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Homemeasure on a Boolean algebra
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measure on a Boolean algebra
Let $A$ be a Boolean algebra. A measure on $A$ is a nonnegative extended realvalued 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 nondecreasing: $m(a)\leq m(b)$ for $a\leq b$
If $a\leq b$, then $c=ba$ 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, $m(a)\leq m(b)$.

$m$ is subadditive: $m(a\vee b)\leq m(a)+m(b)$.
Since $a\vee b=(ab)\vee b$, and $ab$ and $b$ are disjoint, we have that $m(a\vee b)=m((ab)\vee b)=m(ab)+m(b)$. Since $ab\leq a$, the result follows.
From the three properties above, one readily deduces that $I:=\{a\in A\mid m(a)=0\}$ is a Boolean ideal of $A$.
A measure on $A$ is called a twovalued measure if $m$ maps onto the twoelement set $\{0,1\}$. Because of the existence of an element $a\in A$ with $m(a)=1$, it follows that $m(1)=1$. Consequently, the set $F:=\{a\in A\mid m(a)=1\}$ is a Boolean filter. In fact, because $m$ is twovalued, $F$ is an ultrafilter (and correspondingly, the set $\{a\mid m(a)=0\}$ is a maximal ideal).
Conversely, given an ultrafilter $F$ of $A$, the function $m:A\to\{0,1\}$, defined by $m(a)=1$ iff $a\in F$, is a twovalued 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 $a^{{\prime}},b^{{\prime}}\in F$, so $a^{{\prime}}\wedge b^{{\prime}}\in F$, 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
$m(\bigvee K)=\sum_{{a\in K}}m(a).$ 
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