Sigma Algebras ... Axler, Page 26 ...

  • #1
Math Amateur
Gold Member
1,067
47
Summary:
I need help in order to fully understand the implications of Axler's definition of a ##\sigma##-algebra ... ...
I am reading Sheldon Axler's book: Measure, Integration & Real Analysis ... and I am focused on Chapter 2: Measures ...

I need help in order to fully understand the implications of Axler's definition of a ##\sigma##-algebra ... ...

The relevant text reads as follows:




Axler - Sigma Algebres ... Page 26 .png





Now in the above text Axler implies that the set of all subsets of ##\mathbb{R}## is not a ##\sigma##-algebra ... ...

... BUT ... which of the three bullet points of the definition of a ##\sigma##-algebra is violated by the set of all subsets of ##\mathbb{R}## ... and how/why is it violated ...



Help will be much appreciated ...

Peter
 

Answers and Replies

  • #2
member 587159
Summary:: I need help in order to fully understand the implications of Axler's definition of a ##\sigma##-algebra ... ...

I am reading Sheldon Axler's book: Measure, Integration & Real Analysis ... and I am focused on Chapter 2: Measures ...

I need help in order to fully understand the implications of Axler's definition of a ##\sigma##-algebra ... ...

The relevant text reads as follows:




View attachment 267217




Now in the above text Axler implies that the set of all subsets of ##\mathbb{R}## is not a ##\sigma##-algebra ... ...

... BUT ... which of the three bullet points of the definition of a ##\sigma##-algebra is violated by the set of all subsets of ##\mathbb{R}## ... and how/why is it violated ...



Help will be much appreciated ...

Peter

Axler does not say that. The collection of all subsets is a ##\sigma##-algebra (trivially). Axler says that we cannot define Lebesgue-measure on this ##\sigma##-algebra and that's why we define Lebesgue measure on Borel sets.
 
Last edited by a moderator:
  • Like
Likes Math Amateur
  • #3
90
50
Now in the above text Axler implies that the set of all subsets of ##\mathbb{R}## is not a ##\sigma##-algebra
No, he doesn't. The set of all subsets of ##\mathbb{R}## is obviously a ##\sigma##-algebra. It's just that we can't extend the notion of length on all the subsets of ##\mathbb{R}## without violating some highly desirable properties we want it to have (like countable additivity). So we force ourselves to give up on the idea of using all the subsets of ##\mathbb{R}## as the domain of our measure, but we still want this set of subsets to satisfy the properties in the definition of ##\sigma##-algebra.
 
Last edited:
  • Like
Likes Math Amateur
  • #4
Math Amateur
Gold Member
1,067
47
Thanks to Dragon27 and Math_QED for clarifying the issue ...

Much appreciate your help ...

Peter
 

Related Threads on Sigma Algebras ... Axler, Page 26 ...

Replies
5
Views
520
  • Last Post
Replies
1
Views
2K
Replies
2
Views
532
Replies
3
Views
2K
  • Last Post
Replies
1
Views
999
Replies
3
Views
4K
  • Last Post
Replies
1
Views
2K
Top