Homework Help: Sigma Algebras

1. Sep 8, 2009

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.

$$A_i \epsilon$$A. 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?

Last edited: Sep 8, 2009
2. Sep 9, 2009

Billy Bob

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?)