Let (X, F) be a measurable space and let f be a function from X to Y . Let Abe a collection of subsets of Y such that f −1(E) ∈ F for every E ∈ A. Showthat f −1(D) ∈ F for every set D which belongs to the σ-algebra generated byA.
Question
Let (X, F) be a measurable space and let f be a function from X to Y . Let Abe a collection of subsets of Y such that f −1(E) ∈ F for every E ∈ A. Showthat f −1(D) ∈ F for every set D which belongs to the σ-algebra generated byA.
Solution
To prove this, we need to show that the pre-image under f of any set in the σ-algebra generated by A is in F.
The σ-algebra generated by A, denoted by σ(A), is the smallest σ-algebra that contains A. It contains A, the empty set, the complements of sets in A, and countable unions of sets in A.
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If D is in A, then by assumption, f^(-1)(D) is in F.
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If D is the empty set, then f^(-1)(D) is the empty set, which is in F because F is a σ-algebra.
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If D is the complement of a set in A, then D = Y \ E for some E in A. Then f^(-1)(D) = f^(-1)(Y \ E) = X \ f^(-1)(E). Since f^(-1)(E) is in F, its complement is also in F because F is a σ-algebra.
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If D is a countable union of sets in A, then D = union(E_i) for some countable collection {E_i} of sets in A. Then f^(-1)(D) = f^(-1)(union(E_i)) = union(f^(-1)(E_i)). Since each f^(-1)(E_i) is in F, their countable union is also in F because F is a σ-algebra.
Therefore, for any set D in the σ-algebra generated by A, f^(-1)(D) is in F. This completes the proof.
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