Valueless Measures on Pointless Spaces

Abstract

On our ordinary representations of space, space is composed of indivisible, dimensionless points; extended regions are understood as infinite sets of points. Region-based theories of space reverse this atomistic picture, by taking as primitive several relations on extended regions, and recovering points as higher-order abstractions from regions. Over the years, such theories have focused almost exclusively on the topological and geometric structure of space. We introduce to region-based theories of space a new primitive binary relation (‘qualitative probability’) that is tied to measure. It expresses that one region is smaller than or equal in size to another. Algebraic models of our theory are separation σ-algebras with qualitative probability: $(\mathbb {B}, \ll , \preceq )$, where $\mathbb {B}$ is a Boolean σ-algebra, ≪ is a separation relation on $\mathbb {B}$, and ≼ is a qualitative probability on $\mathbb {B}$. We show that from algebraic models of this kind we can, in an interesting class of cases, recover a compact Hausdorff topology X, together with a countably additive measure μ on a σ-field of Borel subsets of that topology, and that $(\mathbb {B}, \ll , \preceq )$ is isomorphic to a ‘standard model’ arising out of the pair (X, μ). It follows from one of our main results that any closed ball in Euclidean space, $\mathbb {R}^{n}$, together with Lebesgue measure arises in this way from a separation σ-algebra with qualitative probability.

Appendix A: Proof of Theorem 18

In this appendix, we prove Theorem 18: the extension of Villegas’s representation theorem (Theorem 17) to abstract Boolean σ-algebras. We begin by recalling the Loomis-Sikorski representation theorem [17] and [23]:

Theorem 121

If $\mathbb {B}$ is a Boolean σ-algebra, then there is a σ-field of sets, $\mathbb {F}$, and a σ-ideal I of $\mathbb {F}$ such that $\mathbb {B}$ is isomorphic to the quotient $\mathbb {F} \slash I$.

Throughout this appendix, let $\mathbb {B}$ be a Boolean σ-algebra, and let ≼ be a monotonely continuous, bottomless qualitative probability on $\mathbb {B}$. By Theorem 120, we know that there is a σ-field $\mathbb {F}$ of subsets of a set Ω, a σ-ideal I of $\mathbb {F}$ and an isomorphism:

$$\varphi: \mathbb{B} \to \mathbb{F} \slash I$$

We denote by upper-case letters A, B, C, … elements of the σ-field $\mathbb {F}$, and by |A| the equivalence class in $\mathbb {F} \slash I$ containing the set A. Note that since $\mathbb {F}$ is a σ-field and I is a σ-ideal, the map $|\cdot |: \mathbb {F} \to \mathbb {F} \slash I$ defined by A↦|A| is a Boolean σ-homomorphism.

Define the relation $\preceq ^{\prime }$ on $\mathbb {F} \slash I$ by putting: $|A| \preceq ^{\prime } |B|$ iff φ^− 1(|A|) ≼ φ^− 1(|B|). Then clearly $\preceq ^{\prime }$ is a monotonely continuous, bottomless qualitative probability on $\mathbb {F} \slash I$. We now define the relation $\preceq _{\mathbb {F}}$ on $\mathbb {F}$ by putting:

$$A \preceq_{\mathbb F} B \text{ iff } |A| \preceq^{\prime} |B|$$

Note that for any $A, B \in \mathbb {F}$, $A \prec _{\mathbb {F}} B$ iff $|A| \prec ^{\prime } |B|$.

Lemma 122

$\preceq _{\mathbb F}$ is a monotonely continuous, bottomless qualitative probability on $\mathbb {F}$.

Proof

The reader can verify that $\preceq _{\mathbb F}$ is a qualitative probability on $\mathbb {F}$. To see that $\preceq _{\mathbb F}$ is monotonely continuous, suppose that $A_{1} \subseteq A_{2} \subseteq A_{3} \dots$ and $A_{n} \preceq _{\mathbb F} B$ for all $n \in \mathbb N$. Since $|\cdot |: \mathbb {F} \to \mathbb {F} \slash I$ is a Boolean homomorphism, $|A_{1}| \leq |A_{2}| \leq |A_{3}| \dots$, and by definition of $\preceq _{\mathbb {F}}$, $|A_{n}| \preceq ^{\prime } |B|$ for all $n \in \mathbb N$. So by monotone continuity of $\preceq ^{\prime }$, $\bigvee _{n \in \mathbb {B}} |A_{n}| \preceq ^{\prime } |B|$. But since |⋅| is a Boolean σ-homomorphism, $\bigvee _{n \in \mathbb {B}} |A_{n}|= |\bigcup _{n \in \mathbb N} A_{n} |$. Therefore $\bigcup _{n \in \mathbb N} A_{n} \preceq _{\mathbb {F}} B$.

To see that $\preceq _{\mathbb {F}}$ is bottomless, suppose that $A \in \mathbb {F}$ and $\emptyset \prec _{\mathbb {F}} A$. Then $|\emptyset | \prec ^{\prime } |A|$. Since $\preceq ^{\prime }$ is bottomless, there exists $b \in \mathbb {F} \slash I$ such that b ≤|A| and $|\emptyset | \prec ^{\prime } b \prec ^{\prime } |A|$. Let b = |B|. We can assume WLOG that $B \subseteq A$. (If not, let $B^{\prime }=B \cap A$. Then clearly $B^{\prime } \subseteq A$. Since |B|≤|A|, we have: $b=|B| = |B| \wedge |A| = |B \cap A| = |B^{\prime }|$.) Since $|\emptyset | \prec ^{\prime } |B| \prec ^{\prime } |A|$, we have $\emptyset \prec _{\mathbb {F}} B \prec _{\mathbb {F}} A$. □

Lemma 123

There is a unique probability measure on $\mathbb {F}$ that represents $\preceq _{\mathbb {F}}$, and it is countably additive.

Proof

Immediate from Lemma 121 and Theorem 17. □

Let $m_{\mathbb {F}}$ be the unique probability measure on $\mathbb {F}$ that represents $\preceq _{\mathbb {F}}$. Note that for any A ∈ I, $m_{\mathbb {F}}(A)=0$. Indeed, if A ∈ I, then |A| = |∅|, so $|A| \preceq ^{\prime } |\emptyset |$, and therefore $A \preceq _{\mathbb {F}} \emptyset$. Since $m_{\mathbb {F}}$ represents $\preceq _{\mathbb {F}}$, $m_{\mathbb {F}}(A) \leq m_{\mathbb {F}}(\emptyset ) =0$.

Define the function $m^{\prime }$ on $\mathbb {F} \slash I$ by putting:

$$ m^{\prime}(|A|)= m_{\mathbb{F}}(A) $$

(3)

Note that $m^{\prime }$ is well-defined, because if $|A|=|A^{\prime }|$, then $A \bigtriangleup A^{\prime } \in I$, so $m_{\mathbb {F}}(A \bigtriangleup A^{\prime }) =0$, and $m_{\mathbb {F}}(A) = m_{\mathbb {F}}(A^{\prime })$.

Proposition 124

$m^{\prime }$ is the unique probability measure on $\mathbb {F} \slash I$ that represents $\preceq ^{\prime }$, and it is countably additive.

Proof

We first show that $m^{\prime }$ is a countably additive probability measure that represents $\preceq ^{\prime }$, and then prove the uniqueness claim. Note that $m^{\prime }(1) = m^{\prime }(|{\Omega }|) = m_{\mathbb {F}}({\Omega })=1$, since $m_{\mathbb {F}}$ is a probability measure on $\mathbb {F}$. Clearly also $m^{\prime }$ is real-valued and non-negative, since $m_{\mathbb {F}}$ is. Suppose that $\{a_{n} | n \in \mathbb N\}$ is a pairwise disjoint set of elements in $\mathbb {F} \slash I$. Then a_n = |A_n| for some $A_{n} \in \mathbb {F}$. WLOG we can assume that the A_n’s are pairwise disjoint. (If not, let B₀ = A₀ and let $B_{n+1}=A_{n+1} \setminus (A_{0} \cup {\dots } \cup A_{n})$, for $n \in \mathbb N$. Then the B_n’s are pairwise disjoint, and |A_n| = |B_n| for all $n \in \mathbb N$.) So we have:

$$ \begin{array}{@{}rcl@{}} m^{\prime}\left( \bigvee_{n \in \mathbb N} a_{n}\right) &=& m^{\prime}\left( \bigvee_{n \in \mathbb N} |A_{n}|\right) \\ &=& m^{\prime}\left( \left|\bigcup_{n \in \mathbb N} A_{n}\right|\right) \\ &=& m_{\mathbb{F}}\left( \bigcup_{n \in \mathbb N} A_{n}\right) \\ &=& \sum\limits_{n \in \mathbb N} m_{\mathbb{F}}(A_{n}) \qquad\qquad \text{ by countable additivity of } m_{\mathbb{F}}\\ &=& \sum\limits_{n \in \mathbb N} m^{\prime}(|A_{n}|)\\ & =& \sum\limits_{n \in \mathbb N} m^{\prime}(a_{n}) \end{array} $$

Thus $m^{\prime }$ is a countably additive probability measure on $\mathbb {F} \slash I$. To see that $m^{\prime }$ represents $\preceq ^{\prime }$, note that:

$$|A| \preceq^{\prime} |B| \Longleftrightarrow A \preceq_{\mathbb{F}} B \Longleftrightarrow m_{\mathbb{F}}(A) \leq m_{\mathbb{F}}(B) \Longleftrightarrow m^{\prime}(|A|) \leq m^{\prime}(|B|)$$

For the uniqueness claim, suppose that μ is a probability measure on $\mathbb {F} \slash I$ that represents $\preceq ^{\prime }$. We must show that $\mu =m^{\prime }$. Define the function ν on $\mathbb {F}$ by putting: ν(A) = μ(|A|). Then ν is a probability measure on $\mathbb {F}$. Indeed, ν is real-valued and non-negative, since μ is; ν(Ω) = μ(|Ω|) = μ(1) = 1; and finally, if A and B are disjoint elements of $\mathbb {F}$, then |A| and |B| are disjoint elements of $\mathbb {F} \slash I$, so ν(A ∪ B) = μ(|A|∨|B|) = μ(|A|) + μ(|B|) = ν(A) + ν(B). Moreover, ν represents $\preceq _{\mathbb {F}}$, since:

$$A \preceq_{\mathbb{F}} B \Longleftrightarrow |A| \preceq^{\prime} |B| \Longleftrightarrow \mu(|A|) \leq \mu(|B|) \Longleftrightarrow \nu(A) \leq \nu(B)$$

But $m_{\mathbb {F}}$ is the unique probability measure on $\mathbb {F}$ that represents $\preceq _{\mathbb {F}}$. Therefore, $\nu =m_{\mathbb {F}}$. But then $\mu =m^{\prime }$. We have shown that $m^{\prime }$ is the unique probability measure on $\mathbb {F} \slash I$ that represents $\preceq ^{\prime }$. □

Recall that φ is a Boolean isomorphism from $\mathbb {B}$ to $\mathbb {F} \slash I$. Define the function m on $\mathbb {B}$ by putting:

$$m(a) = m^{\prime}(\varphi(a))$$

Since φ is an isomorphism, we have the following.

Proposition 125

m is the unique probability measure on $\mathbb {B}$ that represents ≼, and it is countably additive.

We have shown that if $\mathbb {B}$ is a Boolean σ-algebra and ≼ is a monotonely continuous, bottomless qualitative probability on $\mathbb {B}$, then there is a unique probability measure on $\mathbb {B}$—namely m defined above—that represents ≼, and it is countably additive. Thus Theorem 18 is proved.