Probability help/sigma-algebras

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To prove that F is a sigma-algebra on A, it must be shown that F contains A, is closed under complementation, and closed under countable unions. A is in F since F consists of intersections of A with elements E from the sigma-algebra ε, ensuring A itself is included. The complement of A ∩ E is also in F, as it can be expressed using the properties of complements and intersections. Additionally, the countable union of sets in F remains within F, adhering to the definition of a sigma-algebra. Understanding these properties of sigma-algebras is crucial for completing the proof.
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Let (S, ε, P) be a probability space and let A be an element of ε with P(A)>0. Let F={AπE :E is an element of ε }
-Prove that F is a sigma-algebra on A.



Not sure even where to go with this really. I know that to be a sigma-algebra has to be closed under complementation and countable unions. I'm not very good with proofs, and just a push in the right direction would help me out a ton. Thanks!
 
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So you will need to show that:

- A is in F
- The complement of A cap E is in F
- A countable union U_i A cap E_i is in F.

All you need to know are the properties of sets and that epsilon is a sigma algebra. Try it and show what you have tried.
 
Oh, I was under the impression I had to do it the other way around, such that F is in A, etc..maybe that's why I'm having such a hard time here.

Ok so...

A is in F because F=A intersect E, and since an intersect of A is in F, then all of A has to be in F (not sure how to write this properly)
F=AnE is the same as F=(A^c U E^c)^c (^c=complement)
not really sure how to go about the last one
 
FTaylor244 said:
A is in F because F=A intersect E, and since an intersect of A is in F, then all of A has to be in F (not sure how to write this properly)
F=AnE is the same as F=(A^c U E^c)^c (^c=complement)
not really sure how to go about the last one

F is a set of subsets of A, namely the set of A cap E for E in the sigma algebra. Check the definition of a sigma algebra. Try again keeping that in mind.
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...

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