Measure Algebras

TL;DR

This paper characterizes measure algebras as Boolean σ-algebras that are both weakly distributive and uniformly concentrated, providing a precise algebraic condition for measure algebra identification.

Contribution

It offers a new algebraic characterization of measure algebras based on weak distributivity and uniform concentration properties.

Findings

Measure algebras are exactly Boolean σ-algebras that are weakly distributive and uniformly concentrated.

Provides a necessary and sufficient condition for a Boolean σ-algebra to be a measure algebra.

Connects algebraic properties with measure-theoretic concepts in Boolean algebras.

Abstract

A Boolean $\sigma$-algebra $B$ is a measure algebra if and only if it is weakly distributive and uniformly concentrated.

Equations

$$\underset{n}{\mathrm{limsup}}{a}_n=\underset{n=1}{\overset{\infty }{\bigwedge }}\underset{k=n}{\overset{\infty }{\bigvee }}{a}_k\text{ and }\underset{n}{\mathrm{liminf}}{a}_n=\underset{n=1}{\overset{\infty }{\bigvee }}\underset{k=n}{\overset{\infty }{\bigwedge }}{a}_k,$$

$$\mu (b)=\inf \{\underset{n}{\lim }m(b\cap u_n)\}$$

$$G=\{\sum _{i=1}^{i=m}{\alpha }_i{K}_{c_i}:c_i\in C,0\le {\alpha }_i\le 1,\sum {\alpha }_i=1\}$$

$$Q=\{\alpha (g-\kappa )+f:g\in G,\alpha \ge 0,f\ge 0\}.$$

$$U-Q=\{h-\alpha (g-\kappa )-f:||h||<\delta ,g\in G,\alpha \ge 0,f\ge 0\}.$$

$$p(v)=\inf \{\gamma >0:\frac{v}{\gamma }\in U-Q\}.$$

$$|\bigcup _{i\in I}{A}_i|>|I|.$$

$$\Pi =\sum _{n=3}^{n=k}\left(\genfrac{}{}{0ex}{}{m}{n}\right)\left(\genfrac{}{}{0ex}{}{p}{n}\right)\frac{n^{3n}}{p^{3n}}.$$

$$\frac{m^n}{n!}\cdot \frac{p^n}{n!}\cdot \frac{n^{3n}}{p^{3n}}<\frac{e^{2n}{n}^n{m}^n}{p^{2n}}=(e^2\cdot \frac{n}{p}\cdot \frac{m}{p}{)}^n.$$

$$\Pi <\sum _{n=3}^{n=k}{\left(\frac{1}{2}\right)}^n<1.$$

$$b_I=\bigcap \{c_i:i\in I\}\cap \bigcap \{-c_i:i\notin I\}.$$

$$\frac{p}{k}=\frac{p}{p+1}\cdot \frac{p+1}{m}\cdot \frac{m}{k}\ge \frac{3}{4}\cdot \frac{1}{K}\cdot 30K^2\ge 20K$$

$$15\frac{m}{p}\le 15\cdot \frac{p+1}{p}\cdot \frac{m}{p+1}\le 15\cdot \frac{4}{3}\cdot K=20K.$$

$$\bigcup \{b_I:|I|\le k\}=1\text{ and for each }i\in M,c_i=\bigcup \{b_I:|I|\le k\text{ and }i\in I\}.$$

$$a_{ij}=\bigcup \{b_I:|I|\le k,i\in I\text{ and }{f}_I(i)=j\}.$$

$$f(x)-p(x-u)\le p(u).$$

$$f(x)+\gamma \le p(x+u),$$

$$f(x)-\gamma \le p(x-u).$$

$$f(x)+\gamma =f(x)+\underset{y}{\sup }(f(y)-p(y-u))=\underset{y}{\sup }(f(x+y)-p(y-u))\le p(x+u).$$

$$U_{n+1}\vee U_{n+1}\subset U_n$$

$$U_{n_1+1}\vee ...\vee U_{n_k+1}\subset U_{n_1}.$$

$$m(a)=\inf \{r\in D\cup \{1\}:a\in V_r\}.$$

Full text

1 Boolean $\sigma$-algebras

A Boolean algebra is an algebra $B$ of subsets of a given nonempty set $S$, with Boolean operations $a\cup b$, $a\cap b$, $-a=S-a$, and the zero and unit elements $\mathbf{0}=\emptyset$ and $\mathbf{1}=S$. A Boolean $\sigma$-algebra is a Boolean algebra $B$ such that every countable set $A\subset B$ has a supremum $\sup A=\bigvee A$ (and an infimum $\inf A=\bigwedge A$) in the partial ordering of $B$ by inclusion.

Definition 1.1**.**

A measure (a strictly positive $\sigma$-additive probability measure) on a Boolean $\sigma$-algebra $B$ is a real valued function $m$ on $B$ such that

• (i)

$m(\mathbf{0})=0,\ m(a)>0\text{ for }a\not=\mathbf{0}\text{, and }m(\mathbf{1})=1$**

• (ii)

$m(a)\leq m(b)\text{ if }a\subset b$**

• (iii)

$m(a\cup b)=m(a)+m(b)\text{ if }a\cap b=\mathbf{0}$**

• (iv)

$m(\bigvee_{n=1}^{\infty}a_{n})=\sum_{n=1}^{\infty}m(a_{n})\text{ whenever the }a_{n}\text{ are pairwise disjoint.}$**

A measure algebra is a Boolean $\sigma$-algebra that carries a measure.

A function $m$ that satisfies (i), (ii) and (iii) above is called a (strictly positive) finitely additive measure. And $m$ satisfies (iv) if and only if it is continuous:

• (iv)

if $\{a_{n}\}_{n}$ is a decreasing sequence in $B$ with $\bigwedge_{n=1}^{\infty}a_{n}=\mathbf{0}$ then $\lim_{n}m(a_{n})=\mathbf{0}.$

Let $B$ be a Boolean algebra and let $B^{+}=B-\{\mathbf{0}\}$. A set $A\subset B^{+}$ is an antichain if $a\cap b=\mathbf{0}$ whenever $a$ and $b$ are distinct elements of $A$. A partition $W$ (of $\mathbf{1}$) is a maximal antichain, i.e. an antichain with $\bigvee W=\mathbf{1}$. $B$ satisfies the countable chain condition (ccc) if it has no uncountable antichains. Every measure algebra is ccc because $B^{+}=\bigcup_{n=1}^{\infty}\{C_{n}\}_{n}$ where $C_{n}=\{a:m(a)\geq 1/n\}$ and every antichain in $C_{n}$ has size at most $n$.

If $\{a_{n}\}_{n}$ is a sequence in a Boolean $\sigma$-algebra $B$, one defines

[TABLE]

and if $\limsup_{n}a_{n}=\liminf_{n}a_{n}=a$, then $a$ is the limit of the sequence, denoted $\lim_{n}a_{n}$.

A sequence $\{a_{n}\}_{n}$ converges to $\mathbf{0}$ if and only if $\limsup a_{n}=\mathbf{0}$ if and only if there exist $b_{n}\geq a_{n}$, $b_{n}$ decreasing, with $\bigwedge_{n}b_{n}=\mathbf{0}$. A sequence $\{a_{n}\}_{n}$ converges to $\mathbf{1}$ if and only if $\liminf a_{n}=\mathbf{1}$. (Exercise: If $\lim a_{n}=\lim b_{n}=\mathbf{0}$ then $\lim(a_{n}\cup b_{n})=\mathbf{0}$).

If $\{a_{n}\}_{n}$ is an antichain then $\lim a_{n}=\mathbf{0}$.

In a measure algebra, if $\lim a_{n}=\mathbf{0}$ then $\lim m(a_{n})=0$. If $\sum m(a_{n})<\infty$ then $\lim a_{n}=\mathbf{0}$.

Definition 1.2**.**

A Boolean $\sigma$-algebra $B$ is weakly distributive if whenever $\{W_{n}\}_{n}$ is a sequence of countable maximal antichains then each $W_{n}$ has a finite subset $E_{n}$ such that $\lim_{n}\bigcup E_{n}=\mathbf{1}$.

Equivalently, if for every $k$, $\{a_{kn}\}_{n}$ is an increasing sequence with $\bigvee_{n}a_{kn}=\mathbf{1}$ then there exists a function $f:N\to N$ such that $\lim a_{k,f(k)}=\mathbf{1}$.

Or, if for every $k$, $\{a_{kn}\}_{n}$ is a decreasing sequence with $\bigwedge_{n}a_{kn}=\mathbf{0}$ then there exists a function $f:N\to N$ such that $\lim a_{k,f(k)}=\mathbf{0}$.

Definition 1.3**.**

A Boolean $\sigma$-algebra $B$ is uniformly weakly distributive if there exists a sequence of functions $\{F_{n}\}_{n}$ such that for each countable maximal antichain $W$, $F_{n}(W)$ is a finite subset of W, and if $\{W_{n}\}_{n}$ is a sequence of countable maximal antichains then $\lim_{n}\bigcup F_{n}(W_{n})=\mathbf{1}$.

If $B$ is a measure algebra then $B$ is uniformly weakly distributive: For every $n$, if $W$ is a countable maximal antichain, let $F_{n}(W)=E\subset W$ be such that $m(\bigcup E)\geq 1-1/2^{n}$.

Definition 1.4**.**

A Boolean $\sigma$-algebra $B$ is concentrated if for every sequence $\{A_{n}\}$ of finite antichains with $|A_{n}|\geq 2^{n}$ there exist $a_{n}\in A_{n}$ such that $\lim a_{n}=\mathbf{0}$.

$B$* is uniformly concentrated if there exists a function $F$ such that for each finite antichain $A$, $F(A)$ is an element of $A$, and if $A_{n}$ is a sequence of finite antichains with $|A_{n}|\geq 2^{n}$ then $\lim_{n}F(A_{n})=\mathbf{0}$.*

A measure algebra is uniformly concentrated: if $A$ is a finite antichain, let $F(A)$ be an element of $A$ of least measure (then $m(F(A))\leq 1/|A|$.)

Theorem 1.5**.**

A Boolean $\sigma$-algebra $B$ is a measure algebra if and only if it is weakly distributive and uniformly concentrated.

2 Fragmentations

Definition 2.1**.**

A fragmentation of $B$ is a sequence of subsets $C_{1}\subset C_{2}\subset...\subset C_{n}\subset...$ such that $\bigcup_{n}C_{n}=B^{+}$ and for every $n$, if $a\in C_{n}$ and $a\leq b$ then $b\in C_{n}$.

A fragmentation is $\sigma$-finite cc if for every $n$, every antichain $A\subset C_{n}$ is finite.

A fragmentation is $\sigma$-bounded cc if for every $n$ there is a constant $K_{n}$ such that every antichain $A\subset C_{n}$ has size $\leq K_{n}$.

A fragmentation is $G_{\delta}$ if for every $n$, no sequence in $C_{n}$ converges to $\mathbf{0}$.

A fragmentation is tight if whenever $\{a_{n}\}_{n}$ is a sequence such that $a_{n}\notin C_{n}$ for every $n$, then $\lim a_{n}=\mathbf{0}$.

A fragmentation is graded if for every $n$, whenever $a\cup b\in C_{n}$ then either $a\in C_{n+1}$ or $b\in C_{n+1}$.

(The name $G_{\delta}$ comes from the fact that if $B$ is weakly distributive and the fragmentation is $G_{\delta}$ then the set $\{\mathbf{0}\}$ is a $G_{\delta}$ set in the convergence topology.) A $G_{\delta}$ fragmentation is $\sigma$-finite cc.

In a measure algebra, the fragmentation defined by $C_{n}=\{a:m(a)\geq 1/2^{n}\}$ has all the above properties.

Lemma 2.2**.**

(a) If $B$ is uniformly concentrated then $B$ has a tight $\sigma$-bounded cc fragmentation.

(b) If $B$ is weakly distributive and has a tight $\sigma$-finite cc fragmentation then $B$ is uniformly weakly distributive.

Consequently, a weakly distributive, uniformly concentrated $\sigma$-algebra is uniformly weakly distributive.

Proof.

(a) Let $F$ be a function on finite antichains that witnesses that $B$ is uniformly concentrated. For every $n$ let $C_{n}$ be the set of all $a\neq\mathbf{0}$ such that there is no antichain $A$ with $|A|\geq 2^{n}$ and $a\leq F(A)$.

It is easy to see that $C_{n}$ is a fragmentation: If $a\notin\bigcup_{n}C_{n}$ then for every $n$ there exists an antichain $A_{n}$ with $|A_{n}|\geq 2^{n}$ and $a\leq F(A_{n})$. Hence $a\leq\lim F(A_{n})=\mathbf{0}$, and so $a=\mathbf{0}$. The fragmentation is tight because if $a_{n}\notin C_{n}$ then $a_{n}\leq F(A_{n})$ for some $A_{n}$ and so $\lim a_{n}=\mathbf{0}$.

It is $\sigma$-bounded cc because if $A\subset C_{n}$ is an antichain then $F(A)\in C_{n}$ and so $|A|<2^{n}$.

(b) Let $B$ be weakly distributive and let $\{C_{n}\}_{n}$ be a tight $\sigma$-finite cc fragmentation; we shall find the functions $F_{n}$ witnessing that $B$ is uniformly weakly distributive.

Let $n\in N$, and let $W$ be a countable maximal antichain. We claim that there exists a finite set $E\subset W$ such that for every finite $E^{\prime}\subset W-E$, $\bigcup E^{\prime}\notin C_{n}$: otherwise we find an infinite sequence $\{E_{k}\}_{k}$ of pairwise disjoint nonempty subsets of $W$ with $\bigcup E_{k}\in C_{n}$, an infinite antichain in $C_{n}$. We let $F_{n}(W)$ be this $E$.

Now let $W_{n}$, $n\in N$, be countable maximal antichains. We claim that $\lim\bigcup F_{n}(W_{n})=\mathbf{1}$. Since $B$ is weakly distributive, there exist finite sets $E_{n}\subset W_{n}$ such that $\lim\bigcup E_{n}=\mathbf{1}$. For each $n$ let $a_{n}=\bigcup E_{n}-\bigcup F_{n}(W_{n})$. By the claim above, $a_{n}\notin C_{n}$. Because $\{C_{n}\}_{n}$ is tight, we have $\lim a_{n}=\mathbf{0}$, and because $\bigcup F_{n}(W_{n})\geq\bigcup E_{n}\cap(-a_{n})$, we have $\lim\bigcup F_{n}(W_{n})=\lim\bigcup E_{n}=\lim-a_{n}=\mathbf{1}$.

∎

Lemma 2.3**.**

A Boolean $\sigma-$algebra is uniformly weakly distributive if and only if it has a tight $G_{\delta}$ fragmentation.

Proof.

First assume that $B$ is uniformly weakly distributive and let $F_{n}$ be functions witnessing it. For each $n$ we let $C_{n}$ be the set of all $a$ such that for some $k\leq n$ $a\cap\bigcup F_{k}(W)\neq\mathbf{0}$ for every countable maximal antichain $W$.

To show that $\{C_{n}\}_{n}$ is a fragmentation, we show that $\bigcup_{n}C_{n}=B^{+}$: if $a\notin C_{n}$ for all $n$ then for all $k$ there is a $W_{k}$ such that $a\cap\bigcup F_{k}(W_{k})=\mathbf{0}$, and because $\lim(a\cap\bigcup F_{k}(W_{k}))=a$ we have $a=\mathbf{0}$.

To show that $\{C_{n}\}_{n}$ is tight, let $a_{n}\notin C_{n}$ for each $n$. For each $n$ there is a $W_{n}$ such that $a_{n}\cap b_{n}=\mathbf{0}$ where $b_{n}=\bigcup F_{n}(W_{n})$. Since $\lim b_{n}=\mathbf{1}$, we have $\lim-b_{n}=\mathbf{0}$, and because $a_{n}\leq-b_{n}$, it follows that $\lim a_{n}=\mathbf{0}$.

To show that $\{C_{n}\}_{n}$ is $G_{\delta}$, let $n\in N$ and let $\lim a_{k}=\mathbf{0}$; it suffices to find a $k\in N$ such that $a_{k}\notin C_{n}$. We may assume that $\{a_{k}\}_{k}$ is strictly decreasing and let $W$ be the maximal antichain $\{a_{k-1}-a_{k}\}_{k}$ where $a_{0}=\mathbf{1}$. Let $E=F_{1}(W)\cup...\cup F_{n}(W)$. There exists a $k$ large enough so that $a_{k}\cap\bigcup E=\mathbf{0}$. It follows that $a_{k}\notin C_{n}$.

For the converse, let $\{C_{n}\}_{n}$ be a tight $G_{\delta}$ fragmentation. In view of Lemma 2.2 (b) it suffices to show that $B$ is weakly distributive. For every $k$, let $\{a_{kn}\}_{n}$ be a decreasing sequence with $\bigwedge_{n}a_{kn}=\mathbf{0}$. We shall find a function $f:N\to N$ such that $\lim a_{k,f(k)}=\mathbf{0}$. Given $k\in N$, there is some $f(k)$ such that $a_{k,f(k)}\notin C_{k}$, as $\{C_{n}\}_{n}$ is $G_{\delta}$. Because $\{C_{n}\}_{n}$ is tight, $\lim a_{k,f(k)}=\mathbf{0}$ follows.

∎

A tight $G_{\delta}$ fragmentation is essentially unique: if $C_{n}$ and $C^{\prime}_{n}$ are such, then for each $n$ there is a $k$ such that $C_{n}\subset C^{\prime}_{k}$. (If not, there exists an $n$ such that for all $k$ there is some $a_{k}\in C_{n}-C^{\prime}_{k}$ and so $\lim a_{k}=\mathbf{0}$).

In the appendix we use a tight $G_{\delta}$ fragmentation to construct a continuous submeasure, thus showing that $B$ is uniformly weakly distributive if and only if $B$ is a Maharam algebra. The construction of a measure (in the next two chapters) is considerably more difficult.

Lemma 2.4**.**

If $\{C_{n}\}_{n}$ is a $G_{\delta}$ fragmentation of $B$ and if $B$ is concentrated then $\{C_{n}\}_{n}$ is $\sigma-$bounded cc.

Proof.

By contradiction, assume that for some $n$, there exist arbitrarily large finite antichains in $C_{n}$, and for each $k$, let $A_{k}$ be an antichain in $C_{n}$ of size $\geq 2^{k}$. Since $B$ is concentrated, there exist $a_{k}\in A_{k}$ with $\lim a_{k}=\mathbf{0}$, a contradiction. ∎

Lemma 2.5**.**

If $\{C_{n}\}_{n}$ is a tight $G_{\delta}$ fragmentation then for every $n$ there exists a $k>n$ such that for every $c\in C_{n}$, if $c=a\cup b$ then either $a\in C_{k}$ or $b\in C_{k}$.

Thus if $B$ has a tight $G_{\delta}$ fragmentation then $B$ has one that is also graded (i.e. $a\cup b\in C_{n}$ implies that either $a\in C_{n+1}$ or $b\in C_{n+1}$.)

Proof.

Otherwise, for every $k>n$ there exist $c_{k}=a_{k}\cup b_{k}\in C_{n}$ such that $a_{k}\notin C_{k}$ and $b_{k}\notin C_{k}$. By tightness, $\lim a_{k}=\lim b_{k}=\mathbf{0}$ and so $\lim c_{k}=\mathbf{0}$, a contradiction. ∎

In conclusion, we proved in this chapter that a weakly distributive uniformly concentrated Boolean $\sigma-$algebra has a graded $\sigma-$bounded cc fragmentation. In the next two chapters we construct a measure on $B$ under the assumption that $B$ is weakly distributive and has a graded $\sigma-$bounded cc fragmentation.

3 Kelley’s Theorem

In this Section we introduce Kelley’s condition for the existence of finitely additive measure on a Boolean algebra. But first we show how the measure problem reduces to finitely additive measures.

Theorem 3.1**.**

(Pinsker, Kelley) A Boolean $\sigma$-algebra $B$ carries a measure if an only if it is weakly distributive and carries a finitely additive measure.

Proof.

Let $m$ be a finitely additive measure on $B$. We let

[TABLE]

where the infimum is taken over all increasing sequences $\{u_{n}\}_{n}$ with $\bigvee_{n}u_{n}=\mathbf{1}$.

We show that $\mu$ is a $\sigma$-additive measure, and if $B$ is weakly distributive then $\mu$ is strictly positive.

First show that $\mu(a\cup b)=\mu(a)+\mu(b)$ if $a\cap b=\mathbf{0}$. If $s=\{u_{n}\}_{n}$, let $\mu_{s}(x)=\lim_{n}m(x\cap u_{n})$. Clearly, $\mu_{s}(a\cup b)=\mu_{s}(a)+\mu_{s}(b)$, and so $\mu(a)+\mu(b)\leq\mu_{s}(a\cup b)$. Hence $\mu(a)+\mu(b)\leq\mu(a\cup b)$.

For each $\varepsilon>0$ there is a sequence $s=s_{a}=\{u_{n}\}_{n}$ such that $\mu_{s}(a)\leq\mu(a)+\varepsilon$, and similarly $s_{b}=\{v_{n}\}_{n}$. Let $s=\{u_{n}\cap v_{n}\}_{n}$. Then $\mu(a\cup b)\leq\mu_{s}(a\cup b)=\mu_{s}(a)+\mu_{s}(b)\leq\mu_{s_{a}}(a)+\mu_{s_{b}}(b)\leq\mu(a)+\mu(b)+2\varepsilon$, and the equality follows.

To show the continuity of $\mu$, let $a_{n}$ be a decreasing sequence with $\bigwedge_{n}a_{n}=\mathbf{0}$; we show that $\lim\mu(a_{n})=0$. Let $\varepsilon>0$.

Let $M=\lim m(a_{n})$, and let $K$ be such that $m(a_{K})-M<\varepsilon$. Let $s=\{u_{n}\}_{n}$ where $u_{n}=-a_{n}$. As for all $k,n\geq K$, $m(a_{k}-a_{n})<\varepsilon$, we have, for every $k\geq K$, $\mu_{s}(a_{k})=\lim_{n}m(a_{k}\cap u_{n})=\lim_{n}m(a_{k}-a_{n})\leq\varepsilon$, and hence $\mu(a_{k})\leq\varepsilon$.

Finally, assume that $B$ is weakly distributive, and let $b\in B$ be such that $\mu(b)=0$; we show that $b=\mathbf{0}$. As $\mu(b)=0$, there is for each $k$ an increasing sequence $\{u_{kn}\}_{n}$ with $\bigvee_{n}u_{kn}=\mathbf{1}$ such that $\lim_{n}m(b\cap u_{kn})<1/k$.

By weak distributivity there is a function $f$ such that $\lim u_{k,f(k)}=\mathbf{1}$. Hence $\bigvee_{n}\bigwedge_{k\geq n}u_{k,f(k)}=\mathbf{1}$. For each $k$ let $a_{k}=b\cap u_{k,f(k)}$. We have $\bigvee_{n}\bigwedge_{k\geq n}a_{k}=b$. But because $m(a_{k})<1/k$, it follows that $\bigwedge_{k\geq n}a_{k}=\mathbf{0}$, and so $b=\bigvee_{n}\bigwedge_{k\geq n}a_{k}=\mathbf{0}$. ∎

In order to state and prove Kelley’s Theorem, we now work with Boolean set algebras, and use the term “finitely additive” for measures that are not necessarily strictly positive - when we need the condition $m(a)>0$ if $a\neq\mathbf{0}$ we call $m$ strictly positive.

Let $B$ be a Boolean set algebra, $B\subset P(S)$ for some set $S$.

Definition 3.2**.**

Let $C$ be a subset of $B^{+}$. For every finite sequence $s=\langle c_{1},...c_{n}\rangle$ in $C$, let $\kappa_{s}=k/n$ where $k$ is the largest size of a subset $J\subset\{1,...,n\}$ such that $\bigcap_{i\in J}c_{i}$ is nonempty. The intersection number of $C$ is the infimum $\kappa=\inf\kappa_{s}$ over all finite sequences $s$ in $C$.

The sequences $s$ do not have to be nonrepeating.

Note that for any $n_{0}$, the infimum $\inf\kappa_{s}$ taken over all sequences $s$ of length $n\geq n_{0}$ is still $\kappa$: if $s$ is a sequence of length $n<n_{0}$, let $t$ be such that $t\cdot n\geq n_{0}$, and let $s^{*}$ be a sequence we get when repeating each term of $s$ $t$-times. Then $\kappa_{s^{*}}=\kappa_{s}$.

To better understand the significance of the intersection number, assume that $m$ is a finitely additive measure on $B$, and let $C\subset B^{+}$. Let $M$ be such that $m(c)\geq M$ for all $c\in C$. We show that the intersection number $\kappa$ of $C$ is at least $M$.

Let $s=\langle c_{1},...c_{n}\rangle$ be a sequence in $C$. For each $i\leq n$, let $K_{i}$ be the characteristic function of $c_{i}$, i.e. $K_{i}(x)=1$ if $x\in c_{i}$ and $=0$ if $x\notin c_{i}$. Let $g=\sum_{i=1}^{n}K_{i}$ and consider $I_{s}=\int g\,dm$, the area below the graph of $g$. Since $\int K_{i}=m(c_{i})$, we have $I_{s}=\sum_{i}m(c_{i})\geq M\cdot n$ and so (because $m(S)=1$) $||g||=\max_{x\in S}g(x)\geq I_{s}\geq M\cdot n$.

Thus there exists some $x\in S$ such that $\sum_{i}K_{i}(x)\geq M\cdot n$; in other words, $x$ belongs to at least $M\cdot n$ members of the sequence. Hence $\kappa_{s}\geq M\cdot n/n=M$ and consequently $\kappa\geq M$.

Theorem 3.3**.**

(Kelley) Let $C\subset B^{+}$ have a positive intersection number $\kappa$. Then there exists a finitely additive measure $m$ on $B$ such that $m(c)\geq\kappa$ for all $c\in C$.

Corollary 3.4**.**

If a Boolean algebra $B$ has a fragmentation $\{C_{n}\}$ such that each $C_{n}$ has a positive intersection number, then $B$ carries a strictly positive finitely additive measure.

Proof.

(of Corollary) For each $n$ let $m_{n}$ be positive on $C_{n}$. If we let $m(a)=\sum_{n}m_{n}(a)/2^{n}$, $m$ is a strictly positive, finitely additive measure on $B$. ∎

To prove Kelley’s Theorem and construct a finitely additive measure on $B$ we shall consider the vector space of all bounded functions on $S$ (including all characteristic functions $K_{a}$ for all $a\in B$) and find a linear functional $F$ such that $F(\mathbf{1})=1$, $F(K_{a})\geq 0$, and $F(K_{c})\geq\kappa$ for all $c\in C$. Then we let $m(a)=F(K_{a})$ for all $a\in B$.

To find the linear functional we use the Hahn-Banach Theorem (for a proof, see Appendix):

Theorem 3.5**.**

Let $p$ be a function such that $p(x)\geq 0$ for all $x$, $p(x+y)\leq p(x)+p(y)$, $p(\alpha\,x)=\alpha\,p(x)$ for all $\alpha\geq 0$, and $p(\mathbf{1})\geq 1$.

Then there exists a linear functional $F$ such that $F(\mathbf{1})=1$ and $F(x)\leq p(x)$ for all $x$.

In the rest of this Chapter we give a proof of Kelley’s Theorem:

Proof.

Let $C\subset B^{+}\subset P(S)$ and let $\kappa$ be the intersection number of $C$. Let $V$ be the vector space of all bounded functions on $S$ with the supremum norm $||f||=\sup\{|f(x)|:x\in S\}$. We shall find a linear functional $F$ on $V$ such that $0\leq F(K_{a})\leq 1$ for all $a\in B$, $F(\mathbf{1})=1$ and $F(K_{c})\geq\kappa$ for all $c\in C$.

Consider the convex hull of the set $\{K_{c}:c\in C\}$:

[TABLE]

Lemma 3.6**.**

For every $g\in G$, $||g||\geq\kappa$.

Proof.

First consider rational coefficients $\alpha_{i}$: for each $i\leq m$, $\alpha_{i}=l_{i}/n$ with $\sum_{i}l_{i}=n$, and $g(x)=f(x)/n$ where $f=\sum_{i}^{m}l_{i}\cdot K_{c_{i}}$.

Consider the sequence $s$ in $C$ of length $n$ where each $c_{i}$ is repeated $l_{i}$ times. By definition of $\kappa$ there are $k$ terms of $s$ with nonempty intersection such that $k/n\geq\kappa$. Let $x$ be a point in the intersection; it follows that $f(x)\geq k$. Hence $g(x))\geq k/n\geq\kappa$.

For arbitrary $\alpha_{i}$ let $\varepsilon>0$. There are rational approximations $\beta_{i}$ of the $\alpha_{i}$ such that $\kappa\leq||\sum\beta_{i}K_{c_{i}}||\leq||\sum\alpha_{i}K_{c_{i}}||+\varepsilon$. Hence $||g||\geq\kappa-\varepsilon$, and so $||g||\geq\kappa$. ∎

Now let

[TABLE]

The set $Q$ contains all $K_{b}$, $b\in B$, (because $K_{b}\geq\mathbf{0}$) and is convex: if $f$ and $g$ are in $Q$ and $\alpha+\beta=1$ ($\alpha,\,\beta>0$) then $\alpha f+\beta g\in Q$. Clearly, it suffices to verify this for $f$ and $g$ in $G$ and that is easy.

Let $\delta=1-\kappa$, and let $U$ be the open ball $\{h:||h||<\delta\}$ of radius $\delta$. Using the vector space convention $A-B=\{a-b:a\in A,\,b\in B\}$, consider the set

[TABLE]

The set $U-Q$ is convex, and because $\mathbf{0}\in Q$, we have $U\subset U-Q$, and so for every $v\in V$ there exists a positive number $\alpha$ such that $\alpha v\in U\subset U-Q$. Now define

[TABLE]

If $\gamma=\alpha+\beta$ and $\frac{x}{\alpha},\frac{y}{\beta}\in U-Q$ then $\frac{x+y}{\gamma}=\frac{\alpha}{\gamma}(\frac{x}{\alpha})+\frac{\beta}{\gamma}(\frac{y}{\beta})\in U-Q$ by convexity, and so $p(x+y)\leq p(x)+p(y)$.

Clearly, $p(\alpha v)=\alpha p(v)$ for all $\alpha\geq 0$. Finally, let $\gamma\leq 1$; we show that $1/\gamma\notin U-Q$, hence $p(\mathbf{1})\geq 1$. If $1/\gamma\in U-Q$ then $1/\gamma=h-f$ (and $h=f+1/\gamma$) where $h\in U$ and $f\in Q$. Since $f\in Q$, there exists, by the Lemma, some $x\in S$ such that $f(x)\geq 0$, and so $f(x)+1/\gamma\geq 1/\gamma\geq 1$. But $h(x)<\delta\leq 1$.

Now we apply the Hahn-Banach Theorem to this function $p$. Note that for all $f\in Q$, $-f\in U-Q$ and hence $p(-f)\leq 1$.

There exists a linear functional $F$ such that $F(\mathbf{1})=1$ and $F(x)\leq p(x)$ for all $x\in V$. If $f\in Q$, then $F(-f)\leq p(-f)\leq 1$ and so $F(f)\geq-1$. As this is true for all $f\in Q$ and $Q$ is closed under multiples by all $\alpha\geq 0$, it must be the case that $F(f)\geq 0$ for all $f\in Q$. In particular, $F(K_{b})\geq 0$ for all $b\in B$.

Also, if $g\in G$, then $g-\kappa\in Q$ and hence $F(g-\kappa)\geq 0$, i.e. $F(g)\geq\kappa$. Consequently, $F(K_{c})\geq\kappa$ for all $c\in C$.

When we let $m(b)=F(K_{b})$ for all $b\in B$, $m$ is a finitely additive measure on $B$ with $m(c)\geq\kappa$ for all $c\in C$.

∎

4 The Kalton-Roberts Proof

We complete the proof by showing that Kelley’s condition applies:

Theorem 4.1**.**

Let $B$ be a Boolean algebra that has a graded $\sigma$-bounded cc fragmentation $\{C_{n}\}$. Then for every $n$, $C_{n}$ has a positive intersection number.

To prove the theorem, we adapt the Kalton-Roberts proof, an ingenious combinatorial argument that converts finite bounds for the size of antichains into positive intersection numbers.

Lemma 4.2**.**

Let $M$ and $P$ be finite sets with $|M|=m$ and $|P|=p\leq m$, and let $k$, $3\leq k\leq p$ be an integer such that $p/k\geq 15\cdot m/p$. Then there exists an indexed family $\{A_{i}:i\in M\}$ such that each $A_{i}$ is a three point subset of $P$ and such that for every $I\subset M$ with $|I|\leq k$,

[TABLE]

It follows that for every $I\subset M$ with $|I|\leq k$ there exists a one-to-one choice function $f_{I}$ on $\{A_{i}:i\in I\}$.

The last statement of the lemma follows by Hall’s “Marriage Theorem”:

Theorem 4.3**.**

A family $\{A_{1},...,A_{n}\}$ of finite sets has a set of distinct representatives if and only if $|\bigcup_{i\in I}A_{i}|\geq|I|$ for every $I\subset\{1,...,n\}$.

For a proof of Hall’s Theorem, see Appendix.

The proof of the lemma uses a counting argument:

Proof.

Consider the families $\{A_{i}:i\in M\}$ of three point subsets of $P$. Let us call such a family bad if $|\bigcup_{i\in I}A_{i}|\leq|I|$ for some $I\subset M$, $|I|\leq k$. If a family is bad then for some $n$, $3\leq n\leq k$, there exist sets $I\subset M$ and $J\subset P$, $|I|=|J|=n$ such that $A_{i}\subset J$ for every $i\in I$.

There are $\binom{p}{3}$ three-point subsets of $P$ and $\binom{n}{3}$ three-point subsets of $J$. Of the $\binom{p}{3}^{n}$ families $\{A_{i}\}_{i\in I}$ with domain $I$, $\binom{n}{3}^{n}$ are such that $\bigcup_{i\in I}A_{i}\subset J$. The ratio of such families (for $3\leq n\leq p$) is $(\binom{n}{3}/\binom{p}{3})^{n}\leq(n^{3}/p^{3})^{n}=n^{3n}/p^{3n}$ because $\binom{n}{3}/\binom{p}{3}\leq n^{3}/p^{3}$.

Because there are $\binom{m}{n}$ subsets $I\subset M$ of size $n$ and $\binom{p}{n}$ subsets $J\subset P$ of size $n$, the probability that a family $\{A_{i}\}_{i\in M}$ is bad is at most

[TABLE]

We have $\binom{m}{n}\cdot\binom{p}{n}\cdot n^{3n}/p^{3n}\leq m^{n}/n!\cdot p^{n}/n!\cdot n^{3n}/p^{3n}$. Using $e^{x}>x^{n}/n!$ we get $e^{n}n!>n^{n}$, hence $1/n!<e^{n}/n^{n}$, and so

[TABLE]

For $n\leq k$ we have $e^{2}\cdot n/p\cdot m/p\leq e^{2}\cdot k/p\cdot m/p\leq e^{2}/{15}<1/2$ because we assumed $p/k\geq 15m/p$ and because $2e^{2}<15$. Therefore

[TABLE]

Consequently, there exists a family $\{A_{i}:i\in M\}$ that is not bad, and so $|\bigcup_{i\in I}|>|I|$ for every $I\subset M$ of size $\leq k$.

∎

We shall now apply the Kalton-Roberts method to prove Theorem 4.1:

Proof.

Let $\{C_{n}\}$ be a graded $\sigma$-bounded cc fragmentation of a Boolean algebra $B$, and let us fix an integer $n$. We prove that the intersection number of each $C_{n}$ is positive, namely $\geq 1/(30K^{2})$ where $K=K_{n+2}$ is the maximal size of an antichain in $C_{n+2}$.

We show that for every $m\geq 100K^{2}$, and every sequence $\{c_{1},...,c_{m}\}$ in $C_{n}$ there exists some $J\subset m$ of size $\geq m/(30K^{2})$ such that $\bigcap_{i\in J}c_{i}$ is nonempty.

Let $M=\{1,...,m\}$ with $m\geq 100K^{2}$ and let $c_{1},...,c_{m}\in C_{n}$. For each $I\subset M$, let

[TABLE]

The sets $b_{I}$ are pairwise disjoint (some may be empty) and $\bigcup\{b_{I}:I\subset M\}=\mathbf{1}$. Note that for each $i\in M$, $\bigcup\{b_{I}:i\in I\}=c_{i}$. We shall find a sufficiently large set $J\subset M$ with nonempty $b_{J}$.

We shall apply Lemma 4.2. First let $k\geq 3$ be the largest $k$ such that $k/m<1/(30K^{2})$ (there is such because $3/m\leq 3/(100K^{2}).$ We have $k<m$ and $(k+1)/m\geq 1/(30K^{2})$. Then let $p$ be the largest $p\geq k$ such that $p/m<1/K$ (there is such because $k/m<1/K$.)

We verify the assumption of the lemma, $p/k\geq 15m/p$ (using $p/(p+1)\geq 3/4$):

[TABLE]

and

[TABLE]

Now we apply the Lemma: Let $P=\{1,...,p\}$. There exist three point sets $A_{i}\subset P$, $i\in M$, and one-to-one functions $f_{I}$ on all $I\subset M$ of size $\leq k$ with $f_{I}(i)\in A_{i}$ for all $i\in I$.

We shall prove that there exists a $J\subset M$ of size $\geq k+1$ (and hence $\geq m/(30K^{2})$) such that $b_{J}$ is nonempty. By contradiction, assume that there is no such $J$. Then

[TABLE]

For each $i\in M$ and $j\in P$ let

[TABLE]

Note that for each $i\in M$, $c_{i}=a_{i,j_{1}}\cup a_{i,j_{2}}\cup a_{i,j_{3}}$ where $A_{i}=\{j_{1},j_{2},j_{3}\}$.

Let $j\in P$. We claim that the $a_{ij}$, $i\in M$, are pairwise disjoint: If $a_{i_{1},j}\cap a_{i_{2},j}$ is nonempty, then because the $b_{I}$ are pairwise disjoint there is some $I$ such that $i_{1}\in I$ and $i_{2}\in I$, and because $f_{I}(i_{1})=j=f_{I}(i_{2})$ and $f_{I}$ is one-to-one, we have $i_{1}=i_{2}$. Hence the $a_{ij}$, $i\in M$, are pairwise disjoint, and so only at most $K$ of them belong to $C_{n+2}$.

Consequently, at most $p\cdot K$ of the $a_{ij}$ belong to $C_{n+2}$ and because $pK<m$, there exists an $i$ such that $a_{ij}\notin C_{n+2}$ for all (three) $j\in A_{i}$.

But then $c_{i}=a_{i,j_{1}}\cup a_{i,j_{2}}\cup a_{i,j_{3}}\notin C_{n}$ because the fragmentation is graded. This contradicts the assumption that $c_{i}\in C_{n}$.

∎

5 Appendix

Theorem 5.1**.**

Let $V$ be a real vector space and let $p$ be a function on $V$ such that $p(x)\geq 0$ for all $x$, $p(x+y)\leq p(x)+p(y)$, $p(\alpha\,x)=\alpha\,p(x)$ for all $\alpha\geq 0$.

Let $W$ be a subspace of $V$ and $f$ a linear functional on $W$ such that $f(x)\leq p(x)$ on $W$.

Then there exists a linear functional $F$ such that $F(x)=f(x)$ for all $x\in W$, and $F(x)\leq p(x)$ for all $x$.

See Walter Rudin: “Functional Analysis”, Second Edition (1991), pp. 57-58.

Proof.

Using Zorn’s Lemma it suffices to extend $f$ one more dimension: let $u\notin W$ and show that there exists an $F$ extending $f$ on the subspace $\{x+\alpha u:\alpha\in\mathbf{R}\}$.

Let $u\notin W$. For every $x\in W$ we have $f(x)\leq p(x)\leq p(x-u)+p(u)$ and so

[TABLE]

Let $\gamma$ be the supremum of the left hand side of the inequality taken over all $x\in W$. We let $F(u)=\gamma$, and $F(x+\alpha u)=f(x)+\alpha\gamma$, for $x\in W$ and $\alpha\in\mathbf{R}$. $F$ is a linear functional and it remains to show that $F(x+\alpha u)\leq p(x+\alpha u)$. Note that this follows from these two inequalities:

[TABLE]

[TABLE]

(Then use $x\pm\alpha u=\alpha(x/\alpha\pm u)$ for $\alpha>0$).

The second inequatity is immediate. For the first one, note that for every $y\in W$, $f(x+y)\leq p(x+u)+p(y-u)$, hence $f(x+y)-p(y-u)\leq p(x+u)$, and so

[TABLE]

∎

6 Appendix

Theorem 6.1**.**

A family $\{A_{1},...,A_{n}\}$ of finite sets has a set of distinct representatives if and only if $|\bigcup_{i\in I}A_{i}|\geq|I|$ for every $I\subset\{1,...,n\}$.

See Béla Bollobás: “Modern Graph Theory” (1998), pp. 77-78.

Proof.

The condition is clearly necessary; we prove the sufficiency by induction. As $n=1$ is obvious, assume that the theorem is true for all $k<n$ and let $\{A_{1},...,A_{n}\}$ be a sequence of finite sets that satisfy the condition.

Suppose first that for any $k<n$, for any $I\subset\{1,...n\}$ with $|I|=k$, $|\bigcup_{i\in I}A_{i}|\geq k+1$. Then we choose $a_{n}\in A_{n}$ arbitrarily, and apply the induction hypothesis to the family $\{A_{1}-\{a_{n}\},...,A_{n-1}-\{a_{n}\}\}$ to get distinct representatives $a_{1},...a_{n-1}$ for $A_{1},...,A_{n-1}$.

Otherwise, there exists a set $S\subset\{1,...,n\}$ of size $k<n$ such that $|\bigcup_{i\in S}A_{i}|=k$. Let $A=\bigcup_{i\in S}A_{i}$. Consider the family $\{A_{i}-A:i\notin S\}$. This family of $n-k$ sets satisfies the condition of the theorem: if $I\subset\{1,...,n\}-S$ has size $m$ then $|\bigcup_{i\in I}A_{i}-A|\geq m$ because $|\bigcup_{i\in I\cup S}A_{i}|\geq k+m$. Thus the family has a set of distinct representatives, all $\notin A$, and since $\{A_{i}:i\in S\}$ has distinct representatives in $A$, we can combine these two sets.

∎

7 Appendix

Definition 7.1**.**

A a continuous strictly positive submeasure on a Boolean $\sigma$-algebra $B$ is a real valued function $m$ on $B$ such that

• (i)

$m(\mathbf{0})=0,\ m(a)>0\text{ for }a\not=\mathbf{0}\text{, and }m(\mathbf{1})=1$**

• (ii)

$m(a)\leq m(b)\text{ if }a\subset b$**

• (iii)

$m(a\cup b)\leq m(a)+m(b)$**

• (iv)

if $\{a_{n}\}_{n}$ is a decreasing sequence in $B$ with $\bigwedge_{n=1}^{\infty}a_{n}=\mathbf{0}$ then $\lim_{n}m(a_{n})=\mathbf{0}.$

A Maharam algebra is a Boolean $\sigma$-algebra that carries a continuous strictly positive submeasure.

Theorem 7.2**.**

(Balcar-Jech) $B$ is a Maharam algebra if and only if it is uniformly weakly distributive.

Proof.

The proof that a Maharam algebra is uniformly weakly distributive is exacly the same as for a measure algebra. To show that the condition is sufficient let $B$ be a $\sigma-$algebra and assume that $B$ has a graded $G_{\delta}$ fragmentation $\{C_{n}\}$. We shall define a submeasure on $B$.

For each $n$ let $U_{n}=B-C_{n}$ and $U_{0}=B$; we have $U_{0}\supset U_{1}\supset...\supset U_{n}\supset...$ and $\bigcap_{n}U_{n}=\{\mathbf{0}\}$. For $X,Y\subset B$, let $X\vee Y$ denote the set $\{x\cup y:x\in X,y\in Y\}$. As the fragmentation is graded, we have

[TABLE]

for all $n\geq 0$. Cosequently, if $n_{1}<...<n_{k}$ then

[TABLE]

This is proved by induction on $k$.

Let $D$ be the set of all $r=\sum^{k}_{i=1}1/2^{n_{i}}$ where $0<n_{1}<...<n_{k}$. For each $r\in D$ we let $V_{r}=U_{n_{1}}\vee...\vee U_{n_{k}}$ and $V_{1}=U_{0}=B$. For each $a\in B$ define

[TABLE]

Using the above property of the $U_{n}$, it follows that $V_{r}\subset V_{s}$ if $r<s$, and $V_{r}\vee V_{s}\subset V_{r+s}$ when $r+s\leq 1.$ From this we have $m(a)\leq m(b)$ if $a\subset b$, and $m(a\cup b)\leq m(a)+m(b).$

The submeasure $m$ is strictly positive because if $a\neq\mathbf{0}$ then for some $n$, $a\notin U_{n},$ and hence $m(a)\geq 1/2_{n}.$

The submeasure $m$ is continuous because the fragmentation is $G_{\delta}:$ If $\{a_{n}\}_{n}$ is a decreasing sequence in $B$ converging to $\mathbf{0}$ then for every $k,$ eventually all $a_{n}$ are in $U_{k}$, hence $m(a_{n})\leq 1/2^{k}$ for eventually all $n,$ and so $\lim_{n}m(a_{n})=0.$

∎