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Radon-Nikodym derivatives satisfy the chain rule along nu << mu << lambda

Statement

Let $λ$ and $μ$ be sigma-finite positive measures and let $ν$ be a signed measure or a finite complex measure on the same measurable space. Assume there is an increasing measurable exhaustion $(X_{n})_{n \in N}$ with $⋃_{n} X_{n} = X$ , $λ (X_{n}) < + \infty$ , $μ (X_{n}) < + \infty$ , and $∣ ν ∣ (X_{n}) < + \infty$ for every $n$ , and assume $ν ≪ μ ≪ λ$ . Then $\frac{d ν}{d λ} = \frac{d ν}{d μ} \frac{d μ}{d λ} λ -almost everywhere.$

Facts & Assumptions

Given: Measures $λ, μ, ν$ with the common finite-exhaustion hypothesis and $ν ≪ μ ≪ λ$ .

[L1]

A nonnegative density composes through another density: if $η (E) = \int_{E} h d μ$ and $μ (E) = \int_{E} k d λ$ with $h, k \geq 0$ , then $η (E) = \int_{E} h k d λ$ . (Integrating against a density agrees with integrating the product)

[L2]

The Radon-Nikodym density is unique up to almost-everywhere equality. (A sigma-finite signed measure that is absolutely continuous with respect to a sigma-finite positive measure has a unique almost-everywhere density)

[L3]

If a signed measure is absolutely continuous with respect to $μ$ , then its Jordan parts are too. (For signed and complex measures, absolute continuity is equivalent for the measure, its Jordan or real-imaginary parts, and its total variation)

[L4]

Jordan decomposition writes a signed measure as $ν = ν^{+} - ν^{-}$ . (Jordan decomposition of a signed measure into unique mutually singular positive parts)

[L5]

The real and imaginary parts of a finite complex measure are finite signed measures. (The real and imaginary parts of a complex measure are finite signed measures, and nu = Re nu + i Im nu)

Proof

technique · direct

1.1

First assume that $ν$ is a positive measure. Choose the nonnegative representatives $h$ of $d ν / d μ$ and $k$ of $d μ / d λ$ furnished by the positive-measure case of the Radon-Nikodym theorem. For every measurable set $E$ , that theorem gives $ν (E) = \int_{E} h d μ, μ (E) = \int_{E} k d λ .$ Applying [L1] to the nonnegative density $h$ therefore yields $ν (E) = \int_{E} h k d λ (E \in A) .$ Hence [L2] gives $d ν / d λ = h k$ $λ$ -almost everywhere in the positive case.

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2.1

Now assume that $ν$ is a signed measure. By [L4], write $ν = ν^{+} - ν^{-}$ . Because $ν ≪ μ$ , [L3] gives $ν^{\pm} ≪ μ$ . Apply step 1.1 to $ν^{+}$ and $ν^{-}$ separately to obtain nonnegative representatives $h^{+}, h^{-}$ with $\frac{d ν^{+}}{d λ} = h^{+} k, \frac{d ν^{-}}{d λ} = h^{-} k λ -almost everywhere.$ Then $h : = h^{+} - h^{-}$ represents $d ν / d μ$ , while $(h^{+} - h^{-}) k$ represents $d ν / d λ$ . Uniqueness from [L2] therefore gives $\frac{d ν}{d λ} = h k = \frac{d ν}{d μ} \frac{d μ}{d λ} λ -almost everywhere.$

L2L3L4step 1.1algebra

3.1

Finally assume that $ν$ is a finite complex measure, and choose a representative $h = u + i v$ of $d ν / d μ$ . By [L5], the finite signed measures $Re ν$ and $Im ν$ are both absolutely continuous with respect to $μ$ , and the measurable-set identity for $h$ shows that $u$ and $v$ represent $d (Re ν) / d μ$ and $d (Im ν) / d μ$ . Applying step 2.1 to those signed measures gives $\frac{d (Re ν)}{d λ} = u \frac{d μ}{d λ}, \frac{d (Im ν)}{d λ} = v \frac{d μ}{d λ} λ -almost everywhere.$ Therefore, for every measurable set $E$ , $ν (E) = Re ν (E) + i Im ν (E) = \int_{E} u \frac{d μ}{d λ} d λ + i \int_{E} v \frac{d μ}{d λ} d λ = \int_{E} h \frac{d μ}{d λ} d λ .$ So $h (d μ / d λ)$ represents $d ν / d λ$ , and [L2] yields $\frac{d ν}{d λ} = h \frac{d μ}{d λ} = \frac{d ν}{d μ} \frac{d μ}{d λ} λ -almost everywhere.$

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Sources

Richard F. Bass, Real Analysis for Graduate Students, Exercise 13.8 (standard reference, not scraped)
John K. Hunter, Measure Theory, §6.8 (standard reference, not scraped)