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 1
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 is the intersection of all σ-algebras that contain A.
The σ-algebra generated by A contains the empty set, the sets in A, the complements of the sets in A, and the countable unions of the sets in A.
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For the empty set and the sets in A: The pre-image of the empty set under any function is the empty set, which is in F because F is a σ-algebra. For any set E in A, we know that f^(-1)(E) is in F by the given condition.
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For the complements of the sets in A: Let E be a set in A. Then the complement of E in Y, denoted by E^c, is also in σ(A). We need to show that f^(-1)(E^c) is in F. Since f^(-1)(E) is in F and F is a σ-algebra, the complement of f^(-1)(E) in X, which is f^(-1)(E^c), is also in F.
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For the countable unions of the sets in A: Let {E_i} be a countable collection of sets in A. Then the countable union of the E_i's, denoted by ∪E_i, is in σ(A). We need to show that f^(-1)(∪E_i) is in F. Since f^(-1)(E_i) is in F for each i and F is a σ-algebra, the countable union of the f^(-1)(E_i)'s, which is f^(-1)(∪E_i), is also in F.
Therefore, the pre-image under f of any set in the σ-algebra generated by A is in F. This completes the proof.
Solution 2
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 is the intersection of all σ-algebras that contain A.
The σ-algebra generated by A contains the empty set, the sets in A, the complements of the sets in A, and the countable unions of the sets in A.
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For the empty set and the sets in A, we know that their pre-images under f are in F by the given condition.
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For the complement of a set E in A, we have f^(-1)(E^c) = (f^(-1)(E))^c. Since f^(-1)(E) is in F and F is a σ-algebra, the complement of f^(-1)(E) is also in F.
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For a countable union of sets in A, say E1, E2, ..., we have f^(-1)(∪Ei) = ∪f^(-1)(Ei). Since each f^(-1)(Ei) is in F and F is a σ-algebra, the countable union of the f^(-1)(Ei) is also in F.
Therefore, the pre-image under f of any set in the σ-algebra generated by A is in F. This completes the proof.
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