9 Image Measures Given: a measure space (Ω

9 Image Measures

Given: a measure space (Ω,A, µ), a measurable space (Ω⁰,A⁰), and anA-A⁰-measurable mapping f : Ω→Ω⁰.

Lemma 1.

f(µ) :A⁰ →R+∪ {∞}

A⁰ 7→µ(f⁻¹(A⁰)) =µ({f ∈A⁰}) defines a measure on A⁰.

Proof. f(µ) is well-defined, since f⁻¹(A⁰) ∈A for any A⁰ ∈ A⁰. Further, for (A_n)n∈N

disjoint,f⁻¹(A_n) are disjoint, too, which implies the σ–additivity.

Definition 1. f(µ) is called the image measure of µ underf. Example 1. Let

(Ω,A, µ) = (R^k,B_k, λ_k), (Ω⁰,A⁰) = (R^k,B_k).

(i) Fix a∈R^k.Forf(ω) =ω+a we get

f(λ_k)(A⁰) =λ_k(A⁰−a) =λ_k(A⁰),

see Analysis III (‘or’ verify this identity for measurable rectangles and apply Theorem 4.4). Thus

f(λ_k) =λ_k. (ii) Fix r∈R\ {0}. For f(ω) =r·ω we get

f(λ_k)(A⁰) =λ_k(1/r·A⁰) = 1

|r|^k ·λ_k(A⁰),

see Analysis III (‘or’ verify this identity for measurable rectangles and apply Theorem 4.4). Thus

f(λ_k) = 1

|r|^k ·λ_k. Theorem 1 (Transformation ‘Theorem’).

(i) for g ∈Z₊(Ω⁰,A⁰)

Z

Ω⁰

g df(µ) = Z

Ω

g◦f dµ (1)

(ii) for g ∈Z(Ω⁰,A⁰)

g isf(µ)-integrable ⇔ g◦f is µ-integrable, in which case (1) holds.

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Proof. Algebraic induction: For indicator functionsg, both sides are equal by defini- tion; further, both sides obey linearity and monotone convergence in g.

Example 2. Consider open setsU, V ⊂R^k and aC¹-diffeomorphismf :U →V. Let (Ω,A, µ) = (U, U∩B_k, λ_k|U∩B_k), (Ω⁰,A⁰) = (V, V ∩B_k).

Put

ν =λ_k|_V∩B_k. Then

f(µ)(A⁰) = Z

f⁻¹(A⁰)

dµ= Z

A⁰

|detDf⁻¹|dν, see Analysis III for the case of an open set A⁰ ⊂V. Thus

f(µ) =|detDf⁻¹| ·ν, and therefore

Z

U

g◦f dµ= Z

V

g df(µ) = Z

V

g· |detDf⁻¹|dν.

Putg =h◦f⁻¹ and ϕ =f⁻¹ to obtain Z

U

h dµ= Z

V

h◦ϕ· |detDϕ|dν.

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