Probability algebras, arbitrary joins and the countable chain condition

Symmetric difference is symmetric, $A\mathrm{△}A=\mathrm{\varnothing }$, and $A\mathrm{△}C\subseteq (A\mathrm{△}B)\cup (B\mathrm{△}C)$. F1 and F2 therefore give an equivalence relation. Its classes form a set quotient of $\mathrm{\Sigma }$. Replacing either input of a union or intersection by an equivalent set changes the result only inside the union of the input symmetric differences; complement preserves symmetric difference. Hence the operations are well-defined and inherit all F4 identities from set operations. Splitting sets into disjoint differences shows that equivalent sets have equal measure, so $m$ is well-defined. It vanishes exactly on the zero class, and $m(1)=1$, making the algebra nontrivial. The equation $[A]\wedge [B]=[A]$ is equivalent to $\mu (A\setminus B)=0$.

For a sequence $(a_n)$ choose representatives $A_n$ by F6 and put $a=[{\bigcup }_n{A}_n]$. This bounds every $a_n$. If $b=[B]$ is another upper bound, each $A_n\setminus B$ is null; F2 makes their union null, so $a\le b$. Thus $a$ is the supremum. Another sequence of representatives gives the same class, again by F2. For a disjoint sequence of Boolean elements, remove from $A_n$ all earlier $A_j$ to obtain literally disjoint representatives: the removed part is a finite union of null intersections. F1 then proves $m({\bigvee }_n{a}_n)={\sum }_{n}m(a_n)$. An empty sequence has supremum zero.

Given $D\subseteq \mathbb{B}$, let $s$ be the supremum supplied by F7 in $[0,1]$ of $m(\bigvee F)$ over finite $F\subseteq D$, including $F=\mathrm{\varnothing }$. Choose finite $F_n\subseteq D$ with $m(\bigvee F_n)>s-2^{-n}$; when $s=0$ all $F_n$ may be empty. Put $D_0={\bigcup }_n{F}_n$, which is countable by F6, and let $c=\bigvee D_0$ using step 2.1. The finite joins over $F_0\cup \cdots \cup F_n$ increase to $c$. Measurable representatives can be chosen increasing by taking successive finite unions, so F3 gives $m(c)=s$.

Let $A$ be an antichain of nonzero elements. For each positive integer $n$ put $A_n=\{a\in A:m(a)\ge 1\mathrm{/}n\}$. Any $n+1$ distinct members of $A_n$ would have disjoint join of measure at least $(n+1)\mathrm{/}n>1$, contrary to step 2.1. Thus $A_n$ has at most $n$ elements. Strict positivity gives $A={\bigcup }_{n\ge 1}{A}_n$, and F6 makes this union countable. The empty antichain and singleton antichains satisfy the same bound.

For any $d\in D$, the finite joins in step 3.1 with $d$ adjoined still have measure at most $s$. F3 gives $m(c\vee d)\le s=m(c)$. Disjoint additivity then gives $m(d\wedge \mathrm{\neg }c)=m(c\vee d)-m(c)=0$, hence $d\le c$ by strict positivity. Any upper bound of $D$ bounds $D_0$ and thus bounds $c$ by step 2.1. Therefore $c=\bigvee D$, proving completeness and the countable-subfamily assertion. If $D$ is empty the construction gives $c=0$; if it is a singleton its supremum is that element.

Put $u=\bigvee D$ and $v=\bigvee \{c\wedge d:d\in D\}$. Each joined term lies below $c\wedge u$, so $v\le c\wedge u$. Conversely $c\wedge d\le v$ implies $d=(c\wedge d)\vee (\mathrm{\neg }c\wedge d)\le v\vee \mathrm{\neg }c$. Hence $u\le v\vee \mathrm{\neg }c$, and finite distributivity gives $c\wedge u\le c\wedge (v\vee \mathrm{\neg }c)\le v$. This proves the identity, including $c=0$, $c=1$ and $D=\mathrm{\varnothing }$. Only the established Boolean supremum is used, not the possibly nonmeasurable union of an arbitrary family of representatives.