Measure Question: Is m Finitely Additive?

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In summary, the conversation discusses the concept of a measure induced by an ultrafilter in nonstandard analysis. The measure, denoted as m, is defined as assigning a value of 1 to elements in the ultrafilter F and 0 to all other elements. The question arises whether m is finitely additive, but it is pointed out that since elements in F cannot be disjoint, the finitely additivity clause is vacuously satisfied. Therefore, m is a measure.
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Yep its me again, with another dumb question.

Say you have a set I with an ultrafilter F on it. Now I came across the following in a text on nonstandard analysis:

let m be the measure induced by F defined as m(A) =1 if A is an element of F and zero otherwise.

I know this is going to be stupid, but it doesn't seem as though m is even finitely additive. Suppose A and B are elements of F then AUB is in F since if they weren't then their complement A'^B' (A' the complement of A and B' the complement of B and ^ for intersection) would be. But if A'^B' is in F then we could intersect A with A'^B' and get the empty set, which is not an element of F. So anyways, since AUB is in F the "induced measure" m(AUB)=1, but the m(A)+m(B)=2, so what gives?

Thanks
 
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  • #2
A and B are not disjoint.
 
  • #3
Doh! Of course, they can't be if they belong to F. So then the fact that m is finitely additive is vacuously satisfied. There are no pairwise disjoint elements of F. So likewise for the countable additivity clause. So then m is a measure because F has not pairwise...i see.

Thanks Matt Grime!
 

FAQ: Measure Question: Is m Finitely Additive?

What does it mean for m to be finitely additive?

Finitely additive refers to a mathematical property where a function, in this case m, satisfies the condition that the measure of the union of a finite number of disjoint sets is equal to the sum of the measures of each individual set. In other words, the measure of the whole is equal to the sum of its parts.

How does finitely additive differ from countably additive?

Finitely additive and countably additive are both properties of functions used in measure theory. The main difference between the two is that finitely additive functions only need to satisfy the condition for a finite number of sets, while countably additive functions need to satisfy it for an infinite number of sets.

Why is the property of finite additivity important in measure theory?

Finite additivity is important in measure theory because it allows us to define measures on a larger class of sets, including uncountable sets. It also allows us to extend the concept of measure from simple geometric shapes to more complicated sets, making it a fundamental concept in modern mathematics.

Can a measure be both finitely additive and countably additive?

Yes, a measure can be both finitely additive and countably additive. In fact, most commonly used measures, such as the Lebesgue measure, are both finitely and countably additive.

How is finitely additive related to the concept of sigma-additivity?

Sigma-additivity is a stronger version of countable additivity, where the condition must hold for a countable number of sets. Finitely additive measures are always sigma-additive, but the converse is not always true. In other words, all sigma-additive measures are finitely additive, but not all finitely additive measures are sigma-additive.

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