By Pearn W. L., Lin G. H.

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Let µ1 and µ2 denote any two extensions of µ. Let M ≡ {A ∈ σ[C] : µ1 (A) = µ2 (A)} denote the class where they are equal. We will first show that (h) M is a monotone class. Let An be monotone in M. 2. Thus (h) holds. 6 implies that σ[C] ⊂ M. Thus µ1 = µ2 on σ[C] (and possibly on even more sets than this). Thus the claimed uniqueness holds. ] Claim 8: Uniqueness holds when µ is a σ-finite measure (label the sets of the measurable partition as Ωn ). We must again demonstrate the uniqueness. Fix n. We will consider µ, µ1 , µ2 on C, on σ[C] ∩ Ωn , and on σ[C ∩ Ωn ].

1. Consider (14). Now: A ∈A implies X −1 (A ) ∈ A (a) implies X −1 (A c ) = [X −1 (A )]c ∈ A implies A c ∈ A , A n ’s ∈ A (b) implies X −1 (A n )’s ∈ A implies X −1 ( n An ) = nX −1 (An ) ∈ A implies n An ∈A. This gives (14). Consider (13). Now, (c) X −1 (σ[C ] ) = (a σ-field containing X −1 (C )) ⊃ σ[X −1 (C )] . Then (14) shows that (d) A ≡ { A : X −1 (A ) ∈ σ[X −1 (C )] } = (a σ-field containing C ) ⊃ σ[C ] , so that (e) X −1 (σ[C ] ) ⊂ X −1 (A ) ⊂ σ[X −1 (C )] . Combining (c) and (e) gives (13).

1 (Induced measure) Suppose that X : (Ω, A, µ) → (Ω , A ) is a measurable function. 15) that (9) µ (A ) ≡ µX (A ) = µ(X −1 (A )) for all A ∈ A , and µ is a measure on (Ω , A ), called the induced measure of X. 6 (Theorem of the unconscious statistician) First, the induced measure µX (·) of the rv X determines the induced measure µg(X) for all measurable ¯ B¯ ). Second, functions g : (Ω , A ) → (R, (10) g(X(ω)) dµ(ω) = X −1 (A ) g(x) dµX (x) for all A ∈ A , A in the sense that if either side exists, then so does the other and they are equal.

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