Radon–Nikodym Theorem — Mathematics Knowledge System

Overview

The Radon–Nikodym theorem establishes when one measure can be expressed as an integral with respect to another. If nu is absolutely continuous with respect to mu (nu << mu), then there exists a measurable function f — the Radon–Nikodym derivative — such that nu(E) = integral_E f dmu for all measurable E. This function f = dnu/dmu generalises the classical derivative and is fundamental to probability theory (conditional expectations), statistics (likelihood ratios), and functional analysis (dual spaces of L^p).

Intuition

If mu is Lebesgue measure and nu has a density (nu(E) = integral_E f dx), then nu << mu trivially holds. The Radon–Nikodym theorem says the converse: whenever nu << mu (nu cannot assign positive mass to mu-null sets), a density f exists. Think of dnu/dmu as the 'rate of change' of nu with respect to mu — the factor by which nu amplifies or shrinks mu-mass. In probability: if Q << P (two probability measures), then the likelihood ratio dQ/dP records how to reweight P-events to get Q-probabilities.

Formal Definition

Definition

Let (X, M, mu) be a sigma-finite measure space and nu a sigma-finite signed measure on (X, M). We say nu is absolutely continuous with respect to mu (nu << mu) if mu(E) = 0 implies nu(E) = 0. The Radon–Nikodym theorem states: nu << mu if and only if there exists a measurable function f: X -> R (unique mu-a.e.) such that nu(E) = integral_E f dmu for all E in M. The function f is called the Radon–Nikodym derivative of nu with respect to mu, written dnu/dmu.

\nu \ll \mu \iff [\mu(E) = 0 \Rightarrow \nu(E) = 0]

Absolute continuity

\nu(E) = \int_E f \, d\mu \quad \forall E \in \mathcal{M}

Radon–Nikodym representation

f = \frac{d\nu}{d\mu} \quad \text{(Radon–Nikodym derivative, unique } \mu\text{-a.e.)}

Radon–Nikodym derivative

Notation

Notation	Meaning
$\nu \ll \mu$	nu is absolutely continuous with respect to mu
$\dfrac{d\nu}{d\mu}$	Radon–Nikodym derivative of nu with respect to mu
$\nu \perp \mu$	nu and mu are mutually singular

Proofs

Proof via Hilbert space methods (sketch)

$\text{Assume } \mu, \nu \text{ are finite positive measures. Let } \rho = \mu + \nu.$ (rho is a finite positive measure dominating both mu and nu.)
$\text{The functional } L^2(\rho) \ni g \mapsto \int g \, d\nu \text{ is bounded.}$ (By Cauchy–Schwarz: |int g dnu| <= ||g||_{L^2(rho)} * nu(X)^{1/2} < inf.)
$\text{By Riesz representation: } \exists h \in L^2(\rho) \text{ with } \int g \, d\nu = \int g h \, d\rho \text{ for all } g.$ (Riesz representation theorem for Hilbert spaces L^2(rho).)
$0 \leq h \leq 1 \quad \rho\text{-a.e.}$ (Take g = 1_{h>1} and g = 1_{h<0} to show these sets have rho-measure zero.)
$\text{Set } f = h/(1-h) \text{ on } \{h < 1\} \text{ and } f = \infty \text{ on } \{h = 1\}. \text{ Show } \mu(\{h=1\}) = 0 \text{ when } \nu \ll \mu, \text{ and } \nu(E) = \int_E f \, d\mu.$ (Algebraic manipulation using the relation int g dnu = int gh dmu + int gh dnu.)

Theorems

Theorem 1: Radon–Nikodym Theorem

\text{Let } (X, \mathcal{M}, \mu) \text{ be sigma-finite and } \nu \text{ a sigma-finite signed measure on } (X,\mathcal{M}). \text{ Then } \nu \ll \mu \iff \exists\, f \in L^1(\mu) \text{ (if } \nu \text{ is finite) such that } \nu(E) = \int_E f \, d\mu \text{ for all } E. \text{ Moreover, } f \text{ is unique } \mu\text{-a.e.}

Theorem 2: Lebesgue Decomposition Theorem

\text{Any sigma-finite signed measure } \nu \text{ decomposes uniquely as } \nu = \nu_{ac} + \nu_s \text{ where } \nu_{ac} \ll \mu \text{ and } \nu_s \perp \mu.

Theorem 3: Chain Rule

\text{If } \lambda \ll \nu \ll \mu, \text{ then } \dfrac{d\lambda}{d\mu} = \dfrac{d\lambda}{d\nu} \cdot \dfrac{d\nu}{d\mu} \quad \mu\text{-a.e.}

Worked Examples

By inspection, nu(E) = integral_E 2x dlambda(x). This is exactly the Radon–Nikodym representation with f(x) = 2x.
$\frac{d\nu}{d\mu}(x) = 2x$
Verify: nu([a,b]) = integral_a^b 2x dx = b^2 - a^2. Check nu << mu: if lambda(E) = 0 then the integral of 2x over E is 0.

Answer: dnu/dlambda(x) = 2x.

Practice Problems

Difficulty 8/10

Prove the chain rule for Radon–Nikodym derivatives: if lambda << nu << mu, then d(lambda)/d(mu) = d(lambda)/d(nu) * d(nu)/d(mu) mu-a.e.

Difficulty 9/10

In probability theory, if P and Q are equivalent probability measures (P << Q and Q << P), how does the Radon-Nikodym derivative dP/dQ relate to dQ/dP?

Difficulty 7/10

The Radon–Nikodym theorem requires mu to be sigma-finite. Which example shows the theorem fails without this condition?

mu = Lebesgue measure, nu = counting measure on [0,1]: nu << mu fails, so the theorem is not applicablemu = counting measure on [0,1] (not sigma-finite), nu = Lebesgue measure: nu << mu holds but no L^1(mu) density existsmu = Dirac measure at 0, nu = Lebesgue measure: both sigma-finite, theorem appliesmu = nu = Lebesgue measure: trivially dnu/dmu = 1

Common Mistakes

Common Mistake

nu << mu means nu is 'smaller' than mu in some sense.

Absolute continuity nu << mu means nu cannot distinguish sets that mu cannot: if mu(E) = 0 then nu(E) = 0. It is a compatibility condition, not a size comparison.

Common Mistake

The Radon–Nikodym derivative dnu/dmu is an ordinary derivative.

It is a measurable function satisfying nu(E) = integral_E (dnu/dmu) dmu. It coincides with the classical derivative in special cases (e.g. absolutely continuous functions), but it is defined measure-theoretically, not pointwise.

Common Mistake

The theorem holds for any two measures without restriction.

The standard theorem requires mu to be sigma-finite. Without this, the theorem can fail (see the counting measure example).

Quiz

Historical Background

Johann Radon proved a version for measures on R^n in 1913. Otto Nikodym extended the result to abstract measure spaces in 1930. The theorem unified various classical results about differentiation and integration and provided the abstract foundation for modern probability theory.

1913
Radon proves the theorem for measures on R^n
Johann Radon
1930
Nikodym extends the result to abstract sigma-finite measure spaces
Otto Nikodym
1940s
Halmos and Savage use Radon-Nikodym to define conditional expectation rigorously
Paul Halmos, Leonard Savage

Summary

nu << mu (absolute continuity) means every mu-null set is also nu-null.
The Radon–Nikodym theorem: nu << mu iff there exists a unique (a.e.) measurable function f = dnu/dmu with nu(E) = integral_E f dmu.
The Lebesgue decomposition splits any nu into nu_ac << mu and nu_s perp mu uniquely.
The chain rule dλ/dmu = dλ/dnu * dnu/dmu holds when lambda << nu << mu.
Key applications: conditional expectations in probability, likelihood ratios in statistics, dual representation of L^p spaces.

References

BookFolland, G. B. (1999). Real Analysis: Modern Techniques and Their Applications (2nd ed.). Wiley. Chapter 3.
BookRudin, W. (1987). Real and Complex Analysis (3rd ed.). McGraw-Hill. Chapter 6.
WebsiteWikipedia — Radon–Nikodym theorem