measure on a Boolean algebra
there is an such that is a real number (not ),
if , then .
By condition 1, suppose , then , so that .
is non-decreasing: for
If , then and are disjoint () and . So . As a result, .
is subadditive: .
Since , and and are disjoint, we have that . Since , the result follows.
From the three properties above, one readily deduces that is a Boolean ideal of .
A measure on is called a two-valued measure if maps onto the two-element set . Because of the existence of an element with , it follows that . Consequently, the set is a Boolean filter. In fact, because is two-valued, is an ultrafilter (and correspondingly, the set is a maximal ideal).
Conversely, given an ultrafilter of , the function , defined by iff , is a two-valued measure on . To see this, suppose . Then at least one of them, say , can not be in (or else ). This means that . If , then , so that . On the other hand, if , then , so , or . This means that .
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 is a countable set of pairwise disjoint elements in such that exists, then
|Title||measure on a Boolean algebra|
|Date of creation||2013-03-22 17:59:16|
|Last modified on||2013-03-22 17:59:16|
|Last modified by||CWoo (3771)|