Proving A is a Sigma-Algebra on Omega

[azdang]

Let f be a function mapping \Omega to another space E with a sigma-algebra[/tex] E. Let A = {A C \Omega: there exists B \epsilon E with A = f^{-1}(B)}. Show that A is a sigma-algebra on \Omega.

Okay, so I should start by showing that \Omega is in A. I wasn't sure if this was as easy as saying that since A is made up of all subsets of \Omega, then clearly, \Omega must be in A since it is a subset of itself.


Next, I would have to show it is closed under complement. Here is what I tried doing.


A = f^{-1}(B)

A^c = (f^{-1}(B))^c = f^{-1}(B^c). Since E is a sigma-algebra, B^c is in E, thus by the definition of A, f^{-1}(B^c) is in A so it is closed under complement.


The last thing would be to show it is closed under countable union. I'm sort of unsure how to set this up, but here is what I tried doing.


A_i \epsilonA. Then, A_i = f^{-1}(B_i) where B_i \epsilon E. So, \bigcup_{i=1}^{\infty}A_i = \Bigcup_{i=1}^{\infty}f^{-1}(B_i)=f^{-1}(\bigcup_{i=1}^{\infty}B_i). And the union of the B_i's is in E since it is a sigma-algebra. Therefore, can I conclude that A is closed under countable union and thus, a sigma-algebra?

 

[Billy Bob]

Okay, so I should start by showing that \Omega is in A. I wasn't sure if this was as easy as saying that since A is made up of all subsets of \Omega

This is invalid, because A is not necessarily made up of all subsets of \Omega. (If it were, then A would be the power set of \Omega.)

You simply need to display a set B such that \Omega is the inverse image of B. There is really only one possible choice for B, isn't there?

Your complement and countable union steps are very good. (In your final write-up of the complement step, you ought to introduce A and B, similar to what you did in the first step of the countable union proof, to satisfy a picky grader.)

By the way, is it even necessary to prove \Omega is in A? I mean, is that part of your definition? (Wouldn't it follow anyway from the other properties, perhaps with a "non-empty" hypothesis here or there?)