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Measure Theory 14 | Radon-Nikodym theorem and Lebesgue's decomposition theorem [dark version] thumbnail

Measure Theory 14 | Radon-Nikodym theorem and Lebesgue's decomposition theorem [dark version]

5 min read

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TL;DR

Lebesgue measure Λ on ℝ assigns each interval its usual length and serves as the reference measure for comparing other measures.

Briefing

Two foundational results in measure theory—Lebesgue’s decomposition theorem and the Radon–Nikodym theorem—turn complicated measures into simpler pieces: one part behaves like an ordinary density with respect to a reference measure, while the remaining part lives on a “length-zero” set. Together, they explain how any reasonable measure on the real line can be split into an absolutely continuous component and a singular component, and then (for the absolutely continuous part) rewritten as an integral against Lebesgue measure. This matters because it converts abstract measure statements into concrete formulas involving functions, which is essential in probability, analysis, and stochastic modeling.

The setup starts with a measure space (X, 𝒜, μ) and focuses on the standard case X = ℝ, 𝒜 = Borel σ-algebra 𝓑(ℝ), and μ taken relative to Lebesgue measure Λ (the “reference measure”). Lebesgue measure is characterized by assigning each interval its usual length. With Λ in place, two key notions define how a measure μ relates to Λ.

A measure μ is absolutely continuous with respect to Λ (written μ ≪ Λ) if every set that has Λ-measure zero also has μ-measure zero. In other words, μ cannot “see” any set that Lebesgue measure declares negligible. The transcript highlights simple examples: Λ is absolutely continuous with respect to itself, and the zero measure is also absolutely continuous. By contrast, μ is singular with respect to Λ (written μ ⟂ Λ) if there exists a measurable set N such that Λ(N) = 0 while μ(ℝ \ N) = 0. Intuitively, μ concentrates entirely on a Λ-null set. A canonical example is the Dirac measure at 0, denoted Δ₀: it assigns mass 1 to {0} and 0 elsewhere, making it singular relative to Lebesgue measure because {0} has Λ-measure zero.

Lebesgue’s decomposition theorem then states that, under appropriate conditions (notably σ-finiteness), any measure μ on ℝ can be uniquely decomposed into two measures on the same σ-algebra: μ = μ_AC + μ_s, where μ_AC is absolutely continuous with respect to Λ and μ_s is singular with respect to Λ. The subscripts are chosen to reflect the behavior: “AC” for absolutely continuous and “s” for singular. This uniqueness means the split is not just possible—it is determined.

The Radon–Nikodym theorem addresses the absolutely continuous part. When μ_AC ≪ Λ and the relevant σ-finiteness assumptions hold, there exists a measurable function h: ℝ → [0, ∞) (a density) such that for every measurable set A, μ_AC(A) can be written as an integral of h over A with respect to Λ: μ_AC(A) = ∫_A h dΛ. The practical payoff is that specifying μ_AC is equivalent to specifying the density function h, turning measure-theoretic complexity into an ordinary integral.

Finally, the transcript stresses that these theorems do not hold in full generality: σ-finiteness is required for both the measure μ and the reference measure (automatically satisfied for Lebesgue measure). With those conditions in place, the pair of theorems becomes a powerful toolkit, with major applications in areas like stochastic processes where measures and densities are central.

Cornell Notes

Lebesgue’s decomposition theorem splits a measure μ into a sum of two uniquely determined parts relative to Lebesgue measure Λ: an absolutely continuous component μ_AC and a singular component μ_s, so μ = μ_AC + μ_s. Absolute continuity (μ_AC ≪ Λ) means μ_AC ignores every Λ-null set; singularity (μ_s ⟂ Λ) means μ_s concentrates on a set of Λ-measure zero. The Radon–Nikodym theorem then goes further: if μ_AC ≪ Λ (with the needed σ-finiteness), there exists a measurable density h ≥ 0 such that μ_AC(A) = ∫_A h dΛ for all measurable sets A. This converts the abstract absolutely continuous measure into an ordinary integral against Lebesgue measure, which is crucial for applications in analysis and probability.

What does it mean for a measure μ to be absolutely continuous with respect to Lebesgue measure Λ?

μ is absolutely continuous with respect to Λ (μ ≪ Λ) if for every measurable set A, Λ(A) = 0 implies μ(A) = 0. So any set that has “zero length” under Λ must also have zero mass under μ. The transcript notes that this definition is always relative to a chosen reference measure; with Λ as the reference, the wording can omit “with respect to Λ.”

How is singularity (μ ⟂ Λ) different from absolute continuity?

Singularity means μ and Λ live on disjoint “measure-theoretic supports.” Concretely, there must exist a measurable set N such that Λ(N) = 0 while μ(ℝ \ N) = 0. The transcript’s example is the Dirac measure at 0, Δ₀: it assigns mass 1 to {0} and 0 elsewhere, and since {0} has Λ-measure 0, Δ₀ is singular with respect to Lebesgue measure.

What does Lebesgue’s decomposition theorem guarantee about any measure μ (under σ-finiteness)?

It guarantees a unique decomposition μ = μ_AC + μ_s on the same σ-algebra, where μ_AC is absolutely continuous with respect to Λ and μ_s is singular with respect to Λ. The uniqueness matters: there is one and only one way to split μ into these two behaviors relative to Λ.

How does the Radon–Nikodym theorem turn an absolutely continuous measure into an integral?

If μ_AC ≪ Λ and the σ-finiteness conditions hold, then there exists a measurable function h: ℝ → [0, ∞) such that for every measurable set A, μ_AC(A) = ∫_A h dΛ. Here h is the density; giving h is equivalent to specifying μ_AC, because the measure of every set is determined by integrating h over that set.

Why does σ-finiteness show up as a required assumption?

The transcript emphasizes that the theorems are not valid in complete generality; σ-finiteness is needed for both the measure μ and the reference measure. Lebesgue measure is σ-finite on ℝ, so the standard case works, but generalizing to other reference measures requires those measures to be σ-finite as well.

Review Questions

  1. State the definitions of absolute continuity (μ ≪ Λ) and singularity (μ ⟂ Λ) relative to Lebesgue measure.
  2. What does Lebesgue’s decomposition theorem say about the existence and uniqueness of μ_AC and μ_s?
  3. Under what conditions does the Radon–Nikodym theorem produce a density h, and what formula does h satisfy?

Key Points

  1. 1

    Lebesgue measure Λ on ℝ assigns each interval its usual length and serves as the reference measure for comparing other measures.

  2. 2

    A measure μ is absolutely continuous with respect to Λ if Λ(A)=0 always forces μ(A)=0 for measurable sets A.

  3. 3

    A measure μ is singular with respect to Λ if there exists N with Λ(N)=0 and μ(ℝ\N)=0, meaning μ concentrates on a Λ-null set.

  4. 4

    Lebesgue’s decomposition theorem (with σ-finiteness) uniquely splits any measure μ into μ=μ_AC+μ_s, where μ_AC ≪ Λ and μ_s ⟂ Λ.

  5. 5

    The Radon–Nikodym theorem (again under σ-finiteness) converts μ_AC into an integral representation μ_AC(A)=∫_A h dΛ using a measurable density h≥0.

  6. 6

    For absolutely continuous measures, specifying the density function h is equivalent to specifying the measure μ_AC on all measurable sets.

Highlights

Absolute continuity forbids a measure from charging any set that Lebesgue measure declares negligible (Λ-null).
Singularity captures the opposite extreme: the measure can concentrate entirely on a set of Λ-measure zero.
Lebesgue’s decomposition theorem provides a unique split μ=μ_AC+μ_s into absolutely continuous and singular parts relative to Λ.
Radon–Nikodym turns μ_AC into a standard integral against Lebesgue measure via a density h: μ_AC(A)=∫_A h dΛ.
Both results rely on σ-finiteness; without it, the clean decomposition and density representation may fail.

Topics

  • Lebesgue Decomposition
  • Radon-Nikodym
  • Absolute Continuity
  • Singular Measures
  • Density Functions