Question

Let ω be a set and (a,A) a measurable space. For any function x:ω→a, define σ(x) to be the collection of all inverse images of x; i.e., {x−1(b);b∈A}. Prove that σ(x) is a σ-algebra and is the smallest σ-algebra with respect to which x is measurable.

Answer 1

To prove that σ(x) is a σ-algebra, we need to show that it satisfies three properties: non-empty, closed under complements, and closed under countable unions. To prove that σ(x) is the smallest σ-algebra with respect to which x is measurable, we need to show that any σ-algebra T containing all the inverse images of x must also contain σ(x).

Step-by-step explanation:

To prove that σ(x) is a σ-algebra, we need to show that it satisfies three properties:

1. σ(x) is non-empty: Since x is a function from ω to a, the inverse image of any element in a will be non-empty, so σ(x) will also be non-empty.
2. σ(x) is closed under complements: Let B be an element in σ(x). This means that there exists an element b in A such that x−1(b) = B. Taking the complement of both sides, we have (x−1(b))c = Bc. Since Bc is also in σ(x) and Bc is the complement of B, σ(x) is closed under complements.
3. σ(x) is closed under countable unions: Let B1, B2, B3, ... be elements in σ(x). This means that there exist elements b1, b2, b3, ... in A such that x−1(b1) = B1, x−1(b2) = B2, etc. Taking the countable union of all these sets, we have x−1(∪i=1∞bi) = ∪i=1∞x−1(bi). Since the countable union of elements in A is also in A, σ(x) is closed under countable unions.

To prove that σ(x) is the smallest σ-algebra with respect to which x is measurable, we need to show that any σ-algebra T containing all the inverse images of x must also contain σ(x). Let T be such a σ-algebra. Since T contains all the inverse images of x, T must contain σ(x).