) 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 of every 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 is the intersection of all σ-algebras that contain A.
The σ-algebra generated by A contains the following sets:
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The empty set ∅ and the whole space Y, because every σ-algebra must contain these two sets.
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The union of any countable collection of sets in σ(A), because every σ-algebra is closed under countable unions.
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The complement of any set in σ(A), because every σ-algebra is closed under taking complements.
We need to show that the pre-image of each of these types of sets is in F.
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The pre-image of the empty set and the whole space Y are the empty set and the whole space X, respectively. Since F is a σ-algebra on X, it must contain these two sets.
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Let {D_n} be a countable collection of sets in σ(A). Then the pre-image of the union of the D_n's is the union of the pre-images of the D_n's, because f is a function. Since each D_n is in σ(A), its pre-image is in F by assumption. Since F is a σ-algebra, it is closed under countable unions, so the union of the pre-images of the D_n's is in F.
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Let D be a set in σ(A). Then the pre-image of the complement of D is the complement of the pre-image of D, because f is a function. Since D is in σ(A), its pre-image is in F by assumption. Since F is a σ-algebra, it is closed under taking complements, so the complement of the pre-image of D is in F.
Therefore, the pre-image of every set in the σ-algebra generated by A is in F. This completes the proof.
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