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Bounded signed integrals obey the density formula

Statement

Let μ be a positive measure, let p:X→[0,∞) be finite-valued and measurable, and suppose P(A)=∫Ap dμ is a probability measure. If f:X→R is measurable and ∣f∣≤M for some finite M≥0, then f is integrable under P, fp is integrable under μ, and ∫Xf dP=∫Xfp dμ.

Facts & Assumptions

Given: A positive measure μ, a finite nonnegative measurable density p, the probability measure P=p dμ, and a measurable real function f with ∣f∣≤M<∞.

[F1]

The measure with density f relative to μ: P(A)=∫Ap dμ for every measurable A.

[F2]

Integrating against a density agrees with integrating the product: for nonnegative measurable g, ∫g dP=∫gp dμ.

[F3]

Closure properties of measurable functions used by the integral: the positive and negative parts f+,f− of a real measurable f are measurable.

[F4]

Arithmetic and lattice operations preserve measurability whenever they are defined: the product of finite real measurable functions is measurable.

[F5]

Monotonicity and nonnegative homogeneity of the nonnegative integral: the nonnegative integral is monotone and homogeneous for nonnegative scalars, with the integral of the zero function equal to zero.

[F6]

Integrable real and complex functions, and their integrals: f is integrable when ∫∣f∣ is finite, and its integral is ∫f+−∫f−.

[F7]

The Lebesgue integral is linear on L1(μ): integrable functions have finite linear integrals.

Proof

technique · direct
1.1F1F3F4F8given

By [F1] and [F8], ∫p dμ=P(X)=1. The functions f+,f− are measurable by [F3] and satisfy 0≤f+,f−≤M. Products f+p and f−p are measurable by [F4].

2.1F1F5F6F8step 1.1

Monotonicity [F5], [F1], and [F8] give ∫f± dP≤MP(X)=M and ∫f±p dμ≤M∫p dμ=M. Hence f+,f− are integrable under P and f+p,f−p are integrable under μ by [F6].

2.2F2step 1.1

Apply [F2] separately to the nonnegative measurable functions f+ and f−. It gives ∫f+ dP=∫f+p dμ and ∫f− dP=∫f−p dμ.

3.1F6F7step 2.1step 2.2algebra∎

Subtract the second equality in step 2.2 from the first. By [F6] and [F7], both sides are the integrals of the corresponding signed functions, so ∫f dP=∫fp dμ.

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