One of the most fundamental concepts in Euclidean geometry is that of the measure {m(E)} of a solid body {E} in one or more dimensions. In one, two, and three dimensions, we refer to this measure as the length, area, or volume of {E} respectively. In the classical approach to geometry, the measure of a body was often computed by partitioning that body into finitely many components, moving around each component by a rigid motion (e.g. a translation or rotation), and then reassembling those components to form a simpler body which presumably has the same area. One could also obtain lower and upper bounds on the measure of a body by computing the measure of some inscribed or circumscribed body; this ancient idea goes all the way back to the work of Archimedes at least. Such arguments can be justified by an appeal to geometric intuition, or simply by postulating the existence of a measure {m(E)} that can be assigned to all solid bodies {E}, and which obeys a collection of geometrically reasonable axioms. One can also justify the concept of measure on “physical” or “reductionistic” grounds, viewing the measure of a macroscopic body as the sum of the measures of its microscopic components.

With the advent of analytic geometry, however, Euclidean geometry became reinterpreted as the study of Cartesian products {{\bf R}^d} of the real line {{\bf R}}. Using this analytic foundation rather than the classical geometrical one, it was no longer intuitively obvious how to define the measure {m(E)} of a general subset {E} of {{\bf R}^d}; we will refer to this (somewhat vaguely defined) problem of writing down the “correct” definition of measure as the problem of measure. (One can also pose the problem of measure on other domains than Euclidean space, such as a Riemannian manifold, but we will focus on the Euclidean case here for simplicity.)

To see why this problem exists at all, let us try to formalise some of the intuition for measure discussed earlier. The physical intuition of defining the measure of a body {E} to be the sum of the measure of its component “atoms” runs into an immediate problem: a typical solid body would consist of an infinite (and uncountable) number of points, each of which has a measure of zero; and the product {\infty \cdot 0} is indeterminate. To make matters worse, two bodies that have exactly the same number of points, need not have the same measure. For instance, in one dimension, the intervals {A := [0,1]} and {B := [0,2]} are in one-to-one correspondence (using the bijection {x \mapsto 2x} from {A} to {B}), but of course {B} is twice as long as {A}. So one can disassemble {A} into an uncountable number of points and reassemble them to form a set of twice the length.

Of course, one can point to the infinite (and uncountable) number of components in this disassembly as being the cause of this breakdown of intuition, and restrict attention to just finite partitions. But one still runs into trouble here for a number of reasons, the most striking of which is the Banach-Tarski paradox, which shows that the unit ball {B := \{ (x,y,z) \in {\bf R}^3: x^2+y^2+z^2 \leq 1 \}} in three dimensions can be disassembled into a finite number of pieces (in fact, just five pieces suffice), which can then be reassembled (after translating and rotating each of the pieces) to form two disjoint copies of the ball {B}. (The paradox only works in three dimensions and higher, for reasons having to do with the property of amenability; see this blog post for further discussion of this interesting topic, which is unfortunately too much of a digression from the current subject.)

Here, the problem is that the pieces used in this decomposition are highly pathological in nature; among other things, their construction requires use of the axiom of choice. (This is in fact necessary; there are models of set theory without the axiom of choice in which the Banach-Tarski paradox does not occur, thanks to a famous theorem of Solovay.) Such pathological sets almost never come up in practical applications of mathematics. Because of this, the standard solution to the problem of measure has been to abandon the goal of measuring every subset {E} of {{\bf R}^d}, and instead to settle for only measuring a certain subclass of “non-pathological” subsets of {{\bf R}^d}, which are then referred to as the measurable sets. The problem of measure then divides into several subproblems:

  1. What does it mean for a subset {E} of {{\bf R}^d} to be measurable?
  2. If a set {E} is measurable, how does one define its measure?
  3. What nice properties or axioms does measure (or the concept of measurability) obey?
  4. Are “ordinary” sets such as cubes, balls, polyhedra, etc. measurable?
  5. Does the measure of an “ordinary” set equal the “naive geometric measure” of such sets? (e.g. is the measure of an {a \times b} rectangle equal to {ab}?)

These questions are somewhat open-ended in formulation, and there is no unique answer to them; in particular, one can expand the class of measurable sets at the expense of losing one or more nice properties of measure in the process (e.g. finite or countable additivity, translation invariance, or rotation invariance). However, there are two basic answers which, between them, suffice for most applications. The first is the concept of Jordan measure of a Jordan measurable set, which is a concept closely related to that of the Riemann integral (or Darboux integral). This concept is elementary enough to be systematically studied in an undergraduate analysis course, and suffices for measuring most of the “ordinary” sets (e.g. the area under the graph of a continuous function) in many branches of mathematics. However, when one turns to the type of sets that arise in analysis, and in particular those sets that arise as limits (in various senses) of other sets, it turns out that the Jordan concept of measurability is not quite adequate, and must be extended to the more general notion of Lebesgue measurability, with the corresponding notion of Lebesgue measure that extends Jordan measure. With the Lebesgue theory (which can be viewed as a completion of the Jordan-Darboux-Riemann theory), one keeps almost all of the desirable properties of Jordan measure, but with the crucial additional property that many features of the Lebesgue theory are preserved under limits (as exemplified in the fundamental convergence theorems of the Lebesgue theory, such as the monotone convergence theorem and the dominated convergence theorem, which do not hold in the Jordan-Darboux-Riemann setting). As such, they are particularly well suited for applications in analysis, where limits of functions or sets arise all the time. (There are other ways to extend Jordan measure and the Riemann integral, but the Lebesgue approach handles limits better than the other alternatives, and so has become the standard approach in analysis.)

In the rest of the course, we will formally define Lebesgue measure and the Lebesgue integral, as well as the more general concept of an abstract measure space and the associated integration operation. In the rest of this post, we will discuss the more elementary concepts of Jordan measure and the Riemann integral. This material will eventually be superceded by the more powerful theory to be treated in the main body of the course; but it will serve as motivation for that later material, as well as providing some continuity with the treatment of measure and integration in undergraduate analysis courses.

— 1. Elementary measure —

Before we discuss Jordan measure, we discuss the even simpler notion of elementary measure, which allows one to measure a very simple class of sets, namely the elementary sets (finite unions of boxes).

Definition 1 (Intervals, boxes, elementary sets) An interval is a subset of {{\bf R}} of the form {[a,b] := \{ x \in {\bf R}: a \leq x \leq b \}}, {[a,b) := \{ x \in {\bf R}: a \leq x < b \}}, {(a,b] := \{ x \in {\bf R}: a < x \leq b \}}, or {(a,b) := \{x \in {\bf R}: a < x < b \}}, where {a \leq b} are real numbers. We define the length {|I|} of an interval {I = [a,b], [a,b), (a,b], (a,b)} to be {|I| := b-a}. (Note we allow degenerate intervals of zero length.) A box in {{\bf R}^d} is a Cartesian product {B := I_1 \times \ldots \times I_d} of {d} intervals {I_1,\ldots,I_d} (not necessarily of the same length), thus for instance an interval is a one-dimensional box. The volume {|B|} of such a box {B} is defined as {|B| := |I_1| \times \ldots \times |I_d|}. An elementary set is any subset of {{\bf R}^d} which is the union of a finite number of boxes.

Exercise 1 (Boolean closure) Show that if {E, F \subset {\bf R}^d} are elementary sets, then the union {E \cup F}, the intersection {E \cap F}, and the set theoretic difference {E \backslash F}, and the symmetric difference {E \Delta F := (E \backslash F) \cup (F \backslash E)} are also elementary. If {x \in {\bf R}^d}, show that the translate {E+x := \{ y+x: y \in E \}} is also an elementary set.

We now give each elementary set a measure.

Lemma 2 (Measure of an elementary set) Let {E \subset {\bf R}^d} be an elementary set.

  1. {E} can be expressed as the finite union of disjoint boxes.
  2. If {E} is partitioned as the finite union {B_1 \cup \ldots \cup B_k} of disjoint boxes, then the quantity {m(E) := |B_1| + \ldots + |B_k|} is independent of the partition. In other words, given any other partition {B'_1 \cup \ldots \cup B'_{k'}} of {E}, one has {|B_1|+\ldots+|B_k| = |B'_1|+\ldots+|B'_{k'}|}.

    We refer to {m(E)} as the elementary measure of {E}. (We occasionally write {m(E)} as {m^d(E)} to emphasise the {d}-dimensional nature of the measure.) Thus, for example, the elementary measure of {(1,2) \cup [3,6]} is {4}.

Proof: We first prove (1.) in the one-dimensional case {d=1}. Given any finite collection of intervals {I_1,\ldots,I_k}, one can place the {2k} endpoints of these intervals in increasing order (discarding repetitions). Looking at the open intervals between these endpoints, together with the endpoints themselves (viewed as intervals of length zero), we see that there exists a finite collection of disjoint intervals {J_1,\ldots,J_{k'}} such that each of the {I_1,\ldots,I_k} are a union of some subcollection of the {J_1,\ldots,J_{k'}}. This already gives (1.) when {d=1}. To prove the higher dimensional case, we express {E} as the union {B_1,\ldots,B_k} of boxes {B_i = I_{i,1} \times \ldots \times I_{i,d}}. For each {j=1,\ldots,d}, we use the one-dimensional argument to express {I_{1,j},\ldots,I_{k,j}} as the union of subcollections of a collection {J_{1,j},\ldots,J_{k'_j,j}} of disjoint intervals. Taking Cartesian products, we can express the {B_1,\ldots,B_k} as finite unions of boxes {J_{i_1,1} \times \ldots \times J_{i_d,d}}, where {1 \leq i_j \leq k'_j} for all {1 \leq j \leq d}. Such boxes are all disjoint, and the claim follows.

To prove (2.) we use a discretisation argument. Observe (exercise!) that for any interval {I}, the length of {I} can be recovered by the limiting formula

\displaystyle  |I| = \lim_{N \rightarrow \infty} \frac{1}{N} \#( I \cap \frac{1}{N} {\bf Z} )

where {\frac{1}{N} {\bf Z} := \{ \frac{n}{N}: n \in {\bf Z}\}} and {\# A} denotes the cardinality of a finite set {A}. Taking Cartesian products, we see that

\displaystyle  |B| = \lim_{N \rightarrow \infty} \frac{1}{N^d} \#( B \cap \frac{1}{N} {\bf Z}^d )

for any box {B}, and in particular that

\displaystyle  |B_1| + \ldots + |B_k| = \lim_{N \rightarrow \infty} \frac{1}{N^d} \#( E \cap \frac{1}{N} {\bf Z}^d ).

Denoting the right-hand side as {m(E)}, we obtain the claim (2.). \Box

Exercise 2 Give an alternate proof of part (2.) of the above lemma by showing that any two partitions of {E} into boxes admit a mutual refinement into boxes that arise from taking Cartesian products of elements from finite collections of disjoint intervals.

Remark 1 One might be tempted to now define the measure {m(E)} of an arbitrary set {E \subset {\bf R}^d} by the formula

\displaystyle  m(E) := \lim_{N \rightarrow \infty} \frac{1}{N^d} \#( E \cap \frac{1}{N} {\bf Z}^d ), \ \ \ \ \ (1)

since this worked well for elementary sets. However, this definition is not particularly satisfactory for a number of reasons. Firstly, one can concoct examples in which the limit does not exist (Exercise!). Even when the limit does exist, this concept does not obey reasonable properties such as translation invariance. For instance, if {d=1} and {E := {\bf Q} \cap [0,1] := \{ x\in {\bf Q}: 0 \leq x \leq 1 \}}, then this definition would give {E} a measure of {1}, but would give the translate {E+\sqrt{2} := \{ x +\sqrt{2}: x \in {\bf Q}; 0 \leq x \leq 1 \}} a measure of zero. Nevertheless, the formula (1) will be valid for all Jordan measurable sets (see Exercise 13). It also makes precise an important intuition, namely that the continuous concept of measure can be viewed as a limit of the discrete concept of (normalised) cardinality. (Another way to obtain continuous measure as the limit of discrete measure is via Monte Carlo integration, although in order to rigorously introduce the probability theory needed to set up Monte Carlo integration properly, one already needs to develop a large part of measure theory, so this perspective, while intuitive, is not suitable for foundational purposes.)

From the definitions, it is clear that {m(E)} is a non-negative real number for every elementary set {E}, and that

\displaystyle  m(E \cup F) = m(E) + m(F)

whenever {E} and {F} are disjoint elementary sets. We refer to the latter property as finite additivity; by induction it also implies that

\displaystyle  m(E_1 \cup \ldots \cup E_k) = m(E_1) + \ldots + m(E_k)

whenever {E_1,\ldots,E_k} are disjoint elementary sets. We also have the obvious degenerate case

\displaystyle  m(\emptyset) = 0.

Finally, elementary measure clearly extends the notion of volume, in the sense that

\displaystyle  m(B) = |B|

for all boxes {B}.

From non-negativity and finite additivity (and Exercise 1) we conclude the monotonicity property

\displaystyle  m(E) \leq m(F)

whenever {E \subset F} are nested elementary sets. From this and finite additivity (and Exercise 1) we easily obtain the finite subadditivity property

\displaystyle  m(E \cup F) \leq m(E) + m(F)

whenever {E, F} are elementary sets (not necessarily disjoint); by induction one then has

\displaystyle  m(E_1 \cup \ldots \cup E_k) \leq m(E_1) + \ldots + m(E_k)

whenever {E_1,\ldots,E_k} are elementary sets (not necessarily disjoint).

It is also clear from the definition that we have the translation invariance

\displaystyle  m(E+x) = m(E)

for all elementary sets {E} and {x \in {\bf R}^d}.

These properties in fact define elementary measure up to normalisation:

Exercise 3 (Uniqueness of elementary measure) Let {d \geq 1}. Let {m': {\mathcal E}({\bf R}^d) \rightarrow {\bf R}^+} be a map from the collection {{\mathcal E}({\bf R}^d)} of elementary subsets of {{\bf R}^d} to the nonnegative reals that obeys the non-negativity, finite additivity, and translation invariance properties. Show that there exists a constant {c \in {\bf R}^+} such that {m'(E) = cm(E)} for all elementary sets {E}. In particular, if we impose the additional normalisation {m'( [0,1)^d ) = 1}, then {m' \equiv m}. (Hint: Set {c := m'([0,1)^d)}, and then compute {m'([0,\frac{1}{n})^d)} for any positive integer {n}.)

Exercise 4 Let {d_1,d_2 \ge 1}, and let {E_1 \subset {\bf R}^{d_1}}, {E_2 \subset {\bf R}^{d_2}} be elementary sets. Show that {E_1 \times E_2 \subset {\bf R}^{d_1+d_2}} is elementary, and {m^{d_1+d_2}(E_1 \times E_2) = m^{d_1}(E_1) \times m^{d_2}(E_2)}.

— 2. Jordan measure —

We now have a satisfactory notion of measure for elementary sets. But of course, the elementary sets are a very restrictive class of sets, far too small for most applications. For instance, a solid triangle or disk in the plane will not be elementary, or even a rotated box. On the other hand, as essentially observed long ago by Archimedes, such sets {E} can be approximated from within and without by elementary sets {A \subset E \subset B}, and the inscribing elementary set {A} and the circumscribing elementary set {B} can be used to give lower and upper bounds on the putative measure of {E}. As one makes the approximating sets {A, B} increasingly fine, one can hope that these two bounds eventually match. This gives rise to the following definitions.

Definition 3 (Jordan measure) Let {E \subset {\bf R}^d} be a bounded set.

  • The inner Jordan measure {m_{*,(J)}(E)} of {E} is defined as

    \displaystyle  m_{*,(J)}(E) := \sup_{A \subset E, A \hbox{ elementary}} m(A).

  • The outer Jordan measure {m^{*,(J)}(E)} of {E} is defined as

    \displaystyle  m^{*,(J)}(E) := \inf_{B \supset E, B \hbox{ elementary}} m(B).

  • If {m_{*,(J)}(E) = m^{*,(J)}(E)}, then we say that {E} is Jordan measurable, and call {m(E) := m_{*,(J)}(E) = m^{*,(J)}(E)} the Jordan measure of {E}. As before, we write {m(E)} as {m^d(E)} when we wish to emphasise the dimension {d}.

By convention, we do not consider unbounded sets to be Jordan measurable (they will be deemed to have infinite outer Jordan measure).

Jordan measurable sets are those sets which are “almost elementary” with respect to outer Jordan measure. More precisely, we have

Exercise 5 (Characterisation of Jordan measurability) Let {E \subset {\bf R}^d} be bounded. Show that the following are equivalent:

  1. {E} is Jordan measurable.
  2. For every {\epsilon > 0}, there exist elementary sets {A \subset E \subset B} such that {m(B \backslash A) \leq \epsilon}.
  3. For every {\epsilon > 0}, there exists an elementary set {A} such that {m^{*,(J)}(A \Delta E) \leq \epsilon}.

As one corollary of this exercise, we see that every elementary set {E} is Jordan measurable, and that Jordan measure and elementary measure coincide for such sets; this justifies the use of {m(E)} to denote both. In particular, we still have {m(\emptyset)=0}.

Jordan measurability also inherits many of the properties of elementary measure:

Exercise 6 Let {E, F \subset {\bf R}^d} be Jordan measurable sets.

  1. (Boolean closure) Show that {E \cup F}, {E \cap F}, {E \backslash F}, and {E \Delta F} are Jordan measurable.
  2. (Non-negativity) {m(E) \geq 0}.
  3. (Finite additivity) If {E, F} are disjoint, then {m(E \cup F) = m(E) + m(F)}.
  4. (Monotonicity) If {E \subset F}, then {m(E) \leq m(F)}.
  5. (Finite subadditivity) {m(E \cup F) \leq m(E) + m(F)}.
  6. (Translation invariance) For any {x \in {\bf R}^d}, {E+x} is Jordan measurable, and {m(E+x)=m(E)}.

Now we give some examples of Jordan measurable sets:

Exercise 7 (Regions under graphs are Jordan measurable) Let {B} be a closed box in {{\bf R}^d}, and let {f: B \rightarrow {\bf R}} be a continuous function.

  1. Show that the graph {\{ (x,f(x)): x \in B \} \subset {\bf R}^{d+1}} is Jordan measurable in {{\bf R}^{d+1}} with Jordan measure zero. (Hint: on a compact metric space, continuous functions are uniformly continuous.)
  2. Show that the set {\{ (x,t): x \in B; 0 \leq t \leq f(x) \} \subset {\bf R}^{d+1}} is Jordan measurable.

Exercise 8 Let {A,B,C} be three points in {{\bf R}^2}.

  1. Show that the solid triangle with vertices {A,B,C} is Jordan measurable.
  2. Show that the Jordan measure of the solid triangle is equal to {\frac{1}{2} |(B-A) \wedge (C-A)|}, where {|(a,b) \wedge (c,d)| := |ad-bc|}.

(Hint: It may help to first do the case when one of the edges, say {AB}, is horizontal.)

Exercise 9 Show that every compact convex polytope in {{\bf R}^d} is Jordan measurable.

Exercise 10

  1. Show that all open and closed Euclidean balls {B(x,r) := \{ y \in {\bf R}^d: |y-x| < r \}}, {\overline{B(x,r)} := \{ y \in {\bf R}^d: |y-x| \leq r \}} in {{\bf R}^d} are Jordan measurable, with Jordan measure {c_d r^d} for some constant {c_d > 0} depending only on {d}.
  2. Establish the crude bounds

    \displaystyle  \left(\frac{2}{\sqrt{d}}\right)^d \leq c_d \leq 2^d.

(An exact formula for {c_d} is {c_d = \frac{1}{d} \omega_d}, where {\omega_d := \frac{2\pi^{d/2}}{\Gamma(d/2)}} is the volume of the unit sphere {S^{d-1} \subset {\bf R}^d} and {\Gamma} is the Gamma function, but we will not derive this formula here.)

Exercise 11 Let {L: {\bf R}^d \rightarrow {\bf R}^d} be a linear transformation.

  1. Show that there exists a non-negative real number {D} such that {m(L(E)) = D m(E)} for every elementary set {E} (note from previous exercises that {L(E)} is Jordan measurable). Hint: apply Exercise 3 to the map {E \mapsto m(L(E))}.
  2. Show that if {E} is Jordan measurable, then {L(E)} is also, and {m(L(E)) = D m(E)}.
  3. Show that {D = |\det L|}. (Hint: Work first with the case when {L} is an elementary transformation, using Gaussian elimination. Alternatively, work with the cases when {L} is a diagonal transformation or an orthogonal transformation, using the unit ball in the latter case, and use the polar decomposition.)

Exercise 12 Define a Jordan null set to be a Jordan measurable set of Jordan measure zero. Show that any subset of a Jordan null set is a Jordan null set.

Exercise 13 Show that (1) holds for all Jordan measurable {E \subset {\bf R}^d}.

Exercise 14 (Metric entropy formulation of Jordan measurability) Define a dyadic cube to be a half-open box of the form

\displaystyle  [\frac{i_1}{2^n}, \frac{i_1+1}{2^n}) \times \ldots \times [\frac{i_d}{2^n}, \frac{i_d+1}{2^n})

for some integers {n, i_1,\ldots,i_d}. Let {E \subset {\bf R}^d} be a bounded set. For each integer {n}, let {{\mathcal E}_*(E, 2^{-n})} denote the number of dyadic cubes of sidelength {2^{-n}} that are contained in {E}, and let {{\mathcal E}^*(E, 2^{-n})} be the number of dyadic cubes of sidelength {2^{-n}} that intersect {E}. Show that {E} is Jordan measurable if and only if

\displaystyle  \lim_{n \rightarrow \infty} 2^{-dn} ({\mathcal E}^*(E,2^{-n}) - {\mathcal E}_*(E,2^{-n})) = 0,

in which case one has

\displaystyle  m(E) = \lim_{n \rightarrow \infty} 2^{-dn} {\mathcal E}_*(E,2^{-n}) = \lim_{n \rightarrow \infty} 2^{-dn} {\mathcal E}^*(E,2^{-n}).

Exercise 15 (Uniqueness of Jordan measure) Let {d \geq 1}. Let {m': {\mathcal J}({\bf R}^d) \rightarrow {\bf R}^+} be a map from the collection {{\mathcal J}({\bf R}^d)} of Jordan-measurable subsets of {{\bf R}^d} to the nonnegative reals that obeys the non-negativity, finite additivity, and translation invariance properties. Show that there exists a constant {c \in {\bf R}^+} such that {m'(E) = cm(E)} for all Jordan measurable sets {E}. In particular, if we impose the additional normalisation {m'( [0,1)^d ) = 1}, then {m' \equiv m}.

Exercise 16 Let {d_1,d_2 \ge 1}, and let {E_1 \subset {\bf R}^{d_1}}, {E_2 \subset {\bf R}^{d_2}} be Jordan measurable sets. Show that {E_1 \times E_2 \subset {\bf R}^{d_1+d_2}} is Jordan measurable, and {m^{d_1+d_2}(E_1 \times E_2) = m^{d_1}(E_1) \times m^{d_2}(E_2)}.

Exercise 17 Let {P, Q} be two polytopes in {{\bf R}^d}. Suppose that {P} can be partitioned into finitely many sub-polytopes which, after being rotated and translated, form a cover of {Q}, with any two of the sub-polytopes in {Q} intersecting only at their boundaries. Conclude that {P} and {Q} have the same Jordan measure. The converse statement is true in one and two dimensions {d=1,2} (this is the Bolyai-Gerwien theorem), but false in higher dimensions (this was Dehn’s negative answer to Hilbert’s third problem).

The above exercises give a fairly large class of Jordan measurable sets. However, not every subset of {{\bf R}^d} is Jordan measurable. First of all, the unbounded sets are not Jordan measurable, by construction. But there are also bounded sets that are not Jordan measurable:

Exercise 18 Let {E \subset {\bf R}^d} be a bounded set.

  1. Show that {E} and the closure {\overline{E}} of {E} have the same outer Jordan measure.
  2. Show that {E} and the interior {E^\circ} of {E} have the same inner Jordan measure.
  3. Show that {E} is Jordan measurable if and only if the topological boundary {\partial E} of {E} has outer Jordan measure zero.
  4. Show that the bullet-riddled square { [0,1]^2 \backslash {\bf Q}^2}, and set of bullets {[0,1]^2 \cap Q^2}, both have inner Jordan measure zero and outer Jordan measure one. In particular, both sets are not Jordan measurable.

Informally, any set with a lot of “holes”, or a very “fractal” boundary, is unlikely to be Jordan measurable. In order to measure such sets we will need to develop Lebesgue measure, which is done in the next set of notes.

Exercise 19 (Carathéodory type property) Let {E \subset {\bf R}^d} be a bounded set, and {F \subset {\bf R}^d} be an elementary set. Show that {m^{*,(J)}(E) = m^{*,(J)}(E \cap F) + m^{*,(J)}(E \backslash F)}.

— 3. Connection with the Riemann integral —

To conclude these notes we briefly discuss the relationship between Jordan measure and the Riemann integral (or the equivalent Darboux integral). For simplicity we will only discuss the classical one-dimensional Riemann integral on an interval {[a,b]}, though one can extend the Riemann theory without much difficulty to higher-dimensional integrals on Jordan measurable sets. (In later notes, this Riemann integral will be superceded by the Lebesgue integral.)

Definition 4 (Riemann integrability) Let {[a,b]} be an interval of positive length, and {f: [a,b] \rightarrow {\bf R}} be a function. A tagged partition {{\mathcal P} = ((x_0,x_1,\ldots,x_n), (x^*_1,\ldots,x^*_n))} of {[a,b]} is a finite sequence of real numbers {a = x_0 < x_1 < \ldots < x_n = b}, together with additional numbers {x_{i-1} \leq x^*_i \leq x_i} for each {i=1,\ldots,n}. We abbreviate {x_i-x_{i-1}} as {\delta x_i}. The quantity {\Delta({\mathcal P}) := \sup_{1 \leq i \leq n} \delta x_i} will be called the norm of the tagged partition. The Riemann sum {{\mathcal R}( f, {\mathcal P} )} of {f} with respect to the tagged partition {{\mathcal P}} is defined as

\displaystyle  {\mathcal R}( f, {\mathcal P} ) := \sum_{i=1}^n f(x^*_i) \delta x_i.

We say that {f} is Riemann integrable on {[a,b]} if there exists a real number, denoted {\int_a^b f(x)\ dx} and referred to as the Riemann integral of {f} on {[a,b]}, for which we have

\displaystyle  \int_a^b f(x)\ dx = \lim_{\Delta({\mathcal P}) \rightarrow 0} {\mathcal R}(f, {\mathcal P})

by which we mean that for every {\epsilon > 0} there exists {\delta > 0} such that {|{\mathcal R}(f, {\mathcal P}) - \int_a^b f(x)\ dx| \leq \epsilon} for every tagged partition {{\mathcal P}} with {\Delta({\mathcal P}) \leq \delta}.

If {[a,b]} is an interval of zero length, we adopt the convention that every function {f: [a,b] \rightarrow {\bf R}} is Riemann integrable, with a Riemann integral of zero.

Note that unbounded functions cannot be Riemann integrable (why?).

The above definition, while geometrically natural, can be awkward to use in practice. A more convenient formulation of the Riemann integral can be formulated using some additional machinery.

Exercise 20 (Piecewise constant functions) Let {[a,b]} be an interval. A piecewise constant function {f: [a,b] \rightarrow {\bf R}} is a function for which there exists a partition of {[a,b]} into finitely many intervals {I_1,\ldots,I_n}, such that {f} is equal to a constant {c_i} on each of the intervals {I_i}. If {f} is piecewise constant, show that the expression

\displaystyle  \sum_{i=1}^n c_i |I_i|

is independent of the choice of partition used to demonstrate the piecewise constant nature of {f}. We will denote this quantity by {\hbox{p.c.} \int_a^b f(x)\ dx}, and refer to it as the piecewise constant integral of {f} on {[a,b]}.

Exercise 21 (Basic properties of the piecewise constant integral) Let {[a,b]} be an interval, and let {f, g: [a,b] \rightarrow {\bf R}} be piecewise constant functions. Establish the following statements:

  1. (Linearity) For any real number {c}, {cf} and {f+g} are piecewise constant, with {\hbox{p.c.} \int_a^b cf(x)\ dx = c \hbox{p.c.} \int_a^b f(x)\ dx} and {\hbox{p.c.} \int_a^b f(x)+g(x)\ dx = \hbox{p.c.} \int_a^b f(x)\ dx + \hbox{p.c.} \int_a^b g(x)\ dx}.
  2. (Monotonicity) If {f \leq g} pointwise (i.e. {f(x) \leq g(x)} for all {x \in [a,b]}) then {\hbox{p.c.} \int_a^b f(x)\ dx \leq \hbox{p.c.} \int_a^b g(x)\ dx}.
  3. (Indicator) If {E} is an elementary subset of {[a,b]}, then the indicator function {1_E: [a,b] \rightarrow {\bf R}} (defined by setting {1_E(x) := 1} when {x \in E} and {1_E(x):=0} otherwise) is piecewise constant, and {\hbox{p.c.} \int_a^b 1_E(x)\ dx = m(E)}.

Definition 5 (Darboux integral) Let {[a,b]} be an interval, and {f: [a,b] \rightarrow {\bf R}} be a bounded function. The lower Darboux integral {\underline{\int_a^b} f(x)\ dx} of {f} on {[a,b]} is defined as

\displaystyle  \underline{\int_a^b} f(x)\ dx := \sup_{g \leq f, \hbox{ piecewise constant}} \hbox{p.c.} \int_a^b g(x)\ dx,

where {g} ranges over all piecewise constant functions that are pointwise bounded above by {f}. (The hypothesis that {f} is bounded ensures that the supremum is over a non-empty set.) Similarly, we define the upper Darboux integral {\overline{\int_a^b} f(x)\ dx} of {f} on {[a,b]} by the formula

\displaystyle  \overline{\int_a^b} f(x)\ dx := \inf_{h \geq f, \hbox{ piecewise constant}} \hbox{p.c.} \int_a^b h(x)\ dx.

Clearly {\underline{\int_a^b} f(x)\ dx \leq \overline{\int_a^b} f(x)\ dx}. If these two quantities are equal, we say that {f} is Darboux integrable, and refer to this quantity as the Darboux integral of {f} on {[a,b]}.

Note that the upper and lower Darboux integrals are related by the reflection identity

\displaystyle  \overline{\int_a^b} -f(x)\ dx = - \underline{\int_a^b} f(x)\ dx.

Exercise 22 Let {[a,b]} be an interval, and {f: [a,b] \rightarrow {\bf R}} be a bounded function. Show that {f} is Riemann integrable if and only if it is Darboux integrable, in which case the Riemann integral and Darboux integrals are equal.

Exercise 23 Show that any continuous function {f: [a,b] \rightarrow {\bf R}} is Riemann integrable. More generally, show that any bounded, piecewise continuous function {f: [a,b] \rightarrow {\bf R}} is Riemann integrable.

Now we connect the Riemann integral to Jordan measure in two ways. First, we connect the Riemann integral to one-dimensional Jordan measure:

Exercise 24 (Basic properties of the Riemann integral) Let {[a,b]} be an interval, and let {f, g: [a,b] \rightarrow {\bf R}} be Riemann integrable. Establish the following statements:

  1. (Linearity) For any real number {c}, {cf} and {f+g} are Riemann integrable, with {\int_a^b cf(x)\ dx = c \cdot \int_a^b f(x)\ dx} and {\int_a^b f(x)+g(x)\ dx = \int_a^b f(x)\ dx + \int_a^b g(x)\ dx}.
  2. (Monotonicity) If {f \leq g} pointwise (i.e. {f(x) \leq g(x)} for all {x \in [a,b]}) then {\int_a^b f(x)\ dx \leq \int_a^b g(x)\ dx}.
  3. (Indicator) If {E} is a Jordan measurable subset of {[a,b]}, then the indicator function {1_E: [a,b] \rightarrow {\bf R}} (defined by setting {1_E(x) := 1} when {x \in E} and {1_E(x):=0} otherwise) is Riemann integrable, and {\int_a^b 1_E(x)\ dx = m(E)}.

Finally, show that these properties uniquely define the Riemann integral, in the sense that the functional {f \mapsto \int_a^b f(x)\ dx} is the only map from the space of Riemann integrable functions on {[a,b]} to {{\bf R}} which obeys all three of the above properties.

Next, we connect the integral to two-dimensional Jordan measure:

Exercise 25 (Area interpretation of the Riemann integral) Let {[a,b]} be an interval, and let {f: [a,b] \rightarrow {\bf R}} be a bounded function. Show that {f} is Riemann integrable if and only if the sets {E_+ := \{ (x,t): x \in [a,b]; 0 \leq t \leq f(x)\}} and {E_- := \{ (x,t): x \in [a,b]; f(x) \leq t \leq 0\}} are both Jordan measurable in {{\bf R}^2}, in which case one has

\displaystyle  \int_a^b f(x)\ dx = m^2( E_+ ) - m^2( E_- ),

where {m^2} denotes two-dimensional Jordan measure. (Hint: First establish this in the case when {f} is non-negative.)

Exercise 26 Extend the definition of the Riemann and Darboux integrals to higher dimensions, in such a way that analogues of all the previous results hold.