Integration Theories

Since I am preparing to teach intro real analysis this fall, it is a good time for me to get reacquainted with Riemann integration and its cousins. These notes are probably not what I will be teaching from: they represent a somewhat higher-level view of the theory, incorporating some ideas from measure theory. The goal is to keep these higher-level view in the back of my head while preparing my lectures so that the presentation I give will be naturally adaptable should the students proceed further in this direction.


Throwing away what we learned in calculus, let's ask anew "What is an integral?"

One answer could be: "An integral is an idealized process that outputs a 'weighted sum' of a function over (uncountably) infinitely many points."

Obviously (in the motivational sense, not the mathematical sense), idealized processes don't always work. So there should be functions which cannot be integrated. Equally obviously we would like this idealized process to agree with certain simple, finite, examples. Finally, for this idealized process to be at all tractable we would like it to be something that we can approximate using the finite approximants.

It therefore may be helpful to think about integration more from the computational and numerical point of view, and construct our idealized process as a suitable limit.

Riemann approximants

Given a real valued function on the interval $[a,b]$, suppose one were asked to compute an estimate of the area under its curve, a reasonable method is by

  1. Divide $[a,b]$ into finitely many smaller intervals, each of whose width we know how to compute.
  2. For each smaller interval, choose a representative point, and use the value of the function at that representative point as the height for our area computation.
  3. Add up the computed areas.

So what do we need to at least get started with a theory of integration for functions $f: X\to Y$ where $X$ and $Y$ are sets?

  1. We would like $Y$ to be real vector space. (So we can multiply $f(x)$ where $x$ is a representative by the "width" of the set [which should be a real number] and then add them and have it make sense.)
  2. We would need a collection $\Sigma \subset 2^X$ of subsets of $X$ whose "width" we know how to compute; we will denote the width mapping by $\mu:\Sigma \to [0,\infty)$.
  3. We also need a mapping $T: \Sigma \to 2^X$ giving the possible choices of representative points for each element of $\Sigma$.

We note that $\Sigma$ should be sufficiently large so that we can take finite partitions of $X$ using elements of $\Sigma$.

The mapping $T$ may send certain elements of $\Sigma$ to the empty set: this indicates that there are no suitable representatives for use with that subset, and therefore we should not use that subset when computing the approximating integral.

We will consider $\Sigma$ and $\mu$ to be given and fixed : for the purposes of this series of posts we can focus on the case where $X = [0,1]$ and $\Sigma$ the set of all non-empty intervals (open, closed, or half-and-half) and $\mu$ assigns to an interval the difference between the endpoints. For the theory to be well-behaved one would need $\Sigma$ and $\mu$ to satisfy some measure theoretic conditions1. Additionally, $Y$ should be endowed with some topological properties to enable us to take limit later, but for the constructing of the approximating sums this is not needed.

  • The pair $(P,\tau)$ is a (finite) tagged partition of $X$ if,
    • $P\subseteq \Sigma$
    • elements of $P$ are pairwise disjoint
    • $P$ is finite
    • $\cup P = X$
    • and $\tau: P \to X$ is a function
  • The tagged partition $(P,\tau)$ is said to be $T$-admissible if $\tau(p) \in T(p)$
  • Given a function $f:X\to Y$, the Riemann sum of $f$ over the tagged partition $(P,\tau)$ is \[ \int_{(P,\tau)} f := \sum_{p\in P} f(\tau(p)) \mu(p). \]

Defining a theory of integration

So far we have discussed the construction of Riemann sums. To actually get an integral, we need to appeal to the intuition that "more fine-grained partitions" gives better approximations.

The set of mapping from $\Sigma\to 2^X$ can be partial-ordered by inclusion. That is, $T_1 \preceq T_2$ if $T_2(S) \subseteq T_1(S)$ for every $S\in \Sigma$.

In particular, this means that if $T_1 \preceq T_2$, then $(P,\tau)$ being $T_2$ admissible implies that it is also $T_1$ admissible.

Now that we have a partial ordering, we can start building nets. From here-on we will assume $Y$ is equipped with a topology.

Definition    [Riemann Integration Theory]
A Riemann integration theory on $(X,\Sigma,\mu)$ is a choice of a directed set $\mathcal{T}$ of mappings from $\Sigma\to 2^X$ ordered by $\preceq$. A function $f:X \to Y$ is said to be $\mathcal{T}$-integrable with integral $\int_{\mathcal{T}} f \in Y$ if, for every open neighborhood $N$ of $\int_{\mathcal{T}} f$, there exists $T\in \mathcal{T}$ such that for every $T$-admissible $(P,\tau)$, $\int_{(P,\tau)} f \in N$.

The integrability condition in the definition can also be stated in the language of nets. Given $\mathcal{T}$, denote by \[ \mathcal{P} := \{ (T, P,\tau) : T\in \mathcal{T} \wedge (P,\tau) \text{ is } T \text{ admissible} \}. \] If we require that $(T_1,P_1,\tau_1) \preceq (T_2, P_2, \tau_2)$ if and only if $T_1 \preceq T_2$, we see that $\mathcal{P}$ is now a directed set. The mapping induced by $f$ that sends \[ \mathcal{P} \ni (T,P,\tau) \mapsto \int_{(P,\tau)} f \] is a net in $Y$. And $f$ being $\mathcal{T}$ integrable is simply the statement that this net converges.

Note that the integrability condition can potentially be vacuously satisfied: if it is the case that our $\mathcal{T}$ contains some $T$ for where there exists no $T$-admissible $(P,\tau)$, then every function is integrable.

Riemann integration

In classical Riemann integration, we have $X = [0,1]$, $\Sigma$ being the set of all subintervals, and $\mu$ being the length function on the intervals.

The directed set $\mathcal{T}_{\text{Riem}}$ consists of the family of functions $T_\delta : \Sigma \to 2^X$ where $\delta \in (0,\infty)$, given by $T_\delta(S) = \emptyset$ if $\mu(S) > \delta$ and $T_\delta(S) = S$ if $\mu(S) \leq \delta$.

Henstock integration

Here we use the formulation of Henstock integrals given by Zhao and Lee. The basic $X, \Sigma, \mu$ are the same as in the Riemann case.

The directed set $\mathcal{T}_{\text{Hens}}$ is now the set of all functions $T: \Sigma \to 2^X$ satisfying the three properties:

  • $T(S) \subseteq S$ (the representative point is within the interval itself)
  • $S_1 \subseteq S_2 \implies S_1 \cap T(S_2) \subset T(S_1)$ (if a point inside a subinterval is representative for a larger interval, than it is also representative for the subinterval)
  • For every $x\in X$, there is a (relative to $X$) open $S\ni x$ such that $x\in T(S)$ (every point is the representative point of some interval)

That $\mathcal{T}_{\text{Hens}}$ is directed requires some checking. What we have is in fact that if $T_1, T_2\in \mathcal{T}_{\text{Hens}}$ then so does \[ T_\cap(S) := T_1(S) \cap T_2(S).\]

We also see that $\mathcal{T}_{\text{Riem}} \subset \mathcal{T}_{\text{Hens}}$. This shows that Riemann integrable functions are also Henstock integrable.

McShane integration

The McShane integral (which is a reformulation of the Lebesgue theory using Riemann sums) is a close variant similar to Henstock. Again with the same $X, \Sigma, \mu$, we can take $\mathcal{T}_{\text{McS}}$ by dropping the requirement that the representative point belongs inside the interval itself. That is to say, only the following two properties are required

  • $S_1 \subseteq S_2 \implies T(S_2) \subset T(S_1)$
  • For every $x\in X$, there is a (relative to $X$) open $S\ni x$ such that $x\in T(S)$

Note that $\mathcal{T}_{\text{McS}}$ and $\mathcal{T}_{\text{Hens}}$ do not mutually include (elements of the latter generally have intervals $S_1 \subset S_2$ with points in $T(S_2)$ that is not in $T(S_1)$). Hence the two theories cannot be directly compared.

Riemann- and Henstock-Stieltjes integration

The Stieltjes variety of Riemann and Henstock integrals keep the same sets $\mathcal{T}$, but modify $\mu$. In place of the difference between endpoints $\sup I - \inf I$ of an interval $I$, an auxiliary function $\varphi: [0,1]\to\mathbb{R}$ is given and we take $\mu(I) = \varphi(\sup I) - \varphi(\inf I)$.

Darboux integration

In the case where $Y = \mathbb{R}$, it is possible to also use the Darboux notion of upper- and lower- sums to define our integration.

We define the upper and lower sums \[ U_{T}(f) = \sup_P \sum_{p\in P} \sup_{x\in T(p)} f(x) \mu(p), \qquad L_{T,P}(f) = \inf_P\sum_{p\in P} \inf_{x\in T(p)} f(x) \mu(p).\] Here the outside most $\sup$ and $\inf$ are taken over partitions $P$. (Observe that if $T$ does not admit any admissible $(P,\tau)$ then $U_{T}(f) = - \infty$, and $L_{T}(f) = + \infty$.) Further define, given $\mathcal{T}$, \[ \mathcal{U}(f) = \inf_{T\in \mathcal{T}} U_{T}(f), \qquad \mathcal{L}(f) = \sup_{T\in\mathcal{T}} L_{T}(f). \]
$f$ is $\mathcal{T}$-integrable if and only if $\mathcal{U}(f) \leq \mathcal{L}(f)$.

Observe that if $\mathcal{T}$ contains some $T$ for which there are no admissible $(P,\tau)$, then any $T' \succeq T$ also has no admissible $(P,\tau)$, and $\mathcal{T}$-integrability holds vacuously. In this case, however, as discussed above we have that $\mathcal{U}(f) = -\infty \leq +\infty = \mathcal{L}(f)$.

Therefore we may suppose that every $T$ has some admissible $(P,\tau)$. We then the monotonicity properties of the upper and lower sums. Given $T_1 \preceq T_2$, if $P$ is compatible with $T_2$ then so it is with $T_1$, and $T_2(p) \subseteq T_1(p)$ for any $p\in P$. So we have that \[ \sum_{p\in P} \sup_{x\in T_2(p)} f(x) \leq \sum_{p\in P}\sup_{x\in T_1(p)} f(x) \] and hence we get for any $T_2$-admissible $(P,\tau)$, that \[ L_{T_1}(f) \leq L_{T_2}(f) \leq \int_{(P,\tau)} f \leq U_{T_2}(f) \leq U_{T_1}(f). \] On the other hand, for any $\epsilon > 0$ there also exists $T_2$-admissible $(P, \tau_U)$ and $(P,\tau_L)$ such that \[ \int_{(P,\tau_L)} \leq L_{T_2}(f) + \epsilon, \qquad U_{T_2}(f) - \epsilon \leq \int_{(P,\tau_U)} f.\] From these standard arguments show that $\mathcal{T}$-integrability is equivalent to $\mathcal{U}(f) = \mathcal{L}(f)$.

  1. For example, with $X$ having a topology, being a "$\sigma$-finite quasi-Radon outer regular measure space" (see, e.g. this article of Fremlin). ^
Willie WY Wong
Assistant Professor

My research interests include partial differential equations, geometric analysis, fluid dynamics, and general relativity.