From this point in, I will define a “set function” as a function 
whose domain is some collection of subsets 
. It’s important to note here that is not defined on points of the set 
, but on subsets of . For some reason, a lot of people find that confusing at first.
We’re primarily concerned with set functions which take their values in the “extended real numbers” 
. That is, the value of 
is either a real number, or 
, or 
, with the latter two being greater than all real numbers and less than all real numbers, respectively.
We say that such a set function 
is “additive” if whenever we have disjoint sets 
and 
in 
with disjoint union 
also in , then we have
Similarly, we say that is finitely additive if for every finite, pairwise disjoint collection 
whose union is also in we have
And we say that is countably additive of for every pairwise-disjoint sequence 
of sets in whose union is also in , we have
Now we can define a “measure” as an extended real-valued, non-negative, countably additive set function defined on an algebra 
, and satisfying 
. With this last assumption, we can show that a measure is also finitely additive. Indeed, given a collection 
, just define 
for 
to get a sequence. Then we find
If is a measure on , we say a set 
has finite measure if 
. We say that 
has “
-finite” measure if there is a sequence of sets 
of finite measure (
) so that 
. If every set in has finite (or -finite) measure, we say that is finite (or -finite) on .
Finally, we say that a measure is “complete” if for every set of measure zero, also contains all subsets of . That is, if , 
, and 
, then 
. At first, this might seem to be more a condition on the algebra than on the measure , but it really isn’t. It says that to be complete, a measure can only assign 
to a set if all of its subsets are also in .
March 19, 2010
Posted by John Armstrong |
Analysis, Measure Theory |
17 Comments
