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.


$$A_i \epsilon$$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?

 

[LeonhardEuler]

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.

But A is not made up of all subsets of $$\Omega$$, just those that are the inverse image of some B$$\epsilon$$ E. However, we know that E is a sigma algebra, so can you think of a set whose inverse image is $$\Omega$$?

Everything else you did looks ok.

 

[azdang]

Ohh okay! So, wouldn't $$f^{-1}(E)= \Omega$$? And E is in E, so I think this works.

 

[LeonhardEuler]

Ohh okay! So, wouldn't $$f^{-1}(E)= \Omega$$? And E is in E, so I think this works.

That works!