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?

 

[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!