Radon-Nikodym Derivative and Lebesgue Decomposition

A measure $\nu$ is absolutely continuous with respect to $\mu$ (written $\nu \ll \mu$) if $\mu(A) = 0$ implies $\nu(A) = 0$.

Measures $\mu$ and $\nu$ are mutually singular (written $\mu \perp \nu$) if there exists $A \in \mathcal{F}$ such that $\mu(A) = 0$ and $\nu(A^c) = 0$.

Proposition 9.1 (Basic Properties). Let $\mu, \nu, \lambda$ be measures on $(X, \mathcal{F})$.

• If $\nu \ll \mu$ and $\mu \ll \lambda$, then $\nu \ll \lambda$.
• If $\nu \ll \mu$ and $\nu \perp \mu$, then $\nu = 0$.
• If $\nu \ll \mu$ and $\mu$ is $\sigma$-finite, then $\nu$ is $\sigma$-finite.

Theorem 9.2 (Radon-Nikodym Theorem). Let $(X, \mathcal{F}, \mu)$ be a $\sigma$-finite measure space and $\nu$ a $\sigma$-finite signed measure with $\nu \ll \mu$. Then there exists a unique (a.e.) measurable function $f : X \to \mathbb{R}$ such that

This function $f$ is denoted $d\nu/d\mu$ and called the Radon-Nikodym derivative of $\nu$ with respect to $\mu$.

Proof. The proof maximises a density over the class it must fit into, then uses absolute continuity to kill the leftover. Its structure; maximise, then show the leftover is both singular and absolutely continuous, hence zero; is von Neumann’s, and the same shape proves the Lebesgue decomposition below.

Step 1: the class and its supremum. Let $\mathcal{C}$ be the set of measurable $h \geq 0$ with

$\mathcal{C}$ is non-empty ($h = 0$) and, $\nu$ being finite, bounded above by $\nu(X)$ in integral. Let

and choose $h_n \in \mathcal{C}$ with $\int h_n\,d\mu \to \alpha$. Set $f = \sup_n h_n$ (pointwise supremum). Claim: $f \in \mathcal{C}$ with $\int f\,d\mu = \alpha$. For each $n$, $\max(h_1, \dots, h_n) \in \mathcal{C}$: the set $A$ partitions into the pieces where each $h_k$ attains the maximum, and on the piece where $h_k$ attains it, $\int \max \le \nu$ follows from $h_k \in \mathcal{C}$. The $f_n$ then increase pointwise to $f$, so by monotone convergence $\int_A f\,d\mu = \lim_n \int_A f_n\,d\mu \leq \nu(A)$; $f \in \mathcal{C}$; and $\int f\,d\mu = \alpha$ by continuity from below.

Step 2: the leftover measure. Define

Since $f \in \mathcal{C}$, $\nu_0$ is a positive measure: countable additivity is $\nu(A) - \int_A f$ against $\nu(A) - \int_A f\mu$, both measures. Claim: $\nu_0 \perp \mu$. For each $n$, let $P_n \sqcup N_n$ be a Hahn decomposition of the signed measure $\nu_0 - \frac{1}{n}\mu$; its positive set $P_n$ is where the leftover exceeds $\frac{1}{n}\mu$.

If $\mu(P_n) > 0$ for some $n$: then $f + \frac{1}{n}\mathbf{1}_{P_n} \in \mathcal{C}$, because on $P_n$ the insertion adds exactly $\frac{1}{n}\mu$, which $\nu_0$ affords there by definition of the Hahn-positive set:

and off $P_n$ the insertion only adds where $\nu_0 \le \frac1n\mu$, so the inequality $\int \le \nu$ persists on all of $A$. But then

contradicting the supremum. Hence $\mu(P_n) = 0$ for every $n$.

If $\nu_0$ is not the zero measure, some such $P_n$ has positive $\nu_0$-mass: let $N = \bigcup_n P_n$, so $\mu(N) = 0$. On $N^c = \bigcap_n N_n$, the definition of the Hahn-negative sets gives $\nu_0(E) \leq \frac{1}{n}\mu(E)$ for every $E \subseteq N^c$ and every $n$; letting $n \to \infty$, $\nu_0(E) = 0$ for all $E \subseteq N^c$. So $\nu_0$ is concentrated on $N$: $\nu_0 \perp \mu$. $\blacksquare$-for-the-claim

Step 3: absolute continuity kills the leftover. Since $f\mu \leq \nu$ pointwise on the class $\mathcal{C}$, $\nu_0 \leq \nu$; and $\nu \ll \mu$ by hypothesis, so $\nu_0 \ll \mu$. A measure both singular to and absolutely continuous with $\mu$ is zero: singularity concentrates $\nu_0$ on a set $N$ with $\mu(N) = 0$, so $\nu_0(N^c) = 0$; absolute continuity $\nu_0 \ll \mu$ then gives $\nu_0(N) \leq \nu_0(\text{anything})$… directly: $\mu(N) = 0$ and $\nu_0 \ll \mu$ force $\nu_0(N) = 0$. Both pieces vanish, so $\nu_0 = 0$, and $\nu(A) = \int_A f\,d\mu$ for every $A$: $f$ represents $\nu$. $\blacksquare$

Step 4: $\sigma$-finite $\nu$. Decompose $\nu$ into finite pieces by the $\sigma$-finiteness of $\mu$, apply Steps 1–3 on each piece, and paste; the pieces’ densities agree on overlaps by the uniqueness below, which is where a.e.-uniqueness is consumed. $\blacksquare$

Step 2: uniqueness. If $f$ and $g$ both represent $\nu$, then $\int_A (f - g)\,d\mu = 0$ for every $A$; taking $A = \{f > g\}$ and $A = \{f < g\}$ in turn forces $f = g$ $\mu$-a.e. $\blacksquare$

Step 3: $\sigma$-finite $\nu$. Decompose $\nu$ into finite pieces by the $\sigma$-finiteness of $\mu$, apply Step 1 on each piece, and paste; the pieces’ densities agree on overlaps by uniqueness, which is where the a.e.-uniqueness of Step 2 is consumed. $\blacksquare$

The proof’s own structure explains why the theorem needs $\sigma$-finiteness twice: once to slice $B$ for the squeeze, once to paste the finite pieces. On non-$\sigma$-finite spaces the theorem genuinely fails, and the failure is the reason the Radon–Nikodym property below is a property of the space rather than a universal truth.

Proof sketch. For the finite case, consider the set of functions $g$ with $\int_A g\, d\mu \leq \nu(A)$ for all $A$. Let $\alpha = \sup\{\int_X g\, d\mu\}$ and choose a maximizing sequence $g_n$. The pointwise supremum $f = \sup_n g_n$ gives the desired derivative. Extend to $\sigma$-finite case by partitioning $X$ into sets of finite measure.

Proposition 9.3 (Linearity). If $\nu_1, \nu_2 \ll \mu$ and $a, b \in \mathbb{R}$, then:

Proposition 9.4 (Chain Rule). If $\lambda \ll \nu$ and $\nu \ll \mu$, then $\lambda \ll \mu$ and:

Proposition 9.5 (Change of Variables). If $\nu \ll \mu$ and $f$ is $\nu$-integrable, then:

Example 9.1. If $\nu$ is absolutely continuous with respect to Lebesgue measure $m$ on $\mathbb{R}$, then $d\nu/dm$ is the Radon-Nikodym derivative. For a probability distribution with density $p(x)$, we have $\nu(A) = \int_A p(x)\, dx$, so $d\nu/dm = p$.

Example 9.2. The Dirac measure $\delta_0$ is not absolutely continuous with respect to Lebesgue measure: $\delta_0 \ll m$ would require $\delta_0(\{0\}) = 1 = \int_{\{0\}} f\, dm = 0$, a contradiction. In fact, $\delta_0 \perp m$ (take $A = \{0\}$, then $m(A) = 0$, $\delta_0(A^c) = 0$).

Theorem 9.6 (Lebesgue Decomposition). Let $\mu$ and $\nu$ be $\sigma$-finite measures on $(X, \mathcal{F})$. Then there exist unique measures $\nu_a$ and $\nu_s$ such that:

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

Proof sketch. Let $\lambda = \mu + \nu$. Apply Radon-Nikodym to $\nu \ll \lambda$ to get $f = d\nu/d\lambda$. Then set $\nu_a(A) = \int_A f\, d\mu$ and $\nu_s = \nu - \nu_a$. Show that $\nu_s$ is singular with respect to $\mu$ by considering the set where $f = 0$ or $f \geq 1$ and using the properties of the derivative.

Example 9.3. The Cantor function $F : [0, 1] \to [0, 1]$ is continuous, monotonically increasing, and has $F(0) = 0$, $F(1) = 1$. The associated measure $\mu_F$ (the Cantor measure or “Devil’s staircase” measure) is singular with respect to Lebesgue measure: $\mu_F \perp m$. By Lebesgue decomposition, $\mu_F = \mu_F + 0$ with $\mu_F \perp m$.

Example 9.4 (Absolutely Continuous Part of a Measure). Let $\nu$ be a measure on $\mathbb{R}$ defined by $\nu(A) = m(A \cap [0,1]) + \sum_{q \in \mathbb{Q} \cap [0,1]} 2^{-q} \delta_q(A)$. Then the Lebesgue decomposition of $\nu$ with respect to $m$ is: $\nu_a(A) = m(A \cap [0,1])$, $\nu_s(A) = \sum_{q \in \mathbb{Q} \cap [0,1]} 2^{-q} \delta_q(A)$.

Example 9.5 (Conditional Expectation). In probability theory, the conditional expectation $\mathbb{E}[X | \mathcal{G}]$ can be defined via the Radon-Nikodym derivative. Given a sub-$\sigma$-algebra $\mathcal{G} \subseteq \mathcal{F}$, define $\nu(G) = \int_G X\, dP$ for $G \in \mathcal{G}$. Then $\nu \ll P|_\mathcal{G}$, and $\mathbb{E}[X | \mathcal{G}] = d\nu/d(P|_\mathcal{G})$.

Application: Differentiation of Measures. The Radon-Nikodym theorem is essential for the differentiation of measures on $\mathbb{R}^n$. The Lebesgue differentiation theorem states that for a locally integrable function $f$:

This is intimately connected with the Radon-Nikodym derivative of the measure $\nu(A) = \int_A f\, dm$.

Definition. A Banach space $X$ has the Radon-Nikodym property if for every finite measure space $(\Omega, \mathcal{F}, \mu)$ and every vector measure $\nu : \mathcal{F} \to X$ that is absolutely continuous with respect to $\mu$ and has bounded variation, there exists $f \in L^1(\mu, X)$ such that $\nu(A) = \int_A f\, d\mu$.

Proposition 9.7. Every separable dual space has the Radon-Nikodym property. In particular, $\ell^1$, $\ell^\infty$, and $L^\infty$ do not have this property.

Problem 1. Let $\mu$ be Lebesgue measure on $[0,1]$ and $\nu(A) = \int_A x^2\, d\mu$. Find $d\nu/d\mu$.

Solution. By definition, $\nu(A) = \int_A f\, d\mu$ with $f(x) = x^2$, so $d\nu/d\mu(x) = x^2$. $\blacksquare$

Problem 2. Decompose $\nu = \delta_0 + m$ (where $m$ is Lebesgue measure on $\mathbb{R}$) into absolutely continuous and singular parts with respect to $m$.

Solution. $\nu_a = m$ (since $m \ll m$) and $\nu_s = \delta_0$ (since $\delta_0 \perp m$). Indeed, $\nu = m + \delta_0$ with $m \ll m$ and $\delta_0 \perp m$. $\blacksquare$

1. Prove that if $\nu \ll \mu$ and $\mu \ll \nu$, then $d\nu/d\mu > 0$ $\mu$-a.e. and $d\mu/d\nu = (d\nu/d\mu)^{-1}$.
2. Show that the Radon-Nikodym derivative is unique up to $\mu$-null sets.
3. Find the Lebesgue decomposition of $\nu(A) = \int_A e^{-x^2}\, dx + \sum_{n=1}^\infty 2^{-n} \delta_n(A)$ with respect to Lebesgue measure.
4. Prove that if $\mu$ and $\nu$ are $\sigma$-finite and $\nu \ll \mu$, then $\int f\, d\nu = \int f (d\nu/d\mu)\, d\mu$ for all measurable $f \geq 0$.

• Lebesgue Outer Measure and Caratheodory Extension: Constructs the Lebesgue measure that serves as the reference measure for the Radon-Nikodym derivative.

• Lebesgue Measurable Sets and Non-Measurable Sets: Establishes the $\sigma$-algebra of measurable sets needed for the decomposition of measures.

• Fubini and Tonelli Theorems: Product measures and integration techniques used in applications of the Radon-Nikodym derivative.

• Quantum Mechanics

• Graph Theory

• Classical Mechanics

• Electromagnetism

• Statistical Learning

• Statistical Mechanics

The Radon-Nikodym derivative generalises the concept of a density function. If one measure is absolutely continuous with respect to another, meaning it assigns zero to every set that the reference measure does, then the Radon-Nikodym theorem guarantees the existence of a derivative function that converts one measure into the other via integration. This is analogous to the fundamental theorem of calculus: just as differentiation recovers a function from its integral, the Radon-Nikodym derivative recovers the density from a measure. The Lebesgue decomposition theorem shows that any measure can be uniquely split into an absolutely continuous part and a singular part, like decomposing a signal into a smooth component and a spike.

Mistake 1: Assuming the Radon-Nikodym derivative exists without absolute continuity The Radon-Nikodym theorem requires $\nu \ll \mu$ (absolute continuity). If $\nu$ is not absolutely continuous with respect to $\mu$, the derivative $d\nu/d\mu$ does not exist as a function. The Dirac measure $\delta_0$ is not absolutely continuous with respect to Lebesgue measure, so it has no Radon-Nikodym derivative with respect to $m$.

Mistake 2: Confusing mutual singularity with absolute continuity Absolute continuity ($\nu \ll \mu$) and mutual singularity ($\nu \perp \mu$) are not opposites. A measure can be neither absolutely continuous nor singular with respect to another — the Lebesgue decomposition theorem shows that any $\sigma$-finite measure decomposes uniquely into an absolutely continuous part and a singular part. Students sometimes think these are the only two possibilities.

Mistake 3: Forgetting that the Radon-Nikodym derivative is unique only up to $\mu$-null sets The function $f = d\nu/d\mu$ is determined $\mu$-almost everywhere, not pointwise. Two functions that differ on a $\mu$-null set both serve as Radon-Nikodym derivatives. Students sometimes treat the derivative as a pointwise-defined function, which matters when evaluating it at specific points or when composing with other functions.

• $L^p$ Spaces
• Fubini and Tonelli Theorems
• Lebesgue Integration
• Lebesgue Measurable Sets and Non-Measurable Sets