Given a [measure space](/page/Measure%20Space) $(X, \mathcal{A}, \mu)$ and a second measure $\nu$ on the same $\sigma$-algebra $\mathcal{A}$, one frequently needs to express integrals with respect to $\nu$ in terms of integrals with respect to $\mu$. For instance, if $\mu = \mathcal{L}^1$ is Lebesgue measure on $\mathbb{R}$ and $\nu$ is a probability measure with a density, we write $d\nu = f \, d\mu$ for some nonnegative measurable function $f: \mathbb{R} \to [0, \infty)$. But when does such a representation exist? And when it does, in what sense is the "density" $f$ unique?

The answer is governed by a single structural relationship between measures: [absolute continuity](/page/Absolutely%20Continuous%20Measures). The Radon--Nikodym theorem asserts that absolute continuity $\nu \ll \mu$ is not only necessary but, under a $\sigma$-finiteness hypothesis, also *sufficient* for the existence of a density. This result is one of the central pillars of modern analysis. It connects abstract measure theory to concrete computation: it underpins conditional expectation in probability, duality of [Lp spaces](/page/L%5Ep%20Spaces), the change-of-variables formula, and the classical Fundamental Theorem of Calculus.

To appreciate why the theorem requires care, consider a first attempt that fails.

[example: Failure Without Absolute Continuity] Let $\mu = \mathcal{L}^1$ be Lebesgue measure on $(\mathbb{R}, \mathcal{B}(\mathbb{R}))$, and let $\nu = \delta_0$ be the Dirac measure concentrated at the origin, defined by $\delta_0(A) = \mathbb{1}_A(0)$ for each Borel set $A \in \mathcal{B}(\mathbb{R})$. Suppose, seeking a contradiction, that there exists a measurable function $f: \mathbb{R} \to [0, \infty)$ satisfying $\delta_0(A) = \int_A f \, d\mathcal{L}^1$ for every Borel set $A$. Taking $A = \{0\}$, we need $\int_{\{0\}} f \, d\mathcal{L}^1 = 1$. But $\mathcal{L}^1(\{0\}) = 0$, so $\int_{\{0\}} f \, d\mathcal{L}^1 = 0$ regardless of the choice of $f$. This is a contradiction. No density can exist. The obstruction is structural: $\delta_0$ assigns positive mass to the set $\{0\}$, which has $\mathcal{L}^1$-measure zero. In other words, $\delta_0$ is *not* absolutely continuous with respect to $\mathcal{L}^1$. The Radon--Nikodym theorem identifies absolute continuity as the precise condition that prevents this failure. [/example]

## Motivation

[motivation] ### Why density representations matter Throughout analysis and probability, one repeatedly encounters pairs of measures on the same measurable space and needs to pass between them. Three representative situations illustrate the scope of the problem. **Change of variables.** If $T: \mathbb{R}^n \to \mathbb{R}^n$ is a $C^1$ diffeomorphism and $\mu = \mathcal{L}^n$, then the pushforward measure $\nu = T_\# \mu$ satisfies $\nu(A) = \mathcal{L}^n(T^{-1}(A))$. Computing $\int f \, d\nu$ requires expressing $\nu$ in terms of $\mathcal{L}^n$, which leads to the Jacobian factor $|\det JT^{-1}|$ — a density. **Probability.** Given a random variable $X: (\Omega, \mathcal{F}, \mathbb{P}) \to (\mathbb{R}, \mathcal{B}(\mathbb{R}))$, its distribution $\mu_X = \mathbb{P} \circ X^{-1}$ is a Borel probability measure. The question "does $X$ have a density?" is precisely the question of whether $\mu_X \ll \mathcal{L}^1$. **Duality.** Identifying the dual space of $L^p(X, \mathcal{A}, \mu)$ for $1 \le p < \infty$ requires showing that every bounded linear functional on $L^p$ can be represented as integration against some $g \in L^q$. The key step is constructing $g$ as a Radon--Nikodym derivative. ### What "absolute continuity" captures The naive hope — that any two measures on the same $\sigma$-algebra admit a density — fails spectacularly, as the Dirac measure example above shows. The failure occurs whenever $\nu$ sees sets that $\mu$ considers negligible. Absolute continuity $\nu \ll \mu$ is the condition that $\nu$ is "blind to $\mu$-null sets": whenever $\mu(A) = 0$, we also have $\nu(A) = 0$. This is both necessary and (under $\sigma$-finiteness) sufficient for $d\nu = f \, d\mu$. ### Why $\sigma$-finiteness cannot be dropped The theorem fails without $\sigma$-finiteness on $\mu$. Consider the simplest possible example: $X = \{0\}$ (a single point), $\mathcal{A} = \{\varnothing, \{0\}\}$, with $\mu$ defined by $\mu(\{0\}) = \infty$ and $\nu(\{0\}) = 1$. Note that $\mu$ is *not* $\sigma$-finite: the only cover of $X$ is $\{0\}$ itself, which has infinite $\mu$-measure. Yet $\nu \ll \mu$ holds (the only $\mu$-null set is $\varnothing$, and $\nu(\varnothing) = 0$). Any candidate density $f$ would need $\int_{\{0\}} f \, d\mu = f(0) \cdot \mu(\{0\}) = f(0) \cdot \infty$. If $f(0) > 0$, this integral is $+\infty \ne 1$; if $f(0) = 0$, the integral is $0 \ne 1$. No density exists, despite absolute continuity holding. A more substantial example arises on uncountable spaces. Let $X = [0,1]$ with $\mathcal{A} = \mathcal{B}([0,1])$, let $\mu$ be the counting measure on $[0,1]$ (which assigns to each set its cardinality), and let $\nu = \mathcal{L}^1|_{[0,1]}$. Then $\nu \ll \mu$ (if $\mu(A) = 0$ then $A = \varnothing$, so $\nu(A) = 0$). But $\mu$ is not $\sigma$-finite: $[0,1]$ cannot be covered by countably many sets of finite counting measure, since each such set is finite and a countable union of finite sets is countable. Any candidate density $f$ would need $\int_A f \, d\mu = \sum_{x \in A} f(x) = \nu(A) = \mathcal{L}^1(A)$. Taking $A = \{x\}$ gives $f(x) = \mathcal{L}^1(\{x\}) = 0$ for every $x$, forcing $f \equiv 0$, which cannot reproduce $\nu([0,1]) = 1$. [/motivation]

## Definition

The central object is the Radon--Nikodym derivative, which is defined only when absolute continuity holds.

[definition: Radon-Nikodym Derivative] Let $(X, \mathcal{A})$ be a measurable space, and let $\mu$ and $\nu$ be $\sigma$-finite measures on $\mathcal{A}$ with $\nu \ll \mu$ (that is, $\nu$ is [absolutely continuous](/page/Absolutely%20Continuous%20Measures) with respect to $\mu$). The **Radon--Nikodym derivative** of $\nu$ with respect to $\mu$ is the $\mu$-a.e. unique $\mathcal{A}$-measurable function \begin{align*} \frac{d\nu}{d\mu}: X \to [0, \infty) \end{align*} satisfying \begin{align*} \nu(A) = \int_A \frac{d\nu}{d\mu} \, d\mu \quad \text{for every } A \in \mathcal{A}. \end{align*} The existence and $\mu$-a.e. uniqueness of this function is the content of the Radon--Nikodym theorem stated below. When $\nu$ is a signed measure with $\nu \ll \mu$, the Radon--Nikodym derivative takes values in $\mathbb{R}$ (or $[-\infty, \infty)$) rather than $[0, \infty)$, and is the $\mu$-a.e. unique measurable function satisfying the same integral identity. [/definition]

The notation $\frac{d\nu}{d\mu}$ is suggestive: it behaves like a "ratio" of two measures, much as the classical derivative $\frac{df}{dx}$ is a ratio of infinitesimal increments. This analogy is not merely cosmetic — the chain rule and substitution formulas for Radon--Nikodym derivatives mirror their classical counterparts, as we shall see in the section on calculus of densities.

An equivalent and frequently useful formulation is the **change-of-measure identity**: if $f := \frac{d\nu}{d\mu}$, then for every $\mathcal{A}$-measurable function $g: X \to [0, \infty]$,

\begin{align*} \int_X g \, d\nu = \int_X g \cdot f \, d\mu. \end{align*}

This identity is what makes the Radon--Nikodym derivative a computational tool, not merely an existence statement. It says that integrating against $\nu$ is the same as integrating against $\mu$ with a "reweighting factor" $f$.

## The Radon--Nikodym Theorem

The fundamental question is: does the Radon--Nikodym derivative actually exist? And if so, under what conditions? Two hypotheses are needed — absolute continuity (to prevent the Dirac measure obstruction) and $\sigma$-finiteness (to prevent the counting measure obstruction). Together, they are sufficient.

[quotetheorem:1247]

Several aspects of this theorem warrant careful discussion.

**Necessity of absolute continuity.** The hypothesis $\nu \ll \mu$ is indispensable. If $f$ is any nonnegative measurable function and $\nu(A) = \int_A f \, d\mu$, then whenever $\mu(A) = 0$, the integral $\int_A f \, d\mu = 0$, so $\nu(A) = 0$. Thus the existence of a density *implies* absolute continuity. The theorem's content is the converse: absolute continuity *implies* the existence of a density.

**Necessity of $\sigma$-finiteness.** As demonstrated in the motivation, the counting measure on $[0,1]$ provides a concrete example where $\nu \ll \mu$ holds but no density exists, because $\mu$ fails to be $\sigma$-finite. The $\sigma$-finiteness of $\nu$ can sometimes be relaxed (if $\nu \ll \mu$ and $\mu$ is $\sigma$-finite, then $\nu$ is automatically $\sigma$-finite when $\nu$ is a finite measure), but the $\sigma$-finiteness of $\mu$ is essential. There are extensions to non-$\sigma$-finite settings (e.g., using the concept of a *localizable* measure), but these require substantially different formulations.

**Uniqueness is only $\mu$-a.e.** The density $f$ is determined only up to modification on $\mu$-null sets. On any $\mu$-null set $N$, one can redefine $f$ arbitrarily without affecting any integral $\int_A f \, d\mu$. This is consistent with the general principle that $L^p$ functions are equivalence classes modulo $\mu$-null sets.