Introduction

Over the last several lessons, we have developed two seemingly opposite concepts.

Absolute Continuity

A measure:

$$\nu$$

is absolutely continuous with respect to:

$$\mu$$

if:

$$\nu \ll \mu$$

meaning:

$$\mu(A)=0 \implies \nu(A)=0$$

In this case, the Radon–Nikodym Theorem guarantees the existence of a density:

$$\frac{d\nu}{d\mu}$$

such that:

$$\nu(A)=\int_A \frac{d\nu}{d\mu},d\mu$$

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Singularity

A measure:

$$\nu$$

is singular with respect to:

$$\mu$$

if:

$$\nu \perp \mu$$

meaning the two measures live on disjoint parts of the space.

Examples include:

• Dirac measures
• Cantor measure
• Fractal measures

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A natural question now arises:

Can every measure be broken into an absolutely continuous part and a singular part?

The answer is yes.

This remarkable fact is called the Lebesgue Decomposition Theorem.

It is one of the deepest structural results in all of measure theory.

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The Big Picture

Suppose:

$$\mu$$

is a reference measure.

Think of:

$$\mu=\lambda$$

(Lebesgue measure).

Now suppose:

$$\nu$$

is another measure.

Part of:

$$\nu$$

may behave smoothly relative to:

$$\mu$$

and therefore admit a density.

Another part may be concentrated on exceptional sets and be singular.

The Lebesgue Decomposition Theorem says:

Every measure splits uniquely into these two pieces.

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Statement of the Theorem

Let:

$$\mu$$

and:

$$\nu$$

be σ-finite measures.

Then there exist unique measures:

$$\nu_a$$

and:

$$\nu_s$$

such that:

$$\nu=\nu_a+\nu_s$$

where:

$$\nu_a \ll \mu$$

and:

$$\nu_s \perp \mu$$

This decomposition is unique.

---

Interpretation

The measure:

$$\nu_a$$

is the part that possesses a density.

The measure:

$$\nu_s$$

is the part that lives on a singular set.

Together they reconstruct:

$$\nu$$

exactly.

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Visual Picture

Imagine a probability distribution.

Part is spread smoothly across an interval.

Part is concentrated on a fractal.

The smooth portion belongs to:

$$\nu_a$$

The fractal portion belongs to:

$$\nu_s$$

The Lebesgue decomposition separates these contributions.

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Example 1: Purely Absolutely Continuous

Let:

$$\nu(A)=\int_A x^2,d\lambda$$

Then:

$$\nu \ll \lambda$$

and:

$$\nu_s=0$$

Thus:

$$\nu=\nu_a$$

The entire measure is absolutely continuous.

---

Example 2: Purely Singular

Consider:

$$\nu=\delta_0$$

relative to Lebesgue measure.

Since:

$$\delta_0 \perp \lambda$$

we obtain:

$$\nu_a=0$$

and:

$$\nu=\nu_s$$

The entire measure is singular.

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Example 3: Mixed Measure

Define:

$$\nu=\lambda+\delta_0$$

This measure has two components.

A smooth component:

$$\lambda$$

and a point mass:

$$\delta_0$$

The decomposition becomes:

$$\nu_a=\lambda$$

and:

$$\nu_s=\delta_0$$

Thus:

$$\nu=\nu_a+\nu_s$$

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Why This Example Is Important

This example demonstrates that measures need not be entirely smooth or entirely singular.

Most interesting measures contain a mixture of both behaviors.

The theorem allows us to separate them.

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The Role of Radon–Nikodym

Since:

$$\nu_a \ll \mu$$

the Radon–Nikodym Theorem applies.

Therefore there exists:

$$f=\frac{d\nu_a}{d\mu}$$

such that:

$$\nu_a(A)=\int_A f,d\mu$$

Substituting into the decomposition:

$$\nu(A)=\int_A f,d\mu+\nu_s(A)$$

This formula is often the most useful form of the theorem.

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A Complete Description of a Measure

The theorem tells us:

Every measure consists of:

1. A density component.
2. A singular component.

Nothing else.

This is a profound classification result.

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Sketch of the Proof

The proof is surprisingly elegant.

One considers all measures:

$$\eta$$

satisfying:

$$\eta \le \nu$$

and:

$$\eta \ll \mu$$

Among all such measures, one constructs the largest possible absolutely continuous measure.

Call it:

$$\nu_a$$

Then define:

$$\nu_s=\nu-\nu_a$$

The key step is proving that:

$$\nu_s \perp \mu$$

This uses the maximality of:

$$\nu_a$$

and a contradiction argument.

Uniqueness follows naturally.

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Connection to Jordan Decomposition

Recall the Jordan Decomposition Theorem:

$$\nu=\nu^+-\nu^-$$

for signed measures.

The Lebesgue decomposition is different.

Instead of separating positive and negative parts, it separates:

• absolutely continuous parts
• singular parts

Thus we now have two fundamental decompositions:

Jordan

$$\nu=\nu^+-\nu^-$$

Lebesgue

$$\nu=\nu_a+\nu_s$$

Both play central roles throughout analysis.

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The Probability Perspective

Suppose:

$$P$$

is a probability measure.

Relative to Lebesgue measure:

$$\lambda$$

the theorem states:

$$P=P_a+P_s$$

where:

$$P_a \ll \lambda$$

and:

$$P_s \perp \lambda$$

Thus every probability distribution splits into:

• a density component
• a singular component

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Discrete, Continuous, and Singular Distributions

Probability theory often distinguishes:

Discrete

Example:

$$P(X=0)=1$$

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Continuous

Example:

Normal distributions.

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Singular

Example:

Cantor distributions.

The Lebesgue decomposition explains the mathematical relationship among these cases.

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The Refined Decomposition

In probability theory, one often writes:

$$P=P_d+P_c+P_s$$

where:

• $$P_d$$ is discrete
• $$P_c$$ is absolutely continuous
• $$P_s$$ is singular continuous

This refinement builds directly on the Lebesgue decomposition.

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Why Analysts Love This Theorem

The theorem transforms arbitrary measures into manageable pieces.

When proving results, analysts frequently:

1. Prove the result for densities.
2. Prove the result for singular measures.
3. Combine the pieces.

Without decomposition theorems, many advanced arguments would be impossible.

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Connection to Spectral Theory

Later in functional analysis we will encounter spectral measures.

Spectral measures also decompose into:

• absolutely continuous spectrum
• singular spectrum
• pure point spectrum

This is not a coincidence.

The philosophy originates here.

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Connection to Alain Connes

One of the recurring themes throughout Connes’ work is:

Understand an object by decomposing it into canonical components.

The Lebesgue decomposition is among the earliest examples of this philosophy.

Later we will encounter:

• decomposition of states
• decomposition of representations
• decomposition of von Neumann algebras
• decomposition of spectra

The pattern is always the same:

Break a complicated object into fundamental building blocks.

The Lebesgue Decomposition Theorem is one of the first great examples of this idea.

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Worked Example

Consider:

$$\nu=3\lambda+2\delta_0$$

relative to:

$$\lambda$$

Then:

$$\nu_a=3\lambda$$

because:

$$3\lambda \ll \lambda$$

and:

$$\nu_s=2\delta_0$$

because:

$$\delta_0 \perp \lambda$$

Thus:

$$\nu=\nu_a+\nu_s$$

with:

$$\nu_a(A)=3\lambda(A)$$

and:

$$\nu_s(A)=2\delta_0(A)$$

This is exactly the decomposition predicted by the theorem.

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Key Concepts Learned

By the end of this lesson you should understand:

• Every σ-finite measure admits a unique decomposition:

$$\nu=\nu_a+\nu_s$$

• The absolutely continuous part satisfies:

$$\nu_a \ll \mu$$

• The singular part satisfies:

$$\nu_s \perp \mu$$

• The Radon–Nikodym Theorem applies to:

$$\nu_a$$

• Every measure consists of a density component and a singular component.
• The theorem is one of the central structural results of measure theory.
• It serves as a prototype for many later decomposition theorems in functional analysis and operator algebras.

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Looking Ahead

In the next lesson:

Measure Theory Lesson 21: Product Measures

we begin extending measure theory to higher dimensions. We will learn how to construct measures on spaces such as:

$$X \times Y$$

and eventually prove Tonelli’s and Fubini’s Theorems, which allow multidimensional integration and form the foundation of probability theory, stochastic processes, and modern analysis.