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RA-8

8-12 hours

Advanced

Lebesgue Decomposition & Radon–Nikodym

Master the Lebesgue decomposition theorem, Radon-Nikodym theorem, and signed measures. Understand how measures relate to each other through absolute continuity and singularity.

Learning Objectives

By the end of this course, you will be able to:

Understand signed measures and their decomposition

Master the Lebesgue decomposition theorem

Apply the Radon-Nikodym theorem to find density functions

Understand absolute continuity and singularity of measures

Compute Radon-Nikodym derivatives

Apply these theorems to probability theory and analysis

Prerequisites

Before starting this course, you should be familiar with:

Absolute continuity of functions
Lebesgue measure and integration
Basic measure theory
Convergence theorems

Core Concepts

Measure decomposition and derivatives

Definition 8.1: Signed Measure

A signed measure $\nu$ on a measurable space $(X, \mathcal{M})$ is a function $\nu: \mathcal{M} \to [-\infty, \infty]$ such that:

$\nu(\emptyset) = 0$
$\nu$ is countably additive
$\nu$ takes at most one of the values $+\infty$ or $-\infty$

Definition 8.2: Absolute Continuity of Measures

A signed measure $\nu$ is absolutely continuous with respect to a measure $\mu$ (written $\nu \ll \mu$ ) if:

\mu(E) = 0 \implies \nu(E) = 0

Definition 8.3: Mutual Singularity

Two signed measures $\nu$ and $\mu$ are mutually singular (written $\nu \perp \mu$ ) if there exist disjoint measurable sets $A$ and $B$ such that $\nu$ is concentrated on $A$ and $\mu$ is concentrated on $B$ .

Theorem 8.1: Hahn Decomposition

Let $\nu$ be a signed measure. Then there exists a measurable set $P$ such that:

$\nu(E \cap P) \geq 0$ for all measurable $E$
$\nu(E \cap P^c) \leq 0$ for all measurable $E$

The pair $(P, P^c)$ is called a Hahn decomposition.

Proof of Theorem 8.1:

Proof:

The proof uses the concept of a positive set (a set where $\nu$ is nonnegative on all subsets) and shows that the supremum of measures of positive sets is achieved. The maximizing set serves as $P$ . ∎

∎

Definition 8.4: Jordan Decomposition of Measures

Given a Hahn decomposition $(P, P^c)$ , define:

\nu^+(E) = \nu(E \cap P), \quad \nu^-(E) = -\nu(E \cap P^c)

Then $\nu = \nu^+ - \nu^-$ is the Jordan decomposition of $\nu$ .

Theorem 8.2: Lebesgue Decomposition

Let $\nu$ be a σ-finite signed measure and $\mu$ be a σ-finite positive measure. Then there exist unique signed measures $\nu_a$ and $\nu_s$ such that:

$\nu = \nu_a + \nu_s$
$\nu_a \ll \mu$ (absolutely continuous part)
$\nu_s \perp \mu$ (singular part)

Proof of Theorem 8.2:

Proof:

The proof uses the Radon-Nikodym theorem. First, assume $\nu \ll \mu$ and apply Radon-Nikodym to get the absolutely continuous part.

For the general case, restrict to the case where $\nu$ is finite, then use a limiting argument. The singular part is obtained by subtracting the absolutely continuous part. ∎

∎

Theorem 8.3: Radon-Nikodym Theorem

Let $\nu$ be a σ-finite signed measure and $\mu$ be a σ-finite positive measure such that $\nu \ll \mu$ .

Then there exists a unique (up to a.e. equality) measurable function $f$ such that:

\nu(E) = \int_E f \, d\mu

for all measurable $E$ . The function $f$ is called the Radon-Nikodym derivative and is denoted $\frac{d\nu}{d\mu}$ .

Proof of Theorem 8.3:

Proof:

The proof uses the method of 'differentiation' of measures. For finite measures, consider the set function $\Phi(E) = \nu(E) - \alpha \mu(E)$ and find the right $\alpha$ to construct the derivative.

A more elegant approach uses the Riesz representation theorem on $L^2(\mu)$ or constructs the derivative as a limit of ratios of measures. ∎

∎

Example 8.1: Computing a Radon-Nikodym Derivative

Problem: Let $\nu(E) = \int_E x^2 \, dm(x)$ on $[0,1]$ . Find $\frac{d\nu}{dm}$ .

Solution:

Since $\nu(E) = \int_E x^2 \, dm(x)$ , by definition, $\frac{d\nu}{dm}(x) = x^2$ .

This is the density function that represents $\nu$ with respect to Lebesgue measure.

Corollary 8.1: Chain Rule for Radon-Nikodym Derivatives

If $\nu \ll \mu$ and $\mu \ll \lambda$ , then $\nu \ll \lambda$ and:

\frac{d\nu}{d\lambda} = \frac{d\nu}{d\mu} \cdot \frac{d\mu}{d\lambda}

Theorem 8.4: Hahn Decomposition Construction

Every signed measure $\nu$ on a measurable space $(X, \mathcal{M})$ has a Hahn decomposition: there exist disjoint measurable sets $P$ and $N$ such that $X = P \cup N$ , $\nu(E) \geq 0$ for all $E \subseteq P$ , and $\nu(E) \leq 0$ for all $E \subseteq N$ .

The decomposition is unique up to null sets.

Proof of Theorem 8.4:

Proof:

Define $\alpha = \sup\{\nu(E) : E \in \mathcal{M}\}$ . Since $\nu$ is σ-finite, $\alpha < \infty$ .

Choose a sequence $\{E_n\}$ such that $\nu(E_n) \to \alpha$ . Let $P = \bigcup_n E_n$ and $N = X \setminus P$ .

We claim that $P$ is positive and $N$ is negative. If not, there would be a subset of $P$ with negative measure or a subset of $N$ with positive measure, contradicting the maximality of $\alpha$ . ∎

∎

Theorem 8.5: Computing Radon-Nikodym Derivatives

If $\nu \ll \mu$ and both measures are σ-finite, then the Radon-Nikodym derivative can be computed as a limit:

\frac{d\nu}{d\mu}(x) = \lim_{r \to 0} \frac{\nu(B(x,r))}{\mu(B(x,r))}

for $\\mu$ -almost every $x$ , where the limit is taken over balls (or cubes) centered at $x$ .

Proof of Theorem 8.5:

Proof:

This follows from the Lebesgue differentiation theorem applied to the function $f = \frac{d\nu}{d\mu}$ .

For almost every $x$ , we have

\lim_{r \to 0} \frac{1}{\mu(B(x,r))} \int_{B(x,r)} f \, d\mu = f(x)

But $\int_{B(x,r)} f \, d\mu = \nu(B(x,r))$ , so the result follows. ∎

∎

Example 8.4: Constructing Lebesgue Decomposition

Problem: Decompose the measure $\nu = \delta_0 + m$ on $\mathbb{R}$ (Dirac measure at 0 plus Lebesgue measure) into absolutely continuous and singular parts with respect to Lebesgue measure.

Solution:

The absolutely continuous part is $\nu_{ac} = m$ (Lebesgue measure itself).

The singular part is $\nu_s = \delta_0$ (Dirac measure at 0), since $\delta_0(\{0\}) = 1$ but $m(\{0\}) = 0$ .

We have $\nu = \nu_{ac} + \nu_s$ with $\nu_{ac} \ll m$ and $\nu_s \perp m$ .

The Radon-Nikodym derivative is $\frac{d\nu_{ac}}{dm} = 1$ (the constant function 1).

Example 8.5: Computing Radon-Nikodym Derivative

Problem: Let $\nu(E) = \int_E e^{-x^2} \, dm(x)$ on $\mathbb{R}$ . Compute $\frac{d\nu}{dm}$ .

Solution:

By definition, $\nu(E) = \int_E e^{-x^2} \, dm(x)$ , so

\frac{d\nu}{dm}(x) = e^{-x^2}

This is the density function of $\nu$ with respect to Lebesgue measure. It represents a Gaussian measure (up to normalization).

Example 8.6: Application in Probability Theory

Problem: In probability theory, if $X$ is a random variable with distribution $P_X$ and density $f$ with respect to Lebesgue measure, what is the relationship to Radon-Nikodym derivatives?

Solution:

The probability measure $P_X$ is absolutely continuous with respect to Lebesgue measure, and the probability density function $f$ is exactly the Radon-Nikodym derivative:

f = \frac{dP_X}{dm}

This means $P_X(E) = \int_E f(x) \, dx$ for all Borel sets $E$ .

The Radon-Nikodym theorem provides the rigorous foundation for probability density functions in measure-theoretic probability.

Example 8.7: Singular Measure Example

Problem: Give an example of a measure that is singular with respect to Lebesgue measure.

Solution:

The Dirac measure $\delta_a$ at a point $a$ is singular with respect to Lebesgue measure, since $\delta_a(\{a\}) = 1$ but $m(\{a\}) = 0$ .

More generally, any measure concentrated on a set of Lebesgue measure zero is singular. For example, the measure that assigns to each set the number of rational points it contains (counting measure restricted to rationals) is singular.

Another example: the measure associated with the Cantor function (the Stieltjes measure) is singular with respect to Lebesgue measure, as it is concentrated on the Cantor set, which has measure zero.

Corollary 8.3: Uniqueness of Measure Decomposition

The Lebesgue decomposition of a measure is unique: if $\nu = \nu_{ac} + \nu_s = \nu_{ac}' + \nu_s'$ are two decompositions with $\nu_{ac}, \nu_{ac}' \ll \mu$ and $\nu_s, \nu_s' \perp \mu$ , then $\nu_{ac} = \nu_{ac}'$ and $\nu_s = \nu_s'$ .

This follows from the uniqueness of the Radon-Nikodym derivative and the fact that absolutely continuous and singular parts are mutually singular.

Remark 8.3: Historical Context: Radon-Nikodym Theorem

The Radon-Nikodym theorem is a fundamental result in measure theory:

Johann Radon (1913): Proved the theorem for measures on $\mathbb{R}^n$ .
Otton Nikodym (1930): Extended the result to general measure spaces, giving the theorem its name.
Importance: The theorem provides the rigorous foundation for "densities" and "derivatives" of measures, generalizing the concept of probability density functions.

The theorem is essential in:

Probability theory (conditional expectations, martingales)
Functional analysis (Riesz representation theorem)
Differential geometry (volume forms, densities)
Mathematical physics (change of variables, coordinate transformations)

Remark 8.4: Applications in Probability and Statistics

The Radon-Nikodym theorem is central to modern probability theory:

Conditional expectations: The conditional expectation $E[X|\mathcal{F}]$ is defined using Radon-Nikodym derivatives.
Likelihood ratios: In statistics, the likelihood ratio between two probability measures is a Radon-Nikodym derivative.
Girsanov theorem: In stochastic calculus, this theorem uses Radon-Nikodym derivatives to change probability measures.
Absolute continuity of measures: Two probability measures are equivalent if and only if they are mutually absolutely continuous, which is checked using Radon-Nikodym derivatives.
Bayesian inference: Updating prior distributions to posterior distributions involves Radon-Nikodym derivatives.

These applications demonstrate why the Radon-Nikodym theorem is considered one of the most important results in measure theory.

Example 8.8: Computing Radon-Nikodym Derivative from Definition

Problem: Let $\nu(E) = \int_E \sin(x) \, dm(x)$ on $[0, \\pi]$ . Compute $\frac{d\nu}{dm}$ .

Solution:

By definition, $\nu(E) = \int_E \sin(x) \, dm(x)$ , so

\frac{d\nu}{dm}(x) = \sin(x)

This is the density function representing $\nu$ with respect to Lebesgue measure.

Example 8.9: Lebesgue Decomposition: Mixed Measure

Problem: Decompose the measure $\nu = 2m + \delta_1$ on $[0,2]$ (twice Lebesgue measure plus Dirac at 1) into absolutely continuous and singular parts with respect to Lebesgue measure.

Solution:

The absolutely continuous part is $\nu_{ac} = 2m$ , with Radon-Nikodym derivative $\frac{d\nu_{ac}}{dm} = 2$ .

The singular part is $\nu_s = \delta_1$ , since $\delta_1(\{1\}) = 1$ but $m(\{1\}) = 0$ .

We have $\nu = \nu_{ac} + \nu_s$ with $\nu_{ac} \ll m$ and $\nu_s \perp m$ .

Theorem 8.6: Uniqueness of Radon-Nikodym Derivative

If $\nu \ll \mu$ and $f$ and $g$ are both Radon-Nikodym derivatives of $\nu$ with respect to $\mu$ , then $f = g$ $\mu$ -almost everywhere.

Proof of Theorem 8.6:

Proof:

If $\nu(E) = \int_E f \, d\mu = \int_E g \, d\mu$ for all measurable $E$ , then $\int_E (f - g) \, d\mu = 0$ for all $E$ .

Taking $E = \{f > g\}$ and $E = \{f < g\}$ , we conclude that $f = g$ $\mu$ -almost everywhere. ∎

∎

Corollary 8.4: Radon-Nikodym Derivative of Sums

If $\nu_1, \nu_2 \ll \mu$ , then $\nu_1 + \nu_2 \ll \mu$ and:

\frac{d(\nu_1 + \nu_2)}{d\mu} = \frac{d\nu_1}{d\mu} + \frac{d\nu_2}{d\mu}

This follows from the linearity of the integral and the uniqueness of Radon-Nikodym derivatives.

Remark 8.5: Radon-Nikodym in Banach Spaces

The Radon-Nikodym theorem has been extended to vector-valued measures:

Vector measures: Measures taking values in Banach spaces. The Radon-Nikodym property characterizes Banach spaces where a form of the theorem holds.
Applications: Important in the study of operator-valued measures, stochastic processes, and functional analysis.
Counterexamples: Not all Banach spaces have the Radon-Nikodym property. For example, $c_0$ (the space of sequences converging to 0) does not have this property.

These extensions show the depth and breadth of the Radon-Nikodym theorem in modern mathematics.

Example 8.10: Radon-Nikodym Derivative: Chain Rule Application

Problem: Let $\nu_1(E) = \int_E x \, dm(x)$ and $\nu_2(E) = \int_E x^2 \, dm(x)$ on $[0,1]$ . Compute $\frac{d\nu_2}{d\nu_1}$ .

Solution:

We have $\frac{d\nu_1}{dm} = x$ and $\frac{d\nu_2}{dm} = x^2$ .

By the chain rule (Corollary 8.1),

\frac{d\nu_2}{d\nu_1} = \frac{d\nu_2}{dm} \cdot \frac{dm}{d\nu_1} = x^2 \cdot \frac{1}{x} = x

where we used that $\frac{dm}{d\nu_1} = \frac{1}{x}$ (the reciprocal of the Radon-Nikodym derivative, where it's nonzero).

Example 8.11: Lebesgue Decomposition: Continuous Singular Measure

Problem: Give an example of a measure that is singular with respect to Lebesgue measure but is continuous (no atoms).

Solution:

The measure associated with the Cantor function (the Stieltjes measure $\mu_C$ ) is singular and continuous.

It is singular because it is concentrated on the Cantor set, which has Lebesgue measure zero.

It is continuous because the Cantor function is continuous, so $\mu_C(\{x\}) = 0$ for all $x$ (no point masses).

This shows that singular measures can be very "spread out" even though they're concentrated on a null set.

Example 8.12: Radon-Nikodym: Computing from Limits

Problem: Use Theorem 8.5 to compute the Radon-Nikodym derivative of $\nu(E) = \int_E e^{-x} \, dm(x)$ with respect to Lebesgue measure at $x = 1$ .

Solution:

By Theorem 8.5,

\frac{d\nu}{dm}(1) = \lim_{r \to 0} \frac{\nu(B(1,r))}{m(B(1,r))} = \lim_{r \to 0} \frac{\int_{1-r}^{1+r} e^{-x} \, dx}{2r}

By the fundamental theorem of calculus, this equals $e^{-1}$ , which matches the definition $\frac{d\nu}{dm}(x) = e^{-x}$ .

Lemma 8.1: Absolute Continuity is Transitive

If $\nu \ll \mu$ and $\mu \ll \lambda$ , then $\nu \ll \lambda$ .

This follows from the fact that if $\lambda(E) = 0$ , then $\mu(E) = 0$ (since $\mu \ll \lambda$ ), and then $\nu(E) = 0$ (since $\nu \ll \mu$ ).

Theorem 8.7: Lebesgue Decomposition: Existence and Uniqueness

Every σ-finite signed measure $\nu$ on a measurable space can be uniquely decomposed as $\nu = \nu_{ac} + \nu_s$ where:

$\nu_{ac} \ll \mu$ (absolutely continuous part)
$\nu_s \perp \mu$ (singular part)

The decomposition is unique up to null sets.

Proof of Theorem 8.7:

Proof:

Apply the Radon-Nikodym theorem to the absolutely continuous part of $\nu$ with respect to $\mu$ .

The singular part is then $\nu_s = \nu - \nu_{ac}$ . Uniqueness follows from the uniqueness of Radon-Nikodym derivatives. ∎

∎

Corollary 8.5: Decomposition of Total Variation

If $\nu$ is a signed measure with Lebesgue decomposition $\nu = \nu_{ac} + \nu_s$ , then the total variation measure $|\nu|$ satisfies:

|\nu| = |\nu_{ac}| + |\nu_s|

This follows from the fact that the absolutely continuous and singular parts are mutually singular.

Remark 8.6: Singular Measures and Atoms

Singular measures can be further classified:

Atomic measures: Concentrated on a countable set (like Dirac measures). These are the "simplest" singular measures.
Continuous singular measures: Singular but have no atoms. The Cantor measure is an example.
Pure point measures: Sums of countably many Dirac measures. These are atomic and singular.

Every measure can be decomposed into three parts: absolutely continuous, atomic singular, and continuous singular. This is the Lebesgue decomposition refined further.

Remark 8.7: Radon-Nikodym in Functional Analysis

The Radon-Nikodym theorem has deep connections to functional analysis:

Riesz representation theorem: The Radon-Nikodym theorem is used to prove the Riesz representation theorem for the dual of C(X).
Operator theory: The theorem is used to study operator-valued measures and spectral theory.
Banach space geometry: The Radon-Nikodym property characterizes certain Banach spaces and is related to differentiability of norms.
Martingale theory: In probability, the Radon-Nikodym theorem is fundamental for conditional expectations and martingales.

These connections show that the Radon-Nikodym theorem is not just a measure-theoretic result, but a fundamental tool across mathematics.

Remark 8.1: Key Insights

Key takeaways:

Absolute continuity of measures generalizes absolute continuity of functions
The Radon-Nikodym derivative is the 'density' of one measure with respect to another
Lebesgue decomposition splits any measure into absolutely continuous and singular parts
These theorems are fundamental in probability theory and functional analysis
The Radon-Nikodym derivative satisfies a chain rule analogous to calculus

Practice Quiz

Lebesgue Decomposition & Radon–Nikodym

Questions

Correct

Accuracy

A signed measure

\nu

is absolutely continuous with respect to

\mu

(written

\nu \ll \mu

) if:

Easy

Not attempted

Two measures

\nu

and

\mu

are mutually singular (written

\nu \perp \mu

) if:

Medium

Not attempted

The Lebesgue decomposition theorem states that any

\sigma

-finite signed measure

\nu

can be written as:

Hard

Not attempted

The Radon-Nikodym theorem states that if

\nu \ll \mu

and both are

\sigma

-finite, then:

Hard

Not attempted

The Radon-Nikodym derivative

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

is:

Medium

Not attempted

\nu \ll \mu

and

\mu \ll \lambda

, then:

Hard

Not attempted

A signed measure

\nu

can be decomposed into:

Medium

Not attempted

The Hahn decomposition theorem states that:

Hard

Not attempted

\nu \ll \mu

and

f

is the Radon-Nikodym derivative, then for any measurable function

g

Hard

Not attempted

The total variation of a signed measure

\nu

is:

Medium

Not attempted

Frequently Asked Questions

What is the relationship between absolute continuity of measures and absolute continuity of functions?

A measure $\nu$ is absolutely continuous with respect to $\mu$ if $\nu \ll \mu$. This is analogous to how an absolutely continuous function is an indefinite integral. The Radon-Nikodym theorem makes this precise: $\nu \ll \mu$ if and only if $\nu(E) = \int_E f \, d\mu$ for some density $f$.

What is the Radon-Nikodym derivative?

The Radon-Nikodym derivative $\frac{d\nu}{d\mu}$ is the density function $f$ such that $\nu(E) = \int_E f \, d\mu$. It's analogous to the derivative in calculus, but for measures.

Why is the Lebesgue decomposition important?

The Lebesgue decomposition allows us to split any measure into two parts: one that's 'smooth' with respect to a reference measure (absolutely continuous) and one that's 'concentrated' on a null set (singular). This is fundamental in analysis and probability.

What does it mean for measures to be singular?

Two measures are mutually singular if they are 'supported' on disjoint sets. For example, a discrete measure (concentrated on points) and Lebesgue measure are singular, as Lebesgue measure gives zero to any countable set.

How do I compute a Radon-Nikodym derivative?

In practice, if $\nu$ is given by $\nu(E) = \int_E g \, d\mu$ for some function $g$, then $\frac{d\nu}{d\mu} = g$. More generally, one uses the construction in the proof of the Radon-Nikodym theorem.

What is a signed measure?

A signed measure is like a measure but can take negative values. It must still be countably additive, but $\nu(\emptyset) = 0$ and $\nu$ can be $+\infty$ or $-\infty$ (but not both).

What is the Hahn decomposition?

The Hahn decomposition provides a partition of the space into a positive set (where the signed measure is nonnegative) and a negative set (where it's nonpositive). This is used to define the positive and negative parts of a signed measure.