28 Axioms
28.1 Characterizing Integration
Definition 28.1 (Axiomatic Integration) For any closed interval \(J=[a,b]\) we denote by \(\mathcal{I}(J)\) the set of integrable functions on \(J\). Then a collection of functions \(\mathcal{I}(J)\to \RR\) is an integral, and denoted \[f\mapsto \int_J f\] if it satisfies 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 \[\int_{[a,b]}f = \int_{[a,c]}f +\int_{[c,b]}f\]
Note these axioms do not aim to uniquely specify an integral, but rather to delineate properties that anything worthy of being called an integral must have.
Definition 28.2 (Interesting Integral) An integral is called interesting if \(\mathcal{I}(J)\) contains all continuous functions on \(J\), for all closed intervals \(J\).
Intuition about why we think this should determine the value of all functions…
28.2 The Integral as a Function
Proposition 28.1 (The Integral as a Function) If \(f\in\mathcal{I}([a,b])\) is an integrable function, then there exists a function \(F\colon [a,b]\to\RR\) defined by \[F(x)=\int_{[a,x]}f\]
Proof. This is just subdivision at work: for any \(x\in[a,b]\) we may write \[[a,b]=[a,x]\cup [x,b]\]. Then Axiom III implies that \(f\) is integrable on \([a,x]\), and so the number \(\int_{[a,x]}f\) is defined. This assignment describes a real valued function
\[x\mapsto \int_{[a,x]}f\]
Proposition 28.2 (Integrating over a Degenerate Interval) If \(\{c\}\) is the degenerate closed interval containing a single point, and \(f\) is a function which is integrable on any interval containing \(a\), then \[\int_{\{a\}}f = 0\]
Proof. Let \(f\) be integrable on the interval \([u,v]\) and \(a\in[u,v]\) be a point. Without loss of generality we can in fact take \(a\) to be one of the endpoints of the interval, by subdivision: if \(a\in(u,v)\) then Axiom III implies that \(f\) is integrable on \([u,a]\) and on \([a,v]\) as well.
Thus, we assume \(f\) is integrable on \([a,v]\), and further subdivide this interval as \[[a,v]=[a,a]\cup [a,v]=\{a\}\cup [a,v]\]
By subdivision, we see that \(f\) is integrable on \(\{a\}\) and that
\[\int_{[a,v]}f =\int_{\{a\}}f+\int_{[a,v]}f\]
Subtracting the common integral over \([a,v]\) from both sides yields the result,
\[\int_{\{a\}}f=0\]
Theorem 28.1 If \(f\in\mathcal{I}([a,b])\) is a bounded integrable function, then its integral \(F(x)=\int_{[a,x]}f\) is continuous.
Proof. Let \(f\) be integrable and bounded by \(M\) on \([a,b]\), and set \(F(x)=\int_{[a.x]}f\). Begin by choosing an \(\epsilon>0\). We will prove something even strong than asked - that \(f\) is uniformly continuous by finding a \(\delta>0\) where if \(|y-x|<\delta\) we have \(|F(y)-F(x)|<\epsilon\). Let’s unpack this a bit: if \(x<y\) are two points of \([a,b]\),
\[F(y)-F(x)=\int_{[a,y]}f-\int_{[a,x]}f\]
But subdivision (Axiom III) implies \[\begin{align*} F(y)&=\int_{[a,y]}f\\ &= \int_{[a,x]}f+\int_{[x,y]}f\\ &=F(x)+\int_{[x,y]}f \end{align*}\]
Thus \(F(y)-F(x)\) is just the integral of \(f\) on the subinterval \([x,y]\subset[a,b]\). Because \(f\) is bounded by \(M\) we know \(-M\leq f(x)\leq M\). By subdivsion, \(f\) is then integrable on every sub-interval \(I\subset [a,b]\), and by comparison (Axiom II) this implies \[-M |I|\leq \int_{I}f\leq M|I|\]
So, we choose \(\delta=\epsilon/M\). This immediately yields what we want, as if \(|y-x|<\delta\),
\[-\epsilon = -M\delta < -M|y-x|\leq \int_{[x,y]}f\leq M|y-x|< M\delta =\epsilon\]
Thus \(|F(y)-F(x)|=\Big|\int_{[x,y]}f,dx\Big|<\epsilon\).
Remark 28.1. Of course, the proven result is not really stronger than what was asked, since we began on a closed interval, and we know that continuous on a closed interval implies uniformly continuous.
However, if you look carefully at the proof you see we nowhere used that the original domain was a closed interval! So what we have really proven is that the area function \(F(x)=\int_{[a,b]}f\) is uniformly continuous anytime \(f\) is bounded!
As defined, the integral is only a function of \(x\) for \(x\) greater than the chosen starting endpoint. While this is what is desired in many applications, its also useful to be able to extend the definition to make sense for \(x\) below the starting point as well. This is the integral often met in calculus, which we call the oriented integral as it changes sign when the interval from \(a\) to \(x\) is traced backwards.
Definition 28.3 (The Oriented Integral) Given a function \(f\) which is integrable on the interval between \(a\) and \(b\), the oriented integral of \(f\) is denoted \(\int_a^b f\) and equals \[\int_a^b f=\begin{cases} \int_{[a,b]}f & a\leq b\\ -\int_{[b,a]}f & b<a \end{cases}\]
Corollary 28.1 Given any function \(f\) which is integrable on an interval \([a,b]\), for any \(c\in [a,b]\) the oriented integral defines a function \([a,b] \to\RR\) by \[x\mapsto \int_c^x f\]
28.3 \(\blacklozenge\) Improper Integrals
We have axiomatized the integral for bounded functions on closed intervals, but the definition can be naturally extended to unbounded intervals and (certain) unbounded functions via limits.
Definition 28.4 (Improper Integrals: Unbounded Intervals) The integral of a bounded function \(f\) on a ray \([a,\infty)\) is defined as a limit of its integrals over growing closed intervals \[\int_{[a,\infty)}f :=\lim_{b\to\infty}\int_{[a,b]}f\] with the analogous definition for rays \((-\infty,b]\). The integral over the entire real line is defined by taking each endpoint to \(\pm\infty\) separately
\[\int_{\mathbb{R}}f:=\lim_{\substack{a\to-\infty \\ b\to\infty}}\int_{[a,b]}f\]
That is, both orders \(\lim_{a\to-\infty}\lim_{b\to\infty}\) and \(\lim_{b\to\infty}\lim_{a\to-\infty}\) exist and are equal.
Definition 28.5 (Improper Integrals: Unbounded Functions) If \(f\) is defined on \((a,b]\) and integrable on each subinterval \([t,b]\) for \(t>a\), we define the improper integral on \((a,b]\) as a limit \[\int_{(a,b]}f=\lim_{t\to a^+}\int_{[t,a]}f\] Similarly for functions unbounded on \([a,b)\) but bounded on each \([a,t]\subset [a,b)\), we define \(\int_{[a,b)}f=\lim_{t\to b^-}\int_{[a,t]}f\).
If a function defined on \([a,b]\) is unbounded in a neighborhood of some \(c\in (a,b)\) but is integrable on every subinterval missing \(c\), we say the integral on \([a,b]\) exists if and only if the integral on \([a,c)\) and \((c,b]\) both exist, in which case we define it as equal to their sum.