (both arguments are analogous, all that changes is whether the interval in question is $[c,x]$ or $[x,c]$). Below we proceed under the assumption that $c<x$. In this case, looking at the numerator, we see by subidvision (Axiom III) that

$$\begin{array}{rl}F(x)& ={\int }_{[a,x]}fdx\\ & ={\int }_{[a,c]}fdx+{\int }_{[c,x]}fdx\\ & =F(c)+{\int }_{[c,x]}fdx\end{array}$$ $$⟹F(x)-F(c)={\int }_{[c,x]}fdx$$

Thus the real quantity of interest is this integral over $[c,x]$. Choose $ϵ>0$. Since $f$ is continuous, there is some $\delta >0$ where $|x-c|<\delta$ implies $|f(x)-f(c)|<ϵ$. Equivalently, for all $x\in [c-\delta ,c+\delta ]$ we have $$f(c)-ϵ<f(x)<f(c)+ϵ$$

By subdivison (Axiom III), we know that $f$ is integrable on $[c,x]$, and so by comparison (Axiom II) and the area of rectangles (Axiom I) we have $$(f(c)-ϵ)(x-c)\le {\int }_{[c,x]}fdx\le (f(c)+ϵ)(x-c)$$

Dividing through by $x-c$

$$f(c)-ϵ\le \frac{{\int }_{[c,x]}fdx}{x-c}\le f(c)+ϵ$$

and subtracting $f(c)$ $$-ϵ\le \frac{{\int }_{[c,x]}fdx}{x-c}-f(c)\le ϵ$$

We arrive at the inequality

$$\left|\frac{{\int }_{[c,x]}fdx}{x-c}\right|<ϵ$$

But the numerator here is none other than $F(x)-F(c)$! So, we’ve done it: for all $x>c$ with $|x-c|<\delta$, we have the difference quotient within $ϵ$ of $f(c)$. This implies the limit exists, and that $$F^{\prime }(c)=f(c)$$

Exercise 28.2 Write out the case for $x<c$ following the same logic as above.

This tells us that the area function of $f$ is one of its antiderivatives! The theory of area is the inverse of the theory of rates of change. But which antiderivative? The mean value theorem assures us that the collection of all possible antiderivatives are easy to understand - any two differ by a constant (Corollary 24.3). So to uniquely specify an antiderivative its enough to give its value at one point. And we can do this!

Corollary 28.1 Let $f$ be a continuous function which is integrable on $[a,b]$. Then the function $F(x)={\int }_{[a,x]}fdx$ is uniquely determined as the antiderivative of $F$ such that $F(a)=0$.

This connection of integration with antidifferentiation and the classification of antiderivatives has a useful corollary for computation, which is often called the second fundamental theorem

Theorem 28.3 (FTC Part II) Let $f$ be continuous and integrable on $[a,b]$ and let $F$ be any antiderivative of $f$. Then $${\int }_{[a,b]}fdx=F(b)-F(a)$$

Proof. Denote the area function for $f$ as $A(x)={\int }_{[a,x]}fdx$. Then the quantity we want to compute is $A(b)$.

Now, let $F$ be any antiderivative of $f$. The first part of the fundamental theorem assures us that $A$ is an antiderivative of $f$, and so Corollary 24.3 implies there is some constant $C$ such that $A(x)-F(x)=C$, or $F(x)=A(x)+C$. Now computing,

$$\begin{array}{rl}F(b)-F(a)& =(A(b)+C)-(A(a)+C)\\ & =A(b)-A(a)+(C-C)\\ & =A(b)-A(a)\\ & =A(b)\end{array}$$

Where the last equality comes from the fact that $A(a)={\int }_{\{a\}}fdx=0$ (Proposition 28.1).

We are going to have a lot of endpoint-subtraction going on, so its nice to have a notation for it.

Definition 28.2 Let $[a,b]$ be an interval and $f$ a function. We write $$f{|}_{[a,b]}=f(b)-f(a)$$ as a shorthand for evaluation at the endpoints.

Remark 28.4. It is often convenient when doing calculations to introduce a slight generalization of the integral, which depends on an oriented interval. A natural notation for this is already in use in calculus, using the top and bottom of the integral sign for the locations of the ‘ending’ and ‘starting’ bound respectively:

$${\int }_a^{b}fdx=\{\begin{array}{ll}{\int }_{[a,b]}fdx& a\le b\\ -{\int }_{[b,a]}fdx& a\ge b\end{array}$$

Show that using this notation, we have a clean generalized subdivision rule: for **all points $a,b,c$ irrespective of their orderings, $${\int }_a^{b}fdx={\int }_a^{c}fdx+{\int }_c^{b}fdx$$

This notation helps shorten the computations in the proof of the fundamental theorem (at the expense of adding one new thing to remember).

The fundamental theorem of calculus is a beautiful result for many different reasons. One of course, is that it forges a deep connection between the theory of areas and the theory of derivatives - something missed by the ancients and left undiscovered until the modern advent of the calculus. But second, it shows how incredibly constraining our simple axioms are: we did not prove the fundamental theorem of calculus for any particular definition of the integral (Riemann’s, Lebesgue’s, Darboux’s, etc) but rather showed that if continuous functions are integrable then your theory of integration has no choice whatsoever on how to integrate them!

We’ve seen above that it is possible to construct explicit models of the integration axioms by artificially limiting the domain of integrable functions (to constants, or linear functions for instance). But even these are constrained by the Fundamental theorem: since our example functions were continuous, there was really no choice at all!

The remaining question is of course, is there a theory of integration where all continuous functions are integrable? We will call any definition of integration interesting if it is general enough to include all continuous maps.

Definition 28.3 An integral $\int$ is interesting if $\mathcal{I}(J)$ contains the continuous functions on $J$, for each closed interval $J\subset \mathbb{R}$.

28.3.1 Application: Integration Techniques

Given the fundamental theorem holds for continuous functions, its immediate to build up a strong theory of integration

Theorem 28.4 Let $\int$ be an interesting integral, and $f$ be continuous, $g$ be continuously differentiable on $[a,b]$. Then $${\int }_{[a,b]}(f\circ g)g^{\prime }dx={\int }_{[g(a),g(b)]}fdx$$

Proof. Because $g$ is differentiable it is continuous, so $f\circ g$ is the composition of continuous functions, which is continuous. And as the product of continuous functions is continuous, $f(g(x))g^{\prime }(x)$ is also continuous. Thus this function is integrable on $[a,b]$.

By the Fundamental Theorem, we can evaluate this by antidifferentiation: let $F(x)$ be any antiderivative of $f$, then the chain rule gives $${(F(g(x)))}^{\prime }=F^{\prime }(g(x))g^{\prime }(x)=f(g(x))g^{\prime }(x)$$

Using this antiderivative yields $${\int }_{[a,b]}(f\circ g)g^{\prime }dx=F(g(x)){|}_{[a,b]}=F(g(b))-F(g(a))$$

A crucial but seemingly simple observation is to note this is the same value one would get by evaluating the function $F$ on the endpoints of the interval $[g(a),g(b)]$:

$$F(g(x)){|}_{[a,b]}=F(x){|}_{[g(a),g(b)]}$$

And as $F^{\prime }=f$, this second expression is exactly what one would get from integrating $f$ on the interval $[g(a),g(b)]$ using the Fundamental Theorem.

Similarly without any further theory we can construct the other main integration technique of the calculus: integration by parts!

Theorem 28.5 Let $\int$ be an interesting integral, and $f,g$ be two continuously differentiable functions on $[a,b]$. Then $${\int }_{[a,b]}f{g}^{\prime }dx=fg{|}_{[a,b]}-{\int }_{[a,b]}{f}^{\prime }gdx$$

Exercise 28.3 Prove this using a similar strategy as to what we did above, but using the product rule instead of the chain rule as a starting point.

28.4 The Work to Come

Its pretty incredible that even though we did not set out to uniquely define the integral via our axioms, they manage to completely determine the integral for any function $f$ which is (1) continuous and (2) integrable.

A natural and important question then is which continuous functions are integrable? (Or, in our terminology above, is there an interesting integral at all?). As soon as we know $f$ is integrable, we get the existence of the area function $F$ via subdivision and the proof of the Fundamental Theorem goes through without issue. But how does one construct an integral where one can actually prove all continuous functions are integrable?

28.4.1 Failure of the ‘Calculus Integral’

The example below shows this is actually a difficult problem to answer: one might try to define the integral using a right endpoint Riemann sum (as one would in a calculus course): from this definition one can prove that all continuous functions are integrable, but then when one goes to try and verify the axioms, one finds this is actually not an integral at all!

Definition 28.4 (The “Calculus Integral”) Let $f$ be a function defined on the interval $[a,b]$, and $N$ a natural number. With $\mathrm{\Delta }=(b-a)/N$ we define the (right endpoint) Riemann sum for $f$ with $N$ subintervals is $$\sum _{i=1}^{n}f(a+i\mathrm{\Delta })\mathrm{\Delta }$$

Such a function $f$ is Calculus - integrable if the limit of its Riemann sums exists as the number of subintervals goes to infinity. In this case, the Calculus Integral is defined as the limiting value: $${\int }_{[a,b]}^{\text{Calc}}f(x)dx=\underset{N\to \mathrm{\infty }}{lim}\sum _{i=1}^{N}f(a+i\mathrm{\Delta }x)\mathrm{\Delta }x$$

It turns out that while this definition seems unproblematic when applied to elementary functions seen in a calculus course, it has some rather surprising behavior in general: and taking it as our definition would destroy some of the familiar pillars of integration theory!

To find the trouble, we need to look away from the well behaved functions, and investigate the integrability of some monsters. Here we’ll look at the characteristic function of the rationals.

$$\chi (x)=\{\begin{array}{ll}1& x\in \mathbb{Q}\\ 0& x\notin \mathbb{Q}\end{array}$$

Example 28.3 Let $\chi$ be the above function, equal to $1$ on the rationals and $0$ on the irrationals. Then $f$ is Calculus - Integrable on every interval of the form $[0,a]$ but $$a\in \mathbb{Q}⟹{\int }_{[0,a]}^{\text{Calc}}\chi dx=a$$ $$a\notin \mathbb{Q}⟹{\int }_{[0,a]}^{\text{Calc}}\chi dx=0$$

In fact, its worse than this! As a natural extension of the above, one can show the following:

Exercise 28.4 The function $\chi$ is Calculus-Integrable on any closed interval in $\mathbb{R}$, and the resulting value is:

• The length of the interval, when both endpoints are rational.
• Zero, when one endpoint is rational and the other irrational

This has a very important consequence to our theory: our proposed definition of the integral violates the subdivision rule.

Exercise 28.5 The subdivison rule $${\int }_{[a,b]}^{\text{Calc}}fdx={\int }_{[a,c]}^{\text{Calc}}fdx+{\int }_{[c,b]}^{\text{Calc}}fdx$$ is false for the integral as defined in Definition 28.4.
Hint: look at the interval $[0,2]$, and note $0<\sqrt{2}<2$.