Proving Completeness in Complete Measure Spaces

In summary, if \Omega, \Sigma, \mu is a complete measure space and \mu_{e} is the outer measure defined by \mu, then for any set S, if \mu_{e}(S)=0, then S is measurable in \Sigma. This can be proven by contradiction, assuming that S is not measurable and showing that it leads to a contradiction with the completeness of the measure space.
  • #1
johnson123
17
0
[SOLVED] complete measure space

Homework Statement



Assume that ([tex]\Omega[/tex],[tex]\Sigma[/tex],[tex]\mu[/tex]) is a complete
measure space, let [tex]\mu_{e}[/tex] be the outer measure defined by [tex]\mu[/tex]
. Prove that if [tex]\mu_{e}[/tex](S)=0 [tex]\Rightarrow[/tex] S[tex]\in[/tex][tex]\Sigma[/tex] .

I know that [tex]\mu_{e}[/tex] = [tex]\mu[/tex] when restricted on [tex]\Sigma[/tex]

and that if [tex]\mu[/tex](A)=0, then every subset of A has measure zero from completeness.

I intuitively can see this to be true, since if S where not in [tex]\Sigma[/tex] and
[tex]\mu_{e}[/tex] = 0 , then this would contradict the fact that ([tex]\Omega[/tex],[tex]\Sigma[/tex],[tex]\mu[/tex]) is complete, since the fact that
[tex]\mu_{e}(A_{1})[/tex]=[tex]\mu_{e}(A_{2})[/tex]...=[tex]\mu_{e}(A_{n})[/tex]=0, for [tex]A_{1}\subseteq A_{2}\subseteq...\subseteq A_{n}\subseteq S[/tex]
would imply that [tex]\Sigma[/tex] is incomplete, but not sure how to rigorously prove
it.
 
Last edited:
Physics news on Phys.org
  • #2
Homework EquationsThe Attempt at a Solution Suppose on the contrary that S\notin\Sigma. Hence, S has an outer measure zero but is not measurable. Since S is not measurable, then there exists a subset A of S such that \mu_{e}(A)>0. By completeness, we have that \mu(A)=0. As \mu_{e}(A)>0 and \mu(A)=0, this contradicts the fact that \mu_{e}=\mu when restricted on \Sigma. Therefore, S\in\Sigma.
 

1. What is a complete measure space?

A complete measure space is a mathematical concept used in measure theory to describe a space where all subsets of sets with measure 0 are also measurable. In other words, it is a space that contains all the necessary information to fully measure all its subsets.

2. Why is completeness important in measure theory?

Completeness is important in measure theory because it ensures that all subsets of sets with measure 0 are also measurable. This allows for a more precise and accurate measurement of the space, as well as easier calculations and proofs.

3. How is a complete measure space different from a sigma-finite measure space?

A complete measure space is different from a sigma-finite measure space in that it allows for the measurement of all subsets, while a sigma-finite measure space only allows for the measurement of countable subsets. In other words, a complete measure space is more encompassing than a sigma-finite measure space.

4. Can a complete measure space be incomplete?

No, a complete measure space cannot be incomplete. By definition, a complete measure space contains all the necessary information to measure all its subsets, so it cannot be missing any crucial elements.

5. What are some examples of complete measure spaces?

Some examples of complete measure spaces include the real numbers with the Lebesgue measure, the complex numbers with the Borel measure, and the unit interval with the Lebesgue-Stieltjes measure. These spaces all contain all the necessary information to fully measure all their subsets, making them complete measure spaces.

Similar threads

  • Introductory Physics Homework Help
Replies
2
Views
778
  • Calculus and Beyond Homework Help
Replies
2
Views
2K
Replies
3
Views
985
  • Calculus and Beyond Homework Help
Replies
1
Views
456
  • Calculus and Beyond Homework Help
Replies
7
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
842
  • Calculus and Beyond Homework Help
Replies
1
Views
532
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
450
Back
Top