Introduction
Measurement, in so far as empiricism is concerned, is an assignment of numbers to objects in order to represent properties of the objects in a more fine-grained way than classificatory or comparative methods. The question of “why can we measure with numbers?” is the central philosophical subject of measurement theory.
Intuitively, we know that not all assignments of numbers to objects constitute a meaningful measurement. In mensuration, the classical quantities of measurement are length, area and volume, which can be understood easily by appealing to our spatial intuition. Helmholtz noted that for measurement to be meaningful, there must be some way to combine objects that resembles addition of numbers. Holder’s work, also includes a “solvability condition”, which provides a notion of “subtraction” of objects, by requiring that for any two objects, there exists a third object such that the first can be obtained by combining the second and the third. Moreover, no object is “infinitely greater” than another, akin to the Archimedean property of reals, requiring that for any two objects, there exists a finite number of copies of one that can be combined to exceed the other. These insights from measurement theory suggest an investigation into the familiar measures of length, area and volume, to glean the essential underlying structure.
This article aims to develop these ideas in the context of measure theory, starting with the notion of elementary sets, and then defining the Jordan measure, which serves as a concrete example of a measure that can be extended to the more general Lebesgue measure.
Prerequisites
Definition 1 (Infimum and Supremum) Let \(S\) be a subset of a poset \((P, \leq)\). An upper bound of \(S\) is an element \(u \in P\) such that \(\forall s \in S, s \leq u\). A lower bound of \(S\) is an element \(l \in P\) such that \(\forall s \in S, l \leq s\). The supremum (least upper bound) of \(S\), denoted \(\sup S\), is the least element among the upper bounds of \(S\), if it exists. The infimum (greatest lower bound) of \(S\), denoted \(\inf S\), is the greatest element among the lower bounds of \(S\), if it exists. Often, one speaks of infimum or supremum of an expression over a set. It is useful to list the logical expressions for these cases: \[ L = \sup_{\theta \in \Theta}{f(\theta)} \iff (\forall \theta \in \Theta, f(\theta) \leq L) \land (\forall \epsilon > 0, \exists \theta' \in \Theta | f(\theta') > L - \epsilon) \tag{1}\]
\[ L = \inf_{\theta \in \Theta}{f(\theta)} \iff (\forall \theta \in \Theta, f(\theta) \geq L) \land (\forall \epsilon > 0, \exists \theta' \in \Theta | f(\theta') < L + \epsilon) \tag{2}\]
Remark. An easy way to digest these definitions, is to think of them as “opaque” algorithms. For instance, the knowledge of a supremum \(L\), essentially provides us a \(\theta'\) (by specifying the parameter \(\epsilon\)) such that \(f(\theta')\) is as close as we want to \(L\) from below. Crucially, supremum is “optimal” in the sense that, \(L\) is the largest value for which we may produce \(\theta'\) as described above. The notion transfers to infimum in a similar way.
The following proposition is useful for restricting the domain of supremum expressions.
Proposition 1 (Restriction of Supremum) Let \(f: \Theta \to \mathbb{R}\) be a function such that \(L = \sup_{\theta \in \Theta} f(\theta)\) exists. If \(\forall \theta \in \Theta, \exists \theta' \in \mathcal{A} \subset \Theta\) such that \(f(\theta') \geq f(\theta)\), then: \[ L = \sup_{\theta \in \Theta} f(\theta) = \sup_{\theta \in \mathcal{A}} f(\theta) \]
Proposition 2 (Iterated Suprema) Let \(f: \Theta_1 \times \Theta_2 \to \mathbb{R}\) be a function such that \(L = \sup_{\theta_1 \in \Theta_1, \theta_2 \in \Theta_2} f(\theta_1, \theta_2)\) exists, then: \[ L = \sup_{\theta_1 \in \Theta_1} \sup_{\theta_2 \in \Theta_2} f(\theta_1, \theta_2) = \sup_{\theta_2 \in \Theta_2} \sup_{\theta_1 \in \Theta_1} f(\theta_1, \theta_2) \]
Proof. Let \(T = \sup_{\theta_1 \in \Theta_1} g(\theta_1)\) where \(g(\theta_1) = \sup_{\theta_2 \in \Theta_2} f(\theta_1, \theta_2)\). Clearly, \(f(\theta_1, \theta_2) \leq L\), so \(g(\theta_1) \leq L\) then: \(T \leq L\). Now, \(\forall \epsilon > 0, \exists \theta_1' \in \Theta_1, \theta_2' \in \Theta_2| f(\theta_1', \theta_2') > L - \epsilon \implies g(\theta_1') \geq f(\theta_1', \theta_2') > L - \epsilon\), and thus \(L \leq T\). The claim follows by symmetry.
Proposition 3 (Sum of Suprema) Let \(f, g: \Theta \to \mathbb{R}\) be functions such that \(\sup_{\theta \in \Theta} f(\theta)\) and \(\sup_{\theta \in \Theta} g(\theta)\) exist. Then: \[ \sup_{\theta \in \Theta} (f(\theta) + g(\theta)) \leq \sup_{\theta \in \Theta} f(\theta) + \sup_{\theta \in \Theta} g(\theta) \]
Definition 2 (Sum of a Collection) Let \(I\) be an index set, and let \((a_i)_{i \in I}\) be a collection of non-negative real numbers. The sum of the collection is defined as: \[ \sum_{i \in I} a_i = \sup_{A \subset I, A \text{ finite}} \sum_{i \in A} a_i \]
Proposition 4 (Commutativity of Sum and Supremum) Let \((a_i)_{i \in I},(b_i)_{i \in I}\) be collections of non-negative real numbers indexed by the same set \(I\), then: \[ \sup_{C \subseteq I, C \text{ finite}} \sum_{i \in C}{(a_i + b_i)} = \sup_{A \subseteq I, A \text{ finite}} \sum_{i \in A} a_i + \sup_{B \subseteq I, B \text{ finite}} \sum_{i \in B} b_i \]
Proof. Let \(A, B \subseteq I\) be finite sets, and consider \(s(A, B) = \sum_{i \in A}{a_i} + \sum_{i \in B}{b_i}\). Set \(C = A \cup B\). Due to the non-negativity of the \(a_i\) and \(b_i\), \(s(C, C) \geq s(A, B)\). By definition of supremum, \[ s(A, B) \leq s(C, C) \leq \sup_{D \subseteq I, D \text{ finite}}{\sum_{i \in D} (a_i + b_i)} = L \]
The claim follows by taking iterated supremum over \(A\) and \(B\) on the left hand side (which maintains the inequality since \(L\) is an upper bound) and combining with Proposition 3.
Theorem 1 (Tonelli’s Theorem for Sums) Let \((a_{ij})_{i \in I, j \in J}\) be a collection indexed by two sets \(I\) and \(J\), such that \(a_{ij} \in [0, \infty]\), then: \[ \sum_{(i, j) \in I \times J}{a_{ij}} = \sum_{i \in I} \sum_{j \in J} a_{ij} = \sum_{j \in J} \sum_{i \in I} a_{ij} \]
Proof. It suffices to prove the first equality, since the second will follow by symmetry. Expand both sides of the equality using Definition 2, and apply Proposition 1 to restrict the partial sum on the left-hand side to finite sets of the form \(A \times B\) where \(A \subseteq I\) and \(B \subseteq J\) as the summands are non-negative. The target is now, \[ \sup_{\substack{A \subseteq I, A \text{ finite} \\ B \subseteq J, B \text{ finite}} } \sum_{\substack{i \in A \\ j \in B}} a_{ij} = \sup_{A \subseteq I, A \text{ finite}} \sum_{i \in A} \sup_{B \subseteq J, B \text{ finite}} \sum_{j \in B} a_{ij} \] But, from Proposition 4 (extended to finite sum): \[ \sup_{A \subseteq I, A \text{ finite}} \sum_{i \in A} \sup_{B \subseteq J, B \text{ finite}} \sum_{j \in B} a_{ij} = \sup_{A \subseteq I, A \text{ finite}} \sup_{B \subseteq J, B \text{ finite}} \sum_{i \in A} \sum_{j \in B} a_{ij} \] So, the claim follows by Proposition 2.
Proposition 5 (Countability of Support of a Sum) Let \((a_i)_{i \in I}\) be a collection of non-negative real numbers indexed by a set \(I\), and let \(S = \{i \in I | a_i > 0\}\) be the support of the collection, then: \[ \sum_{i \in I} a_i < \infty \implies S \text{ is countable} \]
Proof. We aim to recover a countable subset of \(I\) that converges to the sum. By Definition 2, we can find for each \(\epsilon_n = \frac{L}{n}\), a finite set \(A_n \subseteq I\) such that \(\sum_{i \in A_n} a_i > L - \epsilon_n\) where \(L = \sum_{i \in I} a_i\). Let \(A = \bigcup_{n=1}^{\infty} A_n\), then \(A\) is countable as a countable union of finite sets. Now, restrict the supremum in the definition of \(\sum_{i \in A}{a_i}\) by Proposition 1 to precisely the finite sets \(A_n\), then clearly \(\sum_{i \in A} a_i = L\).
Elementary Sets
Elementary sets are essentially finite unions of boxes, where a box is a Cartesian product of intervals. We begin with these sets, since the notion of “size”, or “measure” is most intuitive for them. In one dimension, a box is just an interval of either the form \([a, b], (a, b), [a, b), (a, b]\) where \(a < b\). In higher dimensions, a box is a Cartesian product of such intervals. For instance, in two dimensions, a box could be \([a_1, b_1] \times [a_2, b_2]\) which represents a rectangle in the plane. The measure of a box is the product of the lengths of intervals. The measure of two or more non-overlapping boxes is just the sum of their measures. Since this method of assigning numbers to sets is restricted only to boxes and their unions, this is termed as the elementary measure.
Definition 3 (Elementary Measure of an Interval) Let \(I\) be an interval of the form \([a, b], (a, b), [a, b), (a, b]\) where \(a < b\), then the elementary measure of \(I\) is defined as: \[ m(I) = b - a \]
Proposition 6 (Measure as an extension of Count via discretization) Let \(S\) be a box in \(\mathbb{R}^d\), and \(\mathbb{Z}^d\) be the integer lattice in \(\mathbb{R}^d\), then the elementary measure of \(S\) is: \[ m(S) = \lim_{N \to \infty} \frac{1}{N^d} \left| S \cap \frac{1}{N} \mathbb{Z}^d \right| \tag{3}\]
Proof. Let \(I = [a, b]\) be an interval, and \(A_n = I \cap \frac{1}{N} \mathbb{Z}\). Clearly, \(p \in A_n \iff Np \in [aN, bN] \cap \mathbb{Z}\). Let \(m = \lfloor{bN}\rfloor, n = \lceil{aN}\rceil\) so \(|A_n| = m - n + 1\) with \(m \in (bN-1, bN]\) and \(n \in [aN, aN + 1)\). Thus, \(N(b-a) - 1 \leq |A_n| \leq N(b-a) + 1\). By squeezing, \(\lim_{N \to \infty} \frac{1}{N} |A_n| = b - a\). By the definition of box as cartesian product of intervals, the result follows by taking products of limits.
Remark. This proposition can serve as an alternate definition of the elementary measure for boxes, and also provides an intuitive connection between the continuous notion of measure and the discrete notion of counting.
Equation 3 cannot be applied in general to all sets, or even subsets of elementary sets since the limit may not exist. For instance, consider the set \(S = \{\frac{m}{n} \in \mathbb{Q} \bigg | m \leq n, n \text{ is prime}\}\). Then, \(A_n = S \cap \frac{1}{N} \mathbb{Z}\). Now \(f(n) = \frac{1}{n}|A_n|\). Clearly for any prime \(p\), \(f(p) = 1\), but \(f(p^2) = \frac{1}{p}\), so the limit does not exist.
We shall now review some properties of the elementary measure.
Proposition 7 (Properties of Elementary Measure and Elementary Sets) Let \(S, T\) be elementary sets in \(\mathbb{R}^d\), then:
- There exists a finite collection of disjoint boxes \(\{B_i\}_{i\in[n]}\), such that \(S = \bigcup_{i=1}^n B_i\) and \(B_i \cap B_j = \emptyset\) for \(i \neq j\) and \(m(S) = \sum_{i=1}^n m(B_i)\).
- \(T = x + S = \{x + s : s \in S\}\) is an elementary set for any \(x \in \mathbb{R}^d\), and \(m(T) = m(S)\)
- \(U = S \cup T\) is an elementary set. So \(m(U)\) exists.
- \(I = S \cap T\) is an elementary set. So \(m(I)\) exists.
- \(S \triangle T\) is an elementary set, where \(S \triangle T\) is the symmetric difference of \(S\) and \(T\).
- \(S\setminus T\) is an elementary set.
- If \(S \subseteq T\), then \(m(S) \leq m(T)\).
Proof. Three basic facts of boxes are that: (a) the intersection of two boxes is always a box (the empty set is a degenerate box; so, if a set is the union of two boxes, then it is the union of three disjoint boxes), (b) the measure of a box does not change by simply moving it, and (c) the set difference of two boxes is at most a union of \(2^d\) disjoint boxes. By induction, we can conclude that the set difference of a box and an elementary set is an elementary set.
Now, if unions of (n-1) boxes can be written as union of disjoint boxes, and consider a set \(A = B_1 \cup B_2 \cup ... \cup B_n\) where \(B_i\) are boxes. By the induction hypothesis, \(B_1 \cup B_2 \cup ... \cup B_{n-1}\) can be written as a union of disjoint boxes, say \(C_1, C_2, ..., C_m\). Now, \(A = C_1 \cup C_2 \cup ... \cup C_m \cup B_n\). Now \(B_n \setminus \cup _{i=1}^m C_i\) is a union of disjoint boxes, say \(D_1, D_2, ..., D_k\). Then \(A = C_1 \cup C_2 \cup ... \cup C_m \cup D_1 \cup D_2 \cup ... \cup D_k\). This proves (1). (3) follows by definition, (4) follows from basic fact as \(S = \cup{B_i}, T = \cup{B'_i}, S \cap T = \cup_{i,j}{B_i \cap B'_j}\) is a union of disjoint sets when \(B_i, B'_i\) are disjoint. (5), (6) follow from (3), (4). (7) follows from the fact that \(T \setminus S\) and \(S\) are disjoint elementary sets with non-negative measure.
Remark. The algorithm provided to establish (1) can be improved by splitting boxes iteratively to write comprising intervals in each dimension as union of disjoint intervals in \(O(d n \log n)\). This is the standard, elegant approach to the proof. Furthermore, it is noteworthy that only (1) and (2) need to be satisfied for the proof to go through, and no other properties of the elementary measure are used thereafter.
A natural question to ask is: if Definition 3 was defined any differently, how would the elementary measure change?
Lemma 1 (Scaling Property of Elementary Measure) Let \(m: \mathcal{E}(\mathbb{R}^d) \to [0, \infty)\) be a map on the collection of elementary sets in \(\mathbb{R}^d\) satisfying (1), (2) from Proposition 7, and \(\mathbf{l} \in [0, \infty)^d\), then: \[ m([0, l_1) \times [0, l_2) \times ... \times [0, l_d)) = m([0, 1)^d) \prod_{i=1}^d l_i \]
Proof. It is easy to see that the claim holds for \(\mathbf{l} \in \mathbb{Q}_{\geq 0}^d\), we can then argue by density. Suppose \(l_i = \frac{p_i}{q_i}\) where \(p_i, q_i \in \mathbb{Z}_{\geq 0}\), then the set \([0, l_1) \times [0, l_2) \times ... \times [0, l_d)\) can be formed by duplicting and translating the box \([0, \frac{1}{q_1}) \times [0, \frac{1}{q_2}) \times ... \times [0, \frac{1}{q_d})\) as many times as \(p_1 p_2 ... p_d\). Now, had \(p_i = q_i\), we would have \(m([0,1)^d)\), so \(m([0, \frac{1}{q_i}) \times ... \times [0, \frac{1}{q_d})\) itself is \(m([0, 1)^d) \prod_{i=1}^d \frac{1}{q_i}\). Combining we have, \(\forall \mathbf{l} \in \mathbb{Q}_{\geq 0}^d, m([0, l_1) \times [0, l_2) \times ... \times [0, l_d)) = m([0, 1)^d) \prod_{i=1}^d l_i\).
Now by density of \(\mathbb{Q}\) in \(\mathbb{R}\), for any \(\mathbf{l} \in [0, \infty)^d\) and \(\epsilon > 0\), there exists \(\mathbf{l^up}, \mathbf{l^down} \in \mathbb{Q}_{\geq 0}^d\) such that \(l_i < l_i^{up} < l_i + \epsilon\) and \(l_i - \epsilon < l_i^{down} < l_i\). Clearly, \([0, l_1^{down}) \times [0, l_2^{down}) \times ... \times [0, l_d^{down}) \subseteq [0, l_1) \times [0, l_2) \times ... \times [0, l_d) \subseteq [0, l_1^{up}) \times [0, l_2^{up}) \times ... \times [0, l_d^{up})\), so by monotonicity of \(m\) (which follows from (1) and (2) of Proposition 7), we have: \[ m([0, 1)^d) \prod_{i=1}^d l_i^{down} \leq m([0, l_1) \times [0, l_2) \times ... \times [0, l_d)) \leq m([0, 1)^d) \prod_{i=1}^d l_i^{up} \]
So the claim follows by squeezing.
Corollary 1 (Uniqueness of Elementary Measure) Let \(m', m: \mathcal{E}(\mathbb{R}^d) \to \mathbb{R}_{0}^{+}\) be two maps on the collection of elementary sets in \(\mathbb{R}^d\) satisfying (1), (2) from Proposition 7, then: \(m' = C m\) for some constant \(C > 0\).
Proof. It suffices to show that \(m'\) and \(m\) agree upto a constant on boxes cornered at origin. By Lemma 1, \(m'([0, l_1) \times [0, l_2) \times ... \times [0, l_d)) = m'([0, 1)^d) \prod_{i=1}^d l_i\) and \(m([0, l_1) \times [0, l_2) \times ... \times [0, l_d)) = m([0, 1)^d) \prod_{i=1}^d l_i\). So, \(m'([0, l_1) \times [0, l_2) \times ... \times [0, l_d)) = C m([0, l_1) \times [0, l_2) \times ... \times [0, l_d))\) where \(C = \frac{m'([0, 1)^d)}{m([0, 1)^d)}\).
Jordan Measure
Having established the notion of measure for elementary sets, which are finite unions of boxes, we may wish to extend this notion to sets that are “almost” elementary, in the sense that they can be arbitrarily well approximated by elementary sets. This leads us to the concept of Jordan measure for sets in \(\mathbb{R}^d\) that can be approximated from the inside and outside by elementary sets.
Definition 4 (Jordan Measure) Let \(S \subseteq \mathbb{R}^d\) be a bounded set. The Jordan inner measure of \(S\) is defined as: \[ m_{*,J}(S) = \sup_{F \subseteq S, F \in \mathcal{E}(\mathbb{R}^d)}{m(F)} \]
The Jordan outer measure of \(S\) is defined as: \[ m^{*,J}(S) = \inf_{F \supseteq S, F \in \mathcal{E}(\mathbb{R}^d)}{m(F)} \]
The set \(S\) is said to be Jordan measurable with Jordan measure \(m_J(S)\) if \(m_{*,J}(S) = m^{*,J}(S)\).
Intuitively, the outer measure of a set is the limiting value of the volume of all measurable sets enclosing it. Likewise, the inner measure is the limiting value of the volume of all measurable sets contained within it. Yet another way of visualizing the Jordan measure is to imagine the volume being approximated by arbitrarily small voxels. This notion will be made precise later.
Proposition 8 (Characterization of Jordan Measurability) Let \(S \subseteq \mathbb{R}^d\) be a bounded set, then the following are equivalent:
- \(S\) is Jordan measurable.
- \(\forall \epsilon > 0, \exists B, A \in \mathcal{E}(\mathbb{R}^d)\) such that \(B \supset S \supset A\) and \(m(B \setminus A) < \epsilon\).
- \(\forall \epsilon > 0, \exists A \in \mathcal{E}(\mathbb{R}^d)\) such that \(m^{*,J}(A \triangle S) < \epsilon\).
Proof. We will first establish (1)-(2). Suppose \(S\) is Jordan measurable with measure \(m\). Then, \(\forall \epsilon > 0 \exists B, A \in \mathcal{E}(\mathbb{R}^d) | m(B) < m + \frac{\epsilon}{2}, m(A) > m - \frac{\epsilon}{2} \implies m(B \setminus A) < m + \frac{\epsilon}{2} - (m - \frac{\epsilon}{2}) = \epsilon\) by Definition 1. Now, suppose (2) holds. We have (with suitable qualifiers over appropriate domains) \(m(B) < m(A) + \epsilon \leq \sup m(A) + \epsilon \implies \inf m(B) \leq \sup m(A)\). But \(B \supset A \implies m(B) \geq m(A)\), so \(\inf m(B) = \sup m(A)\).