Sigma Algebras generated by Sets

[Bacle]

Hi, All:

A simple question: If A is the sigma- algebra generated by a collection C of subsets

of an ambient set X. Isn't it trivial that the sigma-algebra generated by A is A itself?

One definition is that the sigma algebra generated by a collection S of subsets is the

intersection of all s-algebras containing the collection S . Doesn't this automatically

mean that A is thesigma algebra generated by A? I am trying to avoid using the fact

that a countable union of countable sets is countable (otherwise, each set in A is the

countable union of sets in C, so a countable union of subsets of A is a countable union

of a countable union of sets in C , which is itself a countable union, so it is

contained in A, by def. of A as the s-algebra generated by C --a tongue-twister and proof! )

Thanks.

P.S: I am trying to answer the question for a student who has not yet seen cardinal arithmeti;

that is why I am trying to avoid using that union of countable is countable,

 

[Fredrik]

[strike]The smallest σ-algebra that contains A is {∅,A,A^c,X}.[/strike] (Edit: No it's not. That's the smallest σ-algebra that has the set A as a member, when A is a subset of X. But Bacle asked about the smallest σ-algebra that has the σ-algebra A as a subset. That's obviously A itself).

The definition of a σ-algebra is motivated by the idea that it's supposed to be the domain of a function that assigns a "size" to sets. What properties would we want such a function to have? Let's call the function μ and its domain Σ. Let A and B be two arbitrary members of Σ.

We require that μ(∅)=0. So we need ∅ to be in Σ.

We require μ(A⋃B)=μ(A)+μ(B) whenever A and B are disjoint. So we need A⋃B to be in Σ when A and B are disjoint. But A⋃B=A⋃(B-A), and that's a disjoint union on the right, so if B-A is in Σ, we can conclude that A⋃B is, even when they're not disjoint.

If we don't require that A-B is in Σ, we wouldn't be able to say that μ(A)=μ(A-B)+μ(A-(A-B)). Since this is a property we want μ to have, we require that A-B is in Σ.

Finally, it makes sense to require that X is in Σ. This means that A^c=X-A is too.

These conditions on Σ are exactly what define an "algebra" of sets. A σ-algebra is defined by strengthening the assumption about disjoint unions to apply to countable disjoint unions instead of just finite disjoint unions.

 

[micromass]

Hi Bacle!

Frederik is correct, but I think you mean something else.

You mean to say that A is the sigma-algebra and you want the sigma-algebra generated by A. Well, in that case, I'd say it's pretty trivial that A is indeed the sigma-algebra generated by A. Like you mention, the sigma-algebra generated by A is the intersection of all the sigma-algebra containing A. But A is a sigma-algebra containing A, so the intersection equals A!

(Hint: use [ itex]\mathcal{A}[ /itex] instead of A, it's less confusing that way...)

 

[Fredrik]

Ah, I see now that I didn't read post #1 carefully enough. Yes, I agree that it's trivial. Oh well, I needed to get those arguments straight in my head anyway, and now we have a post we can reference when someone asks why a σ-algebra is defined the way it is.

 

[Bacle]

Thanks, All just a comment.;
:
I wonder what the point of this exercise was (it was taken from Royden, I was told)

Also, methodology-wise, I wonder why the exercise assigned did not require to

first show that there is at least one sigma algebra for a given collection--maybe

because the superset X from where the collection C is drawn would be one.

 

[Fredrik]

I wonder why the exercise assigned did not require to

first show that there is at least one sigma algebra for a given collection--maybe

because the superset X from where the collection C is drawn would be one.

Yes, if C is a collection of subsets of X, the power set of X is a σ-algebra that has C as a subset. This seems like the sort of thing that you shouldn't have to prove because it has (or should have) already been mentioned in the book, right next to the definition of "σ-algebra generated by".