Axiomatic Integration: What the Axioms Determine

They pin the integral down exactly as far as Riemann integrability, and no further.

Second in a series on axiomatizing integration. The axioms, and the fundamental theorem they force, are in the first post. This post measures exactly how much freedom the axioms leave, and the third and fourth each try to remove some of it by adding a new axiom. The measure-zero material below came out of conversations with David Cheng.

Recall the setup from the previous post: an integral assigns to each closed interval $J$ a set $\mathcal{I}(J)$ of integrable functions together with a map $\int_J\colon\mathcal{I}(J)\to\RR$, subject to three axioms - rectangle areas ($\int_{[a,b]}k=k(b-a)$), comparison ($f\leq g\implies\int f\leq\int g$), and subdivision ($f$ is integrable on $[a,b]$ exactly when it is on $[a,c]$ and $[c,b]$, and then the values add). From these alone we proved the fundamental theorem:

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

We read that as a computational tool, since it’s how one actually evaluates an integral. But I want to read it a second way here:

The left hand side mentions an integral. The right hand side does not.

Nothing on the right knows which theory of integration we began with. Antiderivatives are a matter of differentiation alone, and any two of them differ by a constant that cancels in the subtraction. So if two different theories of integration both accept a continuous $f$, they have no choice but to return the same number.

That’s really a statement about theories, not about functions, so let’s give the property a name:

A word about that boundedness hypothesis, which I’ll keep for the rest of this series. Every tool below is built out of suprema and infima of $f$ over little subintervals, so it needs $f$ bounded to say anything at all. Unbounded functions are not a small extension of this story but a genuinely separate one - the axioms give lower bounds on such an integral and no upper bounds whatsoever - and I won’t take them up here.

With that in hand, the fundamental theorem says: every continuous function is axiomatically integrable (continuous functions on a closed interval being bounded automatically). Riemann, Lebesgue, gauge - however differently they are built, they cannot disagree about a continuous function.

So our question for this post is simply: what else?

The fundamental theorem itself is no help here, since it needs continuity in an essential way - FTC I’s proof pins the difference quotient using continuity of $f$ at the point in question - so the moment $f$ has a single bad point that argument has nothing to say. We need a technique which presupposes nothing at all about $f$.

What We Can Learn About a Mystery Integral

Let $\int^?$ be a mystery integral. We know absolutely nothing about it: not how it was constructed, not which functions it accepts. All we know is that it obeys the three axioms, and that it happens to accept some bounded function $f$ on $[a,b]$. How much can we say about the number $\int^?_{[a,b]}f$?

Quite a lot, as it turns out, and the first bound is nearly free. Because $f$ is bounded it has an infimum and a supremum, say $m=\inf_{[a,b]}f\hspace{1cm}M=\sup_{[a,b]}f$ and $f$ is trapped between these two constants: $m\leq f(x)\leq M$ for all $x$. The rectangle axiom says constants are integrable, so comparison applies to both inequalities, and then the rectangle axiom evaluates the outer two:

$m(b-a)\;\leq\;\int_{[a,b]}^?f\;\leq\;M(b-a)$

The mystery integral is pinned inside a box before we have asked a single question about how it was built.

Now subdivision lets us do far better, because the same argument runs on any subinterval. Chop $[a,b]$ into finitely many pieces $P=P_1\cup P_2\cup\cdots\cup P_n$ and record how high and low $f$ goes on each one: $m_i=\inf_{P_i}f\hspace{1cm}M_i=\sup_{P_i}f$ Subdivision guarantees $f$ is still $?$-integrable on each $P_i$, so the bound above applies there: $m_i|P_i|\;\leq\;\int^?_{P_i}f\;\leq\;M_i|P_i|$ and subdivision also tells us these pieces add back up to the whole: $\int_{[a,b]}^?f=\sum_i\int^?_{P_i}f$ Adding the $n$ inequalities therefore bounds the integral over all of $[a,b]$. The two bounds we get are worth naming, since they depend only on $f$ and our choice of chopping: $L(f,P)=\sum_i m_i|P_i|\hspace{1cm}U(f,P)=\sum_i M_i|P_i|$

No theory of integration was involved in writing down $L(f,P)$ and $U(f,P)$ - they are built from infima and suprema of $f$ and the lengths of the pieces, nothing more. Yet the axioms alone force our mystery integral to lie between them.

Upper and Lower Integrals

The proposition is true for every partition at once, and that’s much stronger than it looks. Let’s take the two halves one at a time.

The left inequality says $\int^?_{[a,b]}f$ is an upper bound for the number $L(f,P)$, for every single $P$. So it is an upper bound for the whole collection of lower sums, and hence at least as big as their supremum.

The right inequality says the same thing from the other side: $\int^?_{[a,b]}f$ is a lower bound for every upper sum, so it is at most their infimum.

Both of those quantities are built from $f$ alone, so we may as well name them.

This tells us a lot! The axioms strongly constrain the value of any integral whatsoever: whatever theory you build, and however you build it, the answer must land between these two numbers. In particular, if the two numbers happen to be equal, there is nothing left to decide and $f$ is axiomatically integrable.

What about when they are not equal? Here is the observation that finishes the story: the upper and lower integrals are not merely bounds on integrals - each of them is an integral.

So whenever the upper and lower integrals of $f$ disagree, we have exhibited two genuine integrals, both accepting $f$, which assign it different values. In that case $f$ is not axiomatically integrable.

For a concrete case take $\chi_\QQ$ on $[0,1]$. On any subinterval of positive length the rationals and the irrationals are both dense, so $M_i=1$ and $m_i=0$ on every piece of every partition. Then $U(\chi_\QQ,P)=1$ and $L(\chi_\QQ,P)=0$ for all $P$, giving $\int^{\overline{D}}_{[0,1]}\chi_\QQ = 1 \hspace{1cm} \int^{\underline{D}}_{[0,1]}\chi_\QQ = 0$

Which Functions are Axiomatically Integrable?

Putting the two halves together gives a complete answer, and it is as sharp as one could hope for:

This condition is not a new one. It is exactly the criterion Darboux proposed as a definition of the integral, and the common value is called the Darboux integral of $f$. So the class our axioms determine, which we arrived at without ever choosing a construction, turns out to be a class somebody had already written down.

Two classical theorems now identify it in even more familiar terms. I’ll quote rather than prove them, since neither has anything to do with our axioms:

Chaining these together, the axiomatically integrable functions are precisely those whose discontinuities form a set of measure zero.

So on this whole class of functions, three simple axioms completely determine the theory of integration. Any time you are working in a world where your functions are not discontinuous on a set of positive measure - which is to say, essentially any time - you are free to work purely axiomatically, with no worry of ambiguity at all. You never have to open up a construction, because every construction agrees.

But the class does stop, and $\chi_\QQ$ shows it stops in a completely ordinary place. So:

QUESTION: Are there stronger axioms which determine the value of the integral on a larger class of functions?

The two counterexamples above are a good hint about where to look, because there is something conspicuously wrong with both of them: neither the upper nor the lower integral is linear. The next post asks what happens if we demand that.