Proving A is a Sigma-Algebra on \Omega

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The discussion focuses on proving that the collection A, defined as A = {A ⊆ Ω: there exists B ∈ E with A = f^{-1}(B)}, is a sigma-algebra on the set Ω. The proof involves demonstrating that Ω is in A, showing closure under complements, and verifying closure under countable unions. The participants confirm that Ω is included in A since it can be represented as the inverse image of the sigma-algebra E. Additionally, they establish that A is closed under complements and countable unions, confirming that A meets the criteria of a sigma-algebra.

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

That works!
 

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