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The chain rule for Radon-Nikodym derivatives on $[0, 1]$

Example

Let $μ : = 2 λ$ on $[0, 1]$ , and let $ν (E) : = \int_{E} 2 x χ_{[0, 1]} (x) d λ (x) .$ Then $\frac{d μ}{d λ} = 2, \frac{d ν}{d μ} = x χ_{[0, 1]} (x), \frac{d ν}{d λ} = 2 x χ_{[0, 1]} (x),$ so $\frac{d ν}{d λ} = \frac{d ν}{d μ} \frac{d μ}{d λ} λ -almost everywhere.$

Facts & Assumptions

Given: The measures $μ = 2 λ$ and $ν (E) = \int_{E} 2 x χ_{[0, 1]} (x) d λ (x)$ .

[L1]

For an absolutely continuous signed measure and a sigma-finite positive base satisfying a common finite exhaustion, a Radon--Nikodym derivative is any representative whose measurable-set integrals recover the measure (The Radon-Nikodym derivative as an almost-everywhere equivalence class).

[L3]

Integrals over null sets vanish (A nonnegative integral over a null set vanishes).

[L2]

Under the sigma-finiteness, common finite-exhaustion, and $ν ≪ μ ≪ λ$ hypotheses, Radon--Nikodym derivatives satisfy the chain rule (Radon-Nikodym derivatives satisfy the chain rule along nu << mu << lambda).

Verification

technique · direct

1.1

On the measurable space $[0, 1]$ , the measures $λ$ , $μ$ , and $ν$ are finite, their one-set exhaustion is common, and [L3] gives $ν ≪ μ ≪ λ$ . For every measurable set $E$ , one has $μ (E) = \int_{E} 2 d λ$ and $ν (E) = \int_{E} 2 x χ_{[0, 1]} (x) d λ (x)$ . Therefore [L1] lets us take $d μ / d λ = 2$ and $d ν / d λ = 2 x χ_{[0, 1]}$ .

L1L3given

2.1

The function $x χ_{[0, 1]}$ satisfies $\int_{E} x d μ = \int_{E} 2 x d λ = ν (E)$ , so it represents $d ν / d μ$ . On the measurable space $[0, 1]$ , all three measures are finite, the one-set exhaustion $X_{1} = [0, 1]$ is common, and $ν ≪ μ ≪ λ$ . Applying [L2] now yields $\frac{d ν}{d λ} = \frac{d ν}{d μ} \frac{d μ}{d λ} = x \cdot 2 = 2 x λ -almost everywhere on [0, 1] .$

step 1.1L2algebra∎

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The chain rule for Radon-Nikodym derivatives on [0,1][0,1]