Let $A$ be a Boolean algebra. A measure on $A$ is a non-negative extended real-valued function $m$ defined on $A$ such that

1. there is an $a\in A$ such that $m(a)$ is a real number (not $\infty$ ),
2. if $a\wedge b=0$ , then $m(a\vee b)=m(a)+m(b)$ .

For example, a sigma algebra $\mathcal{B}$ over a set $E$ is a Boolean algebra, and a measure $\mu$ on the measurable space $(\mathcal{B},E)$ is a measure on the Boolean algebra $\mathcal{B}$ .

The following are some of the elementary properties of $m$ :

• $m(0)=0$ .

  By condition 1, suppose $m(a)=r\in \mathbb{R}$ , then $m(a)=m(0\vee a)=m(0)+m(a)$ , so that $m(0)=0$ .

• $m$ is non-decreasing: $m(a)\le m(b)$ for $a\le b$

  If $a\le b$ , then $c=b-a$ and $a$ are disjoint ($c\wedge a=0$ ) and $b=c\vee a$ . So $m(b)=m(c\vee a)=m(c)+m(a)$ . As a result, $m(a)\le m(b)$ .

• $m$ is subadditive: $m(a\vee b)\le m(a)+m(b)$ .

  Since $a\vee b=(a-b)\vee b$ , and $a-b$ and $b$ are disjoint, we have that $m(a\vee b)=m((a-b)\vee b)=m(a-b)+m(b)$ . Since $a-b\le a$ , the result follows.

From the three properties above, one readily deduces that $I:=\lbrace a\in A\mid m(a)=0\rbrace$ is a Boolean ideal of $A$ .

A measure on $A$ is called a two-valued measure if $m$ maps onto the two-element set $\lbrace 0,1\rbrace$ . Because of the existence of an element $a\in A$ with $m(a)=1$ , it follows that $m(1)=1$ . Consequently, the set $F:=\lbrace a\in A \mid m(a)=1\rbrace$ is a Boolean filter. In fact, because $m$ is two-valued, $F$ is an ultrafilter (and correspondingly, the set $\lbrace a\mid m(a)=0\rbrace$ is a maximal ideal).

Conversely, given an ultrafilter $F$ of $A$ , the function $m:A\to \lbrace 0,1\rbrace$ , defined by $m(a)=1$ iff $a\in F$ , is a two-valued measure on $A$ . To see this, suppose $a\wedge b=0$ . Then at least one of them, say $a$ , can not be in $F$ (or else $0=a\wedge b\in F$ ). This means that $m(a)=0$ . If $b\in F$ , then $a\vee b\in F$ , so that $m(a\vee b)=1=m(b)=m(b)+m(a)$ . On the other hand, if $b\notin F$ , then $a',b'\in F$ , so $a'\wedge b'\in F$ , or $a\vee b\notin F$ . This means that $m(a\vee b)=0=m(a)+m(b)$ .

Remark. A measure (on a Boolean algebra) is sometimes called finitely additive to emphasize the defining condition 2 above. In addition, this terminology is used when there is a need to contrast a stronger form of additivity: countable additivity. A measure is said to be countably additive if whenever $K$ is a countable set of pairwise disjoint elements in $A$ such that $\bigvee K$ exists, then $$m(\bigvee K)=\sum_{a\in K} m(a).$$