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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

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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.

  1. 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.

  2. 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.

  3. 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.

This problem has been solved

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.

  1. For the empty set and the sets in A, we know that their pre-images under f are in F by the given condition.

  2. 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.

  3. 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.

This problem has been solved

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