Simple Integration Axioms Do Not Fix All Values

An explicit pair of integrals that satisfies the basic axioms but disagree on a function.

This note is part of the story of sketching out an axiomatic understanding of integration. Recall from previous notes that we may attempt to define ‘an integral’ to be a choice of set of functions $\mathcal{I}(J)$ for each closed interval $J$ together with a real valued map $\int_J\colon\mathcal{I}(J)\to\RR$ satisfying:

Rectangle Areas: If $k\in\RR$ then $f(x)=k$ is an element of $\mathcal{I}([a,b])$ for any interval $[a,b]$ and $\int_{[a,b]}k = k(b-a).$
Comparison: If $f,g\in \mathcal{I}([a,b])$ and $f(x)\leq g(x)$ for all $x\in[a,b]$ then $\int_{[a,b]}f\leq\int_{[a,b]}g$
Subdivision: 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 $\int_{[a,b]}f = \int_{[a,c]}f +\int_{[c,b]}f$

It’s natural to wonder how strongly these axioms constrain the behavior of an integral satisfying them. Its an exciting exercise in a real analysis class to prove that these alone imply the following form of the fundamental theorem for any continuous function which is integrable for the given integral. Thus, if two integrals both agree that a continuous function is integrable, they must also agree on the value of the integral (its given by $F(b)-F(a)$ for $F$ an antiderivative, and all antiderivatives differ from one another by at most the addition of a constant). In a previous note coming from conversations with David Cheng, we took this further and proved that if $f$ is bounded with a measure zero set of discontinuities, than any two integrals which can integrate $f$ must agree on its value.

But trying to press this farther fails! The goal of this note is to provide a simple counterexample, proving

These three axioms do not fix the integral of integrable functions. Precisely, there exists two integrals $\int^1,\int^2$ satisfying the definition and a function $f$ which is integrable for both on $I=[0,1]$ but $\int^1_I f\neq \int^2_I f$

This is also joint work with David Cheng, who checked the details of the specific example given below.