Linearity as an Integration Axiom

Showing its independence from the previous axioms

This note is part of the story of sketching out an axiomatic understanding of integration. Recall the original question: do the following axioms uniquely determine the value of $\int_{I}f$ , when $f$ is integrable?

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 $\int_{[a,b]}f = \int_{[a,c]}f +\int_{[c,b]}f$

This was proven false by finding a specific counterexample: the upper and lower Darboux integrals satisfy these axioms, but disagree on what to assign the characteristic function of the rationals. The purpose of this note is to propose a strengthening of these axioms, which eliminates this counterexample.

Linearity

An integral is linear if for every $f,g$ which are integrable on an interval $I$ , and every pair $c,k\in\mathbb{R}$ of constants, we have $cf+kg$ is also integrable on $I$ , and $\int_{I} (cf+kg)=c\int_I f +k\int_I g$

While this property is familiar from our standard theories of integration, it is not implied by these axioms:

Linearity is independent of the three initial axioms.

It is a standard exercise in real analysis to prove from the construction, that the Darboux integral is linear. Thus there are models of the axioms which are linear. But consider the upper Darboux integral used in this construction. Taken alone this also satisfies the three axioms above, but it fails linearity, as we can check using the characteristic function of the rationals $\chi_\mathbb{Q}$ .

This note proposes adding linearity as an additional axiom. It gives some brief observations on how one could reformulate the problem within this stronger axiomatic framework, and then proposes a logically equivalent (but simpler looking) collection of axioms, where the original first 2 axioms are slightly simplified using the assumption of linearity.

Reformulating the Problem

Simplifying the Axioms

Potential 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:

Normalization: The function $1$ is integrable on any interval $I$ , and returns the length of the interval: $\int_I 1 = |I|$
Nonnegativity: If $f$ is integrable on $I$ and $f\geq 0$ then $\int_I f\geq 0$ .
Subdivision: If $[a,b]$ is an interval and $c\in[a,b]$ , then $f$ is integrable on $[a,b]$ if and only if its integrable on $[a,c]$ and $[c,b]$ . Furthermore, in this case $\int_{[a,b]}f = \int_{[a,c]}f +\int_{[c,b]}f$
Linearity: If $f,g$ are integrable on $I$ then so is $cf+kg$ for any $c,k\in\mathbb{R}$ and $\int_{I}(cf+kg)=c\int_I f+k\int_I g$

Weakening The Rectangle Axiom

If an integral is linear and satisfies $\int_{[a,b]}1 = b-a$ then it satisfies the rectangle axiom.

Let $k$ be any constant. Then since $1$ is integrable on $[a,b]$ so is $k\cdot 1 = k$ and its value is $\int_{[a,b]}k =k\int_{[a,b]}1=k(b-a)$

In fact, we could weaken it further using subdivision: its enough to assume that for any $a$ we have that $1$ is integrable on $[a,0]$ or $[0,a]$ (depending on if $a$ is positive or negative), and that $\int_{[a,0]}1=|a|\hspace{1cm}\int_{[0,a]}1=a$ but this is more clunky, (not simpler!) so we don’t bother.

Weakening the Comparison Axiom

If an integral is linear and for any integrable $f$ with $f\geq 0$ we have $\int_I f\geq 0$ , then it satisfies the comparison axiom.

Thus, we’ve proven what was claimed: the original axioms + linearity and the collection of 4 aixoms stated in the theorem above are logically equivalent, and strictly stronger than the original three alone.