I was with you until 'sigma Algebra'. Here's what ChatGPT says: To prove the existence of a smallest sigma-algebra containing C, we need to construct such an algebra and show that it satisfies the properties of a sigma-algebra and is contained in any other sigma-algebra containing C. Let A be the intersection of all sigma-algebras that contain C. We claim that A is itself a sigma-algebra. To see this, note that: Since X is a subset of any sigma-algebra containing C, X is certainly in A. Similarly, since the empty set is a subset of any set, the empty set is in A. To show that A is closed under complements, let E be any element of A. Then, E is in every sigma-algebra containing C, so its complement X\E is also in every such sigma-algebra. Therefore, X\E is in A as well. Finally, to show that A is closed under countable unions, let E1, E2, E3, ... be a countable collection of sets in A. Then, each Ei is in every sigma-algebra containing C, so their union E = E1 ∪ E2 ∪ E3 ∪ ... is also in every such sigma-algebra. Therefore, E is in A as well. Therefore, A satisfies the properties of a sigma-algebra. To show that it is the smallest sigma-algebra containing C, suppose there is another sigma-algebra B containing C, but not containing some set E in A. Then, B must be missing either E or its complement X\E. But this would mean that B does not contain A, which is the intersection of all sigma-algebras containing C, a contradiction. Therefore, A is the smallest sigma-algebra containing C. Make sense? |
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