Axiomatic Integration: Linearity

Every integral we actually use is linear - but demanding it determines nothing new.

Third in a series on axiomatizing integration. The axioms are set up in the first post, and the second works out exactly which functions they determine. This post takes up the first of two candidate new axioms; the fourth takes up the second.

The previous post measured our uncertainty exactly. If $f$ is bounded on $J=[a,b]$, then every integral satisfying the axioms which can integrate $f$ has to satisfy

$\underline{\int_J}f\;\leq\;\int^?_Jf\;\leq\;\overline{\int_J}f$

and the two Darboux integrals themselves satisfy the axioms, so both endpoints of that interval are actually achieved.

In fact every point of the interval is achieved, and it takes one line to see. For $0\leq\theta\leq1$ put

$\int^\theta_J f=(1-\theta)\underline{\int_J}f+\theta\overline{\int_J}f$

Rectangle areas, comparison and subdivision are all preserved by taking convex combinations, so each $\int^\theta$ is again an integral satisfying the axioms. And as $\theta$ runs from $0$ to $1$ the value $\int^\theta_Jf$ sweeps out the whole interval.

So that interval isn’t just a pair of bounds we happened to be able to prove. It is exactly the set of values the axioms permit. I’ll call it the uncertainty interval of $f$ from here on.

Adding a Fourth Axiom

There is an obvious defect in all of these examples. Unless the two Darboux integrals agree, neither of them is linear - the upper one is only subadditive and the lower one only superadditive - and their convex combinations are generally not linear either. Meanwhile every theory of integration anyone actually uses is linear on the nose.

So the axioms are permitting a freedom that no real theory of integration would ever exercise, and the natural response is to close it off.

This is a real restriction. Not one of the theories $\int^\theta$ satisfies it, so every witness we had for the interval being full is now gone. It also ties a single theory’s answers to one another: once a linear theory has fixed values for $f$ and $g$, it has no remaining choice about $f+g$.

That leaves the question wide open. The squeeze in the previous post never used linearity, so the interval still bounds every linear theory. What we no longer know is whether any value inside it is still reachable. The goal of this post is to show that all of them are.

That is, adding linearity does not shrink the set of possible values for any bounded function, on any interval.

The tool is the ordinary Hahn-Banach theorem. It extends a linear functional from a small space to a big one, provided you can name a ceiling the functional has to stay underneath. Almost all the work is in choosing that ceiling.

Building the Theories

Working on the whole line at once

Here is the plan. We build one linear functional $L$, defined on functions living on all of $\RR$, and then define the integral on an interval $J$ by $\int_J h=L(\tilde h)$ where $\tilde h$ denotes $h$ extended by zero to the rest of the line. Every interval takes its integral from the same $L$.

It would be more natural to work one interval at a time, but that runs straight into subdivision. Subdivision is not a condition on any single interval - it ties the integral on $[a,b]$ to the integrals on $[a,c]$ and $[c,b]$. Building those three separately, we would have to arrange their agreement by hand. Building them all out of one $L$ makes it automatic, because subdivision then becomes a statement about adding functions together, and $L$ is additive.

So let $\mathcal{B}$ be the bounded functions on $\RR$ which vanish outside some bounded set. It’s a vector space, and it contains $\tilde h$ for every bounded $h$ on every closed interval. On the Riemann integrable members of $\mathcal{B}$ we already know what $L$ has to do - it has to return $\int_\RR g$ - and our job is to extend that to everything else.

(A small point we’ll want later: $\tilde h$ is Riemann integrable exactly when $h$ is, since extending by zero adds at most two discontinuities, at the endpoints of $J$.)

Choosing a ceiling

What should the ceiling be? For a given $h$ it should be the largest value we could possibly justify assigning it - and there is an obvious candidate. If some Riemann integrable $g$ lies above $h$, then comparison forces our integral of $h$ to be at most $\int_\RR g$. So the cheapest such $g$ bounds us:

$U(h)=\inf\Big\{\int_\RR g \;:\; g \text{ Riemann integrable},\ g\geq h\Big\}$

This is a real number. A large enough multiple of $\chi_K$ is a majorant, so the set is not empty; and no majorant can have integral below $-\|h\|_\infty|K|$, since $g\geq h\geq-\|h\|_\infty$ on $K$ and $g\geq0$ outside it.

This ceiling is not new. If $h$ lives on an interval $J$ then

$U(\tilde h)=\overline{\int_J}h$

In one direction, the step function built from any partition of $J$ is a Riemann integrable majorant whose integral is that partition’s upper sum, so $U(\tilde h)$ is at most every upper sum. In the other, any Riemann integrable majorant $g\geq\tilde h$ has $\int_\RR g\geq\overline{\int}\tilde h$ by monotonicity of the upper integral over a large enough interval, and that upper integral is $\overline{\int_J}h$ because $\tilde h$ vanishes outside $J$.

So the ceiling is the upper Darboux integral from last time, computed by comparing $h$ against whole functions instead of against partitions. We need three properties of it.

It is sublinear, which is the hypothesis Hahn-Banach needs. Majorants of $h_1$ and of $h_2$ add to a majorant of $h_1+h_2$, so $U(h_1+h_2)\leq U(h_1)+U(h_2)$; and scaling a majorant by $c\geq0$ scales its integral, so $U(ch)=cU(h)$.

Adding a Riemann integrable function just shifts it. For $g$ Riemann integrable, $U(h+g)=U(h)+\int_\RR g$ because adding $g$ matches the majorants of $h$ with the majorants of $h+g$ one for one. Taking $h=0$ gives $U(g)=\int_\RR g$, so the Riemann integral is a linear functional already sitting underneath the ceiling. We will use this identity constantly.

It is asymmetric. $U$ is homogeneous only for nonnegative scalars, so $U(-h)$ is not $-U(h)$. Instead $U(-h)=-\underline{\int}h$ Flipping the sign swaps majorants for minorants, and so swaps the upper Darboux integral for the lower one. This is what handles every negative coefficient below.

The one free choice

Now the construction itself, which is shorter than the setup.

We know what our integral does on Riemann integrable functions, and we want it to send $f$ to $t$. So we enlarge the space by exactly one dimension and simply declare it.

Suppose first that $f$ is not Riemann integrable. Then every function of the form $g+c\tilde f$, with $g$ Riemann integrable and $c$ real, determines $g$ and $c$ uniquely - otherwise $\tilde f$ would be a combination of Riemann integrable functions and hence Riemann integrable itself. So we may define $L(g+c\tilde f)=\int_\RR g+ct$ on that slightly larger space. It is linear, it still agrees with the Riemann integral, and it does what we asked: $L(\tilde f)=t$.

There is exactly one thing to check. Does this declaration already break through the ceiling? We need $L\leq U$.

For $c>0$, the rigid-shift identity gives $U(g+c\tilde f)=\int_\RR g+c\,U(\tilde f)$, so the requirement $L\leq U$ becomes $ct\leq c\,U(\tilde f)$, which is to say $t\leq\overline{\int_a^b}f$

For $c<0$ we need the asymmetry. Writing $c=-d$ with $d>0$, we get $U(g-d\tilde f)=\int_\RR g+d\,U(-\tilde f)=\int_\RR g-d\,\underline{\int_a^b}f$, so the requirement becomes $-dt\leq-d\,\underline{\int_a^b}f$, which is to say $t\geq\underline{\int_a^b}f$

For $c=0$ there is nothing to check, since both sides equal $\int_\RR g$.

So our declaration is consistent with the ceiling exactly when

$\underline{\int_a^b}f\;\leq\;t\;\leq\;\overline{\int_a^b}f$

That deserves a pause. We did not assume the uncertainty interval and then verify it. It fell out, as the precise condition under which our one free choice is compatible with the axioms at all. The interval is not a bound we happened to be able to prove - it is the exact list of decisions we are permitted to make.

(If $f$ is Riemann integrable there is nothing to arrange: the interval is a single point, $t$ has no choice but to be that value, and we simply start from the Riemann integral itself.)

Extending

Hahn-Banach now does its job. A linear functional on a subspace, sitting under a sublinear ceiling, extends to a linear functional on the whole space still sitting under that ceiling. So $L$ extends to all of $\mathcal{B}$ - still linear, still agreeing with the Riemann integral, still sending $\tilde f$ to $t$, and still satisfying $L\leq U$ everywhere.

We never asked for positivity, and we get it anyway. If $h\geq0$ then $-h\leq0$, so the zero function is a Riemann integrable majorant of $-h$ and therefore $U(-h)\leq0$. Then $-L(h)=L(-h)\leq U(-h)\leq0$ so $L(h)\geq0$. Comparison comes out of the ceiling on its own, without our having to arrange it.

Hahn-Banach is not constructive. We are proving that two rival theories of integration exist without ever being able to write down either one.

Reading off the intervals

For each closed bounded interval $J$, declare every bounded function on $J$ to be integrable, and define $\int_J h=L(\tilde h)$

Linearity is immediate, since extension by zero and $L$ are both linear. Comparison follows from positivity applied to $h_2-h_1$. Rectangle areas hold because $\tilde k = k\chi_J$ is Riemann integrable, so $\int_J k=k|J|$.

That leaves subdivision, which is why we went global in the first place. Its domain half is free: a function on $[c,d]$ is bounded exactly when both of its restrictions are. For the values, note that $[c,e]$ and $[e,d]$ share the endpoint $e$, so zero-extending both restrictions counts $h(e)$ twice:

$\widetilde{h|_{[c,e]}}+\widetilde{h|_{[e,d]}}=\tilde h+h(e)\chi_{\{e\}}$

But $\chi_{\{e\}}$ is Riemann integrable with integral zero, so $L$ kills it, and applying $L$ to both sides is exactly the subdivision identity. In the first post we derived that integrals over a single point vanish from subdivision; here we need it in order to get subdivision.

Finally, on our original interval, $\int_{[a,b]}f=L(\tilde f)=t$. That is the theory we set out to build.

What We Have Proved

The original axioms let $f$ take any value in its uncertainty interval. After adding linearity, the same entire interval is still available:

$\Big\{\textstyle\int_{[a,b]}f \;:\; \int \text{ satisfies all four axioms and integrates } f\Big\}=\Big[\underline{\int_a^b}f,\;\overline{\int_a^b}f\Big]$

Containment one way is the squeeze from the previous post; containment the other way is the construction above.

If $f$ is Riemann integrable its two Darboux integrals agree, so every theory is forced onto their common value. If it is not, its uncertainty interval contains more than one point and the theorem builds linear theories realizing different ones.

So the class of determined functions is exactly what it was three axioms ago.

Which raises the obvious next question. Linearity is an algebraic condition - it relates a function to sums and multiples of itself. What if we asked for something geometric instead, and demanded that the integral not care where on the line a function sits? That’s the next post.