proof of Carathéodory’s extension theorem
We show that this is an outer measure (http://planetmath.org/OuterMeasure2). First, it is clearly non-negative. Secondly, if then we can take in (1) to obtain , giving =0. It is also clear that is increasing, so that if then . The only remaining property to be proven is subadditivity. That is, if is a sequence in then
As , equation (1) defining gives
As is arbitrary, this proves subadditivity (2). So, is indeed an outer measure.
The next step is to show that agrees with on . So, choose any . The inequality follows from taking and in (1), and it remains to prove the reverse inequality. So, let be a sequence in covering , and set
As this inequality hold for any sequence covering , equation (1) gives and, by combining with the reverse inequality, shows that does indeed agree with on .
We have shown that extends to an outer measure on the power set of . The final step is to apply Carathéodory’s lemma on the restriction of outer measures. A set is said to be -measurable if the inequality
is satisfied for all subsets of . Carathéodory’s lemma then states that the collection of -measurable sets is a -algebra (http://planetmath.org/SigmaAlgebra) and that the restriction of to is a measure. To complete the proof of the theorem it only remains to be shown that every set in is -measurable, as it will then follow that contains and the restriction of to is a measure.
As and ,
Since is arbitrary, this shows that (3) is satisfied and is -measurable.
|Title||proof of Carathéodory’s extension theorem|
|Date of creation||2013-03-22 18:33:28|
|Last modified on||2013-03-22 18:33:28|
|Last modified by||gel (22282)|