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measure on a Boolean algebra
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(Definition)
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Let $A$ be a Boolean algebra. A measure on $A$ is a non-negative extended real-valued function $m$ defined on $A$ such that
- there is an $a\in A$ such that $m(a)$ is a real number (not $\infty$ ,
- 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 $$m(\bigvee K)=\sum_{a\in K} m(a).$$
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"measure on a Boolean algebra" is owned by CWoo.
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See Also: measure
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measure, two-valued measure, finitely additive, countably additive |
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Cross-references: pairwise disjoint, countable, countable additivity, additivity, stronger, addition, iff, conversely, maximal ideal, ultrafilter, Boolean filter, onto, maps, Boolean ideal, subadditive, disjoint, properties, measurable space, sigma algebra, real number, function, Boolean algebra
There are 55 references to this entry.
This is version 5 of measure on a Boolean algebra, born on 2008-04-11, modified 2008-04-17.
Object id is 10498, canonical name is MeasureOnABooleanAlgebra.
Accessed 2409 times total.
Classification:
| AMS MSC: | 06B99 (Order, lattices, ordered algebraic structures :: Lattices :: Miscellaneous) |
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Pending Errata and Addenda
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