The Fundamental Theorem of Calculus from Integration Axioms

Three simple axioms for integration that force the FTC to be true.

Here are a simple set of axioms that completely characterize integration on continuous functions. I learned this from Peter Clark’s analysis notes and Spivak’s book, and decided to teach integration this way in Real Analysis this year.

I really like the idea of axiomatizing integration - it makes particularly clear what we mean by different theories of integration (Riemann/Darboux vs Lebesgue, etc) they are simply different constructions that satisfy the axioms. It also neatly separates essential properties of the integral from contingent ones giving a clear distinction between things that hold in all integration theories from things that one may hope to improve by building a new integral.

Integration Axioms

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 $\mathcal{I}(J)\to \RR$ is an integral, and denoted $f\mapsto \int_J f$ if it satisfies the following axioms:

If $k\in\RR$ 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 $f(x)\leq g(x)$ for all $x\in[a,b]$ then $\int_{[a,b]}f\leq\int_{[a,b]}g$
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 $\int_{[a,b]}f = \int_{[a,c]}f +\int_{[c,b]}f$

Part of the subtlety of the theory of integration is that the axioms do not specify what functions are integrable. They do insist constants are integrable, but otherwise just apply conditionally. This means we can construct rather uninteresting integration theories if we like - for instance we could declear that only constant functions are integrable, with their integrals given by Axiom I, and then observe that this satisfies II and III so defines an integral. This motivates an additional definition

Interesting Integral

An integral is interesting if all continuous functions on all intervals are integrable.

From now on, we will think only about interesting integrals.

Preliminaries

We begin with a useful edge case of the definition, and look at the integral over a degenerate closed interval $\{a\}=[a,a]$ :

Integral Over a Point

If $\{a\}$ is the degenerate closed interval containing a single point, and $f$ is a function which is integrable on any interval containing $a$ , then $\int_{\{a\}}f\,dx = 0$

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

$\int_{[a,v]}f\,dx =\int_{\{a\}}f\,dx +\int_{[a,v]}f\,dx$

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

$\int_{\{a\}}f\,dx=0$

The fundamental theorem relates a function $f$ to its integral $F(x)=\int_{[a,x]}f$ , so we need to be able to make sense of this expression. The proposition below shows such integration on a variable interval produces a well defined function.

The Integral as a Function

If $f\in\mathcal{I}([a,b])$ is an integrable function, then there exists a function $F\colon [a,b]\to\RR$ defined by $F(x)=\int_{[a,x]}f$

Let $f$ be integrable on the interval $[a,b]$ and $x\in[a,b]$ be a point. Then $[a,b]=[a,x]\cup [x,b]$ , so Axiom III (subdivision) implies that $f$ is integrable on $[a,x]$ (and on $[x,b]$ ) and the number $\int_{[a,x]}f$ is defined. This assignment describes a real valued function

$x\mapsto \int_{[a,x]}f$

The above proposition has a short proof because it did not claim much: we learned nothing about the nature of the area function $F(x)=\int_{[a,x]}f$ . In fact, the axioms constrain the behavior of this function much stronger than it might seem at first: below we prove a fact needed for the Fundamental Theorem but also interesting in its own right - that the integral of bounded functions are continuous.

Integrals are Continuous

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

Let $f$ be integrable and bounded by $M$ on $[a,b]$ , and set $F(x)=\int_{[a,x]}f\,dx$ . 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{align*} F(y)&=\int_{[a,y]}f\\ &= \int_{[a,x]}f+\int_{[x,y]}f\\ &=F(x)+\int_{[x,y]}f \end{align*}

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 $-M\leq f(x)\leq M$ . By subdivsion, $f$ is then integrable on every sub-interval $I\subset [a,b]$ , and by comparison (Axiom II) this implies $-M |I|\leq \int_{I}f\leq M|I|$

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

$-\epsilon = -M\delta < -M|y-x|\leq \int_{[x,y]}f\leq M|y-x|< M\delta =\epsilon$

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

Note: careful readers may have realized that we proved more than asked - the above actually shows that $F$ is uniformly continuous on $[a,b]$ . Of course, this 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!

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!

FTC I

Let $f$ be a continuous function and assume that $f$ is integrable on $[a,b]$ . Denote its area function by $F(x)=\int_{[a,x]}f$

Then $F$ is differentiable, and for all points $x\in(a,b)$ ,

$F^\prime=f$

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

Choose an arbitrary $c\in(a,b)$ . We wish to show that $F^\prime(c)=f(c)$ : that is, we need $\lim_{x\to c}\frac{F(x)-F(c)}{x-c}=f(c)$

In terms of $\epsilon$ s and $\delta$ s, this means for arbitrary $\epsilon$ we need to find a $\delta$ such that if $x$ is within $\delta$ of $c$ , this difference quotient is within $\epsilon$ 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{align*} F(x)&=\int_{[a,x]}f\\ &=\int_{[a,c]}f+\int_{[c,x]}f\\ &= F(c)+\int_{[c,x]}f \end{align*}

$\implies F(x)-F(c)=\int_{[c,x]}f$

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

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 $(f(c)-\epsilon)(x-c)\leq \int_{[c,x]}f \leq (f(c)+\epsilon)(x-c)$

Dividing through by $x-c$

$f(c)-\epsilon \leq \frac{\int_{[c,x]}f}{x-c} \leq f(c)+\epsilon$

and subtracting $f(c)$ $-\epsilon\leq \frac{\int_{[c,x]}f}{x-c}-f(c) \leq \epsilon$

We arrive at the inequality

$\left|\frac{\int_{[c,x]}f}{x-c}\right| < \epsilon$

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 $\epsilon$ of $f(c)$ . This implies the limit exists, and that $F^\prime(c)=f(c)$

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. So to uniquely specify an antiderivative its enough to give its value at one point. And we can do this!

Let $f$ be a continuous function which is integrable on $[a,b]$ . Then the function $F(x)=\int_{[a,x]}f\,dx$ 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

FTC II

Let $f$ be continuous and integrable on $[a,b]$ and let $F$ be any antiderivative of $f$ . Then $\int_{[a,b]}f = F(b)-F(a)$

Denote the area function for $f$ as $A(x)=\int_{[a,x]}f$ . 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 any two antiderivatives of the same function differ by a constant.
Thus there is some constant $C$ such that $A(x)-F(x)=C$ , or $F(x)=A(x)+C$ . Now computing,

\begin{align*} F(b)-F(a)&=\left(A(b)+C\right)-\left(A(a)+C\right)\\ &=A(b)-A(a)+(C-C)\\ &= A(b)-A(a)\\ &=A(b) \end{align*}

Where the last equality comes from the fact that $A(a)=\int_{\{a\}}f\,dx = 0$ which we proved in the preliminaries.

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 for any interesting integral your theory of integration has no choice whatsoever on how to integrate them!

Application: Integration Techniques

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

substitution

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

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 $\left(F(g(x))\right)^\prime =F^\prime(g(x))g^\prime(x)=f(g(x))g^\prime(x)$

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

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 $[g(a),g(b)]$ :

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

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!

Integration by Parts

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

Are We Cheating?

I thought analysis - especially the theory of integration - was supposed to be difficult! But these proof’s aren’t that bad: what’s the trick? The catch is that we’ve derived properties that an integral might have without actually constructing one. Perhaps there are no interesting integrals - if it were impossible to satisfy these axioms on function sets $\mathcal{I}(J)$ containing the continuous functions, then this theoretical work would be true only vacuously, and useful to no one.

So a very important matter still lies ahead, we need to show that there exists an interesting integral. A time-honored means of proving existence thoerems is construction, and so the natural path forward is to attempt and explicitly construct some assignment $\int_{-}$ obeying the axioms on all continuous functions. Any of the standard approaches (Riemann, Darboux, Lesbegue) work fine here, and you can plug-and-play whichever you like from your favorite analysis text.

This is what I appreciate so much about the axiomatic approach. It makes the goal clear before jumping into a construction: we need to find a way to satisfy these axioms. Additionally, it makes clear what doesn’t matter: we (in an undergraduate class) are primarily concerned with the integration of (piecewise) continuous functions, and have already proven that any interesting integrals completely agree on the answer. Thus we needn’t spend time fretting about small details in our definition; the final answers will be completely independent of the underlying construction after all.