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Let f be a function defined on a set X with values in a set Y . For any subsetE of Y , letf −1(E) = {x ∈ X | f (x) ∈ E}.Show that the setFX = {f −1(E) | E ∈ FY }is a σ-algebra of subsets of X if FY is a σ-algebra of subsets of Y

Question

Let f be a function defined on a set X with values in a set Y . For any subsetE of Y , letf −1(E) = {x ∈ X | f (x) ∈ E}.Show that the setFX = {f −1(E) | E ∈ FY }is a σ-algebra of subsets of X if FY is a σ-algebra of subsets of Y

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Solution

To show that FX is a σ-algebra of subsets of X, we need to verify three properties:

  1. X is in FX: Since FY is a σ-algebra, Y is in FY. Then f^(-1)(Y) = X is in FX.

  2. If A is in FX, then its complement X\A is in FX: If A is in FX, then there exists E in FY such that A = f^(-1)(E). Since FY is a σ-algebra, the complement Y\E is in FY. Then f^(-1)(Y\E) = X\A is in FX.

  3. If {A_i} is a countable collection of sets in FX, then the union of {A_i} is in FX: If {A_i} is in FX, then for each i, there exists E_i in FY such that A_i = f^(-1)(E_i). Since FY is a σ-algebra, the union of {E_i} is in FY. Then f^(-1)(union of {E_i}) = union of {A_i} is in FX.

Therefore, FX is a σ-algebra of subsets of X.

This problem has been solved

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