Integration: are the Axioms Enough?

Asking whether the axioms pin down every integral or leave room for choice.

$\newcommand{\RR}{\mathbb{R}}$

This note concerns the axiomatic description of integration:

Integration Axioms

An integral on $\RR$ is 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 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 $\int_{[a,b]}f = \int_{[a,c]}f +\int_{[c,b]}f$

There are many different integrals by this definition (for example the Riemann/Darboux, Lebesgue, and Gauge integrals). As these axioms do not uniquely specify a function, its a natural question how tight of control they exhibit: can two different integrals agree that a certain function $f$ is integrable, but yet disagree on its value? Precisely, we may ask

Big Question: Let $\int^1,\int^2$ both satisfy the integration axioms, and let $f$ be a function such that $f\in \mathcal{I}_1(J)\cap \mathcal{I}_2(J)$ for some interval $J$ . Is it true that $\int^1_J f = \int^2_J f$ ?

This note provides two first steps towards answering this question. First, we can resolve it for continuous functions, thanks to the Fundamental Theorem of Calculus:

All integrals agree when continuous

Let $f$ be a continuous function, and $\int^1,\int^2$ be two integrals for which $f$ is integrable on $[a,b]$ . Then $\int_{[a,b]}^1f = \int_{[a,b]}^2 f$

The fundamental theorem of calculus is provable directly from the axioms for continuous functions: if $\int$ is some integral and $f$ is a continuous function that it can integrate on $[a,b]$ , then $F(x)=\int_{[a,x]}f$ is differentiable and $F^\prime(x)=f(x)$ (In a previous note, on 2024-05-02).

If $\int^1$ and $\int^2$ are two different integrals for which $f$ is integrable, set $F_1=\int^1_{[a,x]}f$ and $F_2(x)=\int^2_{[a,x]}f$ . By the Fundamental theorem, we see $F_1^\prime=F_2^\prime=f$ , and so by a corollary to the Mean Value Theorem, since $F_1$ and $F_2$ have the same derivative they differ by a constant. To compute this constant, we need only find $F_2(x)-F_1(x)$ at some point. But since integrals over a singleton are always zero (proven from the axioms in the same note as the Fundamental Theorem), we see

$F_1(a)=\int_{\{a\}}^1 f =0\hspace{1cm}F_2(a)=\int_{\{a\}}^2 f =0$

Thus $F_2(a)-F_1(a)=0$ and so $F_2(x)=F_1(x)$ for all $x$ . Evaluating at $x=b$ yields the result:

$\int_{[a,b]}^1 f=\int_{[a,b]}^2 f$

The main theorem of this note extends this to functions which are allowed to be discontinuous on a set of measure zero (proven together in discussions with David Cheng on 2024-08-21)

All integrals agree when discontinuities are measure-zero

Let $f$ be bounded and have a measure-zero set of discontinuities. Then if $\int^1,\int^2$ are any two integrals for which $f$ is integrable on $[a,b]$ , $\int_{[a,b]}^1f = \int_{[a,b]}^2 f$

To prove this result, we can reduce to a simpler statement using the following two well known theorems:

Riemann Integrability Criterion

A bounded function $f$ is Riemann Integrable on $[a,b]$ if and only if the set of discontinuities of $f$ is measure zero.

Riemann Darboux Equivalence

A bounded function $f$ is Riemann integrable if and only if it is Darboux integrable.

Using these we can reduce the main work to showing that any integral must agree with the Darboux integral when both are defined. Indeed, let $f$ have a measure-zero set of discontinuities. Then $f$ is Riemann, and hence Darboux integrable, so $\int^{\mathrm{Darboux}}f_{[a,b]}$ is defined. If we can show that for any mystery integral $\int^?$ for which $\int^?_{[a,b]}f$ is defined, that $\int^?_{[a,b]}f=\int^{\mathrm{Darboux}}f_{[a,b]}$ , we can apply this to any two such integrals $\int^1,\int^2$ for which $f$ is integrable, and conclude

$\int^1_{[a,b]}f =\int^{\mathrm{Darboux}}_{[a,b]}f= \int^2_{[a,b]}f$

So we finish the argument by proving this special case, though it is useful to remind ourselves of the definition of the Darboux Integral:

Darboux Integral

Let $\mathcal{P}$ be the set of all partitions of $[a,b]$ , and for any given partition $P=P_1\cup P_2\cup \cdots \cup P_n$ of $[a,b]$ , let $m_i=\inf_{P_i}f$ , $M_i=\sup_{P_i} f$ and $L(f,P)=\sum_i m_i |P_i|\hspace{1cm}U(f,P)=\sum_i M_i|P_i|.$ Then $f$ is Darboux integrable if and only if $\sup_{P\in\mathcal{P}}L(f,P)=\inf_{P\in\mathcal{P}}U(f,P)$ and the Darboux integral is their common value.

Agreeing with the Darboux Integral

If $f$ is Darboux-Integrable on $[a,b]$ and $f$ is also integrable for some other integral $\int^?$ , then the two values agree: $\int^{\mathrm{Darboux}}_{[a,b]}f = \int^?_{[a,b]}f$

Let $P=P_1\cup P_2\cup \cdots \cup P_n$ be an arbitrary partition of $[a,b]$ . Then on each subinterval $P_i$ we know $m_i=\inf_{P_i}f\leq f(x)\leq\sup_{P_i}f=M_i$ By subdivision twice (Axiom III) we know that $f$ is $?$ -integrable on $P_i$ as it started integrable on $[a,b]$ . Because constants are integrable (Axiom I) and we know integration respects inequalities (Axiom II), we can conclude that $\int_{P_i}m_i\leq \int_{P_i}^?f\leq \int_{P_i}^?M_i$

Applying the rest of Axiom I (giving the value of the integral of constants), we see $m_i|P_i|=\int_{P_i}^?m_i\hspace{1cm}M_i|P_i|=\int_{P_i}^?M_i$ Together this yields bounds on $\int^?_{P_i}f$ :

$m_i|P_i|\leq \int_{P_i}^?f\leq M_i|P_i|$

Applying subdivision (Axiom III) inductively to the partition $P=P_1\cup\cdots\cup P_n$ , $\int_{[a,b]}^?f = \sum_i \int_{P_i}^?f.$ Applying the known inequality to each partition gives an inequality on the entire integral: $\sum_i m_i|P_i|\leq \int_{[a,b]}^? f\leq \sum_i M_i|P_i|$ But these two bounds are precisely the upper and lower sums of the Darboux integral on $P$ . Because we have this inequality for all arbitrary partitions $P$ , it follows that $\sup_{P\in \mathcal{P}}\sum_i m_i |P_i|\leq \int_{[a,b]}^?f$ $\int_{[a,b]}^?f\leq \inf_{P\in \mathcal{P}}\sum_i M_i |P_i|$

Using the assumption that $f$ is Darboux Integrable, we know that the infimum and supremum appearing here are actually equal, and give the value $\int^{\mathrm{Darboux}}_{[a,b]}f$ . Thus we’ve squeezed our mystery integral to be exactly this value, as desired:

$\int_{[a,b]}^?f =\int_{[a,b]}^{\mathrm{Darboux}}f$