Saturday, 28 March 2015

Prove that mu is a measure under several conditions.



Suppose that F is a σ-algebra on a set X and μ:F[0,] satisfies the conditions:




  1. μ()=0.

  2. For every pair A and B of disjoint sets in F, μ(AB)=μ(A)+μ(B).


  3. For every decreasing sequence {En} in F (that is En+1En for all n) such that n=1En=, we have limnμ(En)=0.



Prove that μ is a measure on F.



Here's my attempt:



Proof.



Let {En} be a countably infinite collection of sets such that EiEj= for all i,j. Write

E=n=1En


and let
Fn=Enk=1Ek.

for n1.
Then we have
Fn+1=En+1k=1EkEnk=1Ek=Fn

and
n=1Fn=.


Hence, by applying condition (2), we have
μ(Fn)=μ(Enk=1Ek)=μ(E)μ(nk=1Ek)=μ(E)nk=1Ek

and the above holds for all nN. Thus, applying condition (3), we have
μ(E)=limnμ(Fn)+limnnk=1Ek=k=1Ek.

This shows that μ is a measure.



Answer



Your proof is correct.



There are some minor typos in some places, where you wrote Ek instead of μ(Ek), but I am sure you meant the right thing.


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