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

$\nu(A) = \int_A f\, d\mu \quad \text{for all } A \in \mathcal{F}$

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

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:

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

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

$\frac{d\lambda}{d\mu} = \frac{d\lambda}{d\nu} \cdot \frac{d\nu}{d\mu} \quad \mu\text{-a.e.}$

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

$\int f\, d\nu = \int f \frac{d\nu}{d\mu}\, d\mu$

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 $f$:

$\lim_{r \to 0} \frac{1}{m(B_r(x))} \int_{B_r(x)} f\, dm = f(x) \quad \text{a.e.}$

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 $f \geq 0$.