## Introduction

In measure theory, we often want to compare two different ways of assigning "size" to [sets](/page/Set), represented by measures $\mu$ and $\nu$. A natural question arises: if one measure $\nu$ is "controlled" by another measure $\mu$, can we express $\nu$ in terms of $\mu$? The notion of control is captured by **absolute [continuity](/page/Continuity)**, denoted $\nu \ll \mu$, which means that any set considered negligible by $\mu$ (i.e., having $\mu$-measure zero) must also be considered negligible by $\nu$.

For example, if $\mu$ is the Lebesgue measure on $\mathbb{R}$ and $\nu(A) = \int_A f(x) \, dx$ for some non-negative function $f$, then if $\mu(A)=0$, it follows that $\nu(A)=0$. So, $\nu \ll \mu$. The Radon-Nikodym theorem addresses the converse: if we know that $\nu \ll \mu$, can we always find such a "density function" $f$? The theorem answers this affirmatively, providing a powerful tool for relating different measures. It is the measure-theoretic analogue of the [Fundamental Theorem of Calculus](/theorems/632), which relates a function to its derivative through integration.

## Formal Definition and Key Results

We begin with the formal definition of absolute continuity.

[definition] Let $\mu$ and $\nu$ be two $\sigma$-finite Borel measures on $\mathbb{R}^n$. We say that $\nu$ is **absolutely continuous** with respect to $\mu$, written $\nu \ll \mu$, if for every Borel set $A \subseteq \mathbb{R}^n$, the condition $\mu(A) = 0$ implies that $\nu(A) = 0$. [/definition]

The central result connecting this property to integration is the Radon-Nikodym theorem.

[theorem] **(Radon-Nikodym Theorem)** Let $\mu$ and $\nu$ be $\sigma$-finite Borel measures on $\mathbb{R}^n$. If $\nu \ll \mu$, then there exists a non-negative Borel measurable function $f: \mathbb{R}^n \to [0, \infty)$ such that \begin{align*} \nu(A) = \int_A f \, d\mu \end{align*} for every Borel set $A \subseteq \mathbb{R}^n$. The function $f$ is unique up to equality on a set of $\mu$-measure zero. That is, if $g$ is another function with the same property, then $\mu(\{x \in \mathbb{R}^n : f(x) \neq g(x)\}) = 0$. This function $f$ is called the **Radon-Nikodym derivative** of $\nu$ with respect to $\mu$ and is denoted by $\frac{d\nu}{d\mu}$. [/theorem]

The significance of this theorem is that it establishes a precise link between the abstract property of absolute continuity and the concrete existence of a density function. It allows us to "re-weight" the measure $\mu$ using the function $f$ to obtain the measure $\nu$. This is fundamental for changing variables in integration, defining conditional expectations in probability, and understanding the structure of [derivatives](/page/Derivative) in the context of [distributions](/page/Distribution) and weak solutions to PDEs. It essentially guarantees that any measure controlled by another can be expressed as an integral against it.

We now prove the uniqueness part of the theorem.

[proof] **Proof of Uniqueness** Let $f$ and $g$ be two non-negative, measurable [functions](/page/Function) such that for any Borel set $A \subseteq \mathbb{R}^n$, we have \begin{align*} \nu(A) = \int_A f \, d\mu \quad \text{and} \quad \nu(A) = \int_A g \, d\mu. \end{align*} Our goal is to show that $f = g$ holds $\mu$-almost everywhere. **Step 1 (Define the set where f > g)** Let's define the set $E^+ = \{ x \in \mathbb{R}^n : f(x) > g(x) \}$. Our first goal is to show that $\mu(E^+) = 0$. **Step 2 (Use the integral property on E+)** By our assumption, the integral representations for $\nu$ must hold for the specific set $E^+$. Therefore, \begin{align*} \nu(E^+) = \int_{E^+} f \, d\mu \quad \text{and} \quad \nu(E^+) = \int_{E^+} g \, d\mu. \end{align*} Since both expressions equal $\nu(E^+)$, they must be equal to each other: \begin{align*} \int_{E^+} f \, d\mu = \int_{E^+} g \, d\mu. \end{align*} **Step 3 (Show the integral of the difference is zero)** By the linearity of the integral, we can rearrange the equation from Step 2. Since $f$ and $g$ are non-negative, their integrals over $E^+$ are finite if $\nu(E^+) < \infty$. If $\nu(E^+) = \infty$, we use a standard argument based on $\sigma$-finiteness, restricting to subsets where the measure is finite. Let's assume for now that the integrals are finite. \begin{align*} \int_{E^+} f \, d\mu - \int_{E^+} g \, d\mu &= 0 \\ \int_{E^+} (f - g) \, d\mu &= 0. \end{align*} **Step 4 (Deduce that μ(E+) = 0)** Let $h(x) = f(x) - g(x)$. By the definition of the set $E^+$, we have $h(x) > 0$ for all $x \in E^+$. The integral of a strictly positive function over a set can only be zero if the set itself has measure zero. To be formal, let $E^+_k = \{x \in E^+ : f(x) - g(x) > 1/k\}$ for $k \in \mathbb{N}$. Then $E^+ = \bigcup_{k=1}^\infty E^+_k$. We have \begin{align*} 0 = \int_{E^+} (f-g) \, d\mu \ge \int_{E^+_k} (f-g) \, d\mu > \int_{E^+_k} \frac{1}{k} \, d\mu = \frac{1}{k} \mu(E^+_k). \end{align*} This implies $\frac{1}{k} \mu(E^+_k) \le 0$, and since $\mu$ is a non-negative measure, we must have $\mu(E^+_k) = 0$ for all $k \ge 1$. By the countable sub-additivity of measures, \begin{align*} \mu(E^+) = \mu\left(\bigcup_{k=1}^\infty E^+_k\right) \le \sum_{k=1}^\infty \mu(E^+_k) = \sum_{k=1}^\infty 0 = 0. \end{align*} Thus, $\mu(E^+) = 0$. **Step 5 (Repeat for the set where g > f)** An identical argument applies to the set $E^- = \{ x \in \mathbb{R}^n : g(x) > f(x) \}$. Following the same steps, we conclude that $\mu(E^-) = 0$. **Step 6 (Conclude f = g μ-almost everywhere)** The set where $f$ and $g$ are not equal is the union of the sets where they differ: $\{x : f(x) \neq g(x)\} = E^+ \cup E^-$. Using the additivity of the measure $\mu$, \begin{align*} \mu(\{x : f(x) \neq g(x)\}) = \mu(E^+ \cup E^-) \le \mu(E^+) + \mu(E^-) = 0 + 0 = 0. \end{align*} Therefore, the set of points where $f$ and $g$ differ has $\mu$-measure zero, which means $f = g$ $\mu$-almost everywhere. [/proof]

## Applications

The Radon-Nikodym theorem is a foundational result with wide-ranging applications in analysis and probability. Its primary function is to provide a density for any measure that is absolutely continuous with respect to another, which is a key theoretical and practical tool.

### Probability Theory: Change of Measure

In probability, measures are used to assign probabilities to events. Let $(\Omega, \mathcal{F}, P)$ be a probability space. Sometimes it is useful to work with another probability measure $Q$ on the same space. If $Q \ll P$, the Radon-Nikodym theorem guarantees the existence of a random variable $X = \frac{dQ}{dP}$ such that for any event $A \in \mathcal{F}$,

\begin{align*} Q(A) = \int_A X \, dP = E_P[X \mathbf{1}_A], \end{align*}

where $E_P$ denotes expectation with respect to $P$ and $\mathbf{1}_A$ is the indicator function of $A$. This allows us to calculate probabilities and expectations under $Q$ by using the measure $P$. This technique, known as a **change of measure**, is central to mathematical finance for pricing derivatives (via Girsanov's theorem) and in statistical inference.

### Functional Analysis: Dual of $L^p$ spaces

The Radon-Nikodym theorem is an essential ingredient in the proof of the [Riesz Representation Theorem](/theorems/221) for $L^p$ spaces. This theorem identifies the continuous dual space of $L^p(\mu)$ for $1 \le p < \infty$.

[theorem] **(Riesz Representation for $L^p$)** Let $(\Omega, \mathcal{F}, \mu)$ be a $\sigma$-finite measure space and let $1 \le p < \infty$. Let $q$ be the conjugate exponent such that $\frac{1}{p} + \frac{1}{q} = 1$. Then for every continuous linear functional $\phi \in (L^p(\mu))^*$, there exists a unique function $g \in L^q(\mu)$ such that \begin{align*} \phi(f) = \int_\Omega fg \, d\mu \end{align*} for all $f \in L^p(\mu)$. Moreover, $\|\phi\|_{(L^p)^*} = \|g\|_{L^q}$. [/theorem]