Lebesgue decomposition separates a measure into the part visible through a reference measure and the part living on a reference-null set.
Mutual singularity
Singular measures occupy disjoint measurable supports.
Definition. Measures \(\nu\) and \(\mu\) are mutually singular, written \(\nu\perp\mu\), if there is a measurable set \(N\) such that \(\mu(N)=0\) and \(\nu(X\setminus N)=0\).
Example. A point mass \(\delta_x\) is singular with respect to Lebesgue measure, because it is concentrated on the Lebesgue-null set \(\{x\}\).
Lebesgue decomposition theorem
Every sigma-finite measure splits into absolutely continuous and singular parts.
Theorem. If \(\nu\) and \(\mu\) are sigma-finite, then there are unique measures \(\nu_a,\nu_s\) such that \(\nu=\nu_a+\nu_s\), \(\nu_a\ll\mu\), and \(\nu_s\perp\mu\).
Proof. Put \(\rho=\mu+\nu\). Both measures are absolutely continuous with respect to \(\rho\), so Radon-Nikodym gives \[ f=\frac{\mathrm{d}\nu}{\mathrm{d}\rho}, \qquad g=\frac{\mathrm{d}\mu}{\mathrm{d}\rho}. \] Let \(S=\{g=0\}\). Since \[ \mu(S)=\int_Sg\,\mathrm{d}\rho=0, \] the restriction \(\nu_s(A)=\nu(A\cap S)\) is singular with respect to \(\mu\). On \(S^c\), define \(h=f/g\). Then \[ \nu_a(A)=\nu(A\cap S^c) =\int_{A\cap S^c}f\,\mathrm{d}\rho =\int_Ah\,\mathrm{d}\mu, \] so \(\nu_a\ll\mu\), and clearly \(\nu=\nu_a+\nu_s\).
For uniqueness, suppose \(\nu=\alpha_a+\alpha_s=\beta_a+\beta_s\). Then \[ \alpha_a-\beta_a=\beta_s-\alpha_s. \] The left signed measure is absolutely continuous with respect to \(\mu\), while the right is concentrated on the union of two \(\mu\)-null supports and is therefore singular. A signed measure that is both absolutely continuous and singular must vanish, so both parts agree.
Corollary. The absolutely continuous part has the form \[ \nu_a=\frac{\mathrm{d}\nu}{\mathrm{d}\mu}\,\mu, \] where the notation denotes the Radon-Nikodym derivative of \(\nu_a\), not of the whole \(\nu\) unless \(\nu\ll\mu\).
Decomposition of distribution functions
On the line, finite measures correspond to increasing functions.
Proposition. A distribution function decomposes into an absolutely continuous part, a pure jump part, and a singular continuous part.
Proof. Let \(\mu_F\) be the Lebesgue-Stieltjes measure of \(F\). Lebesgue decomposition gives \[ \mu_F=f\,m+\mu_s. \] The set of atoms of a finite measure is countable: for each \(n\), only finitely many atoms can have mass at least \(1/n\). Hence \[ \mu_{\mathrm{pp}}=\sum_x\mu_F(\{x\})\delta_x \] is well-defined, and \(\mu_{\mathrm{sc}}=\mu_s-\mu_{\mathrm{pp}}\) is singular and atomless. Taking cumulative masses on \((-\infty,x]\) yields \[ F=F_{\mathrm{ac}}+F_{\mathrm{jump}}+F_{\mathrm{sc}}, \] up to the constant \(F(-\infty)\).
Example (a mixed law). The probability measure \[ \nu=\frac12\mathcal N(0,1)+\frac14\delta_0+\frac14\mu_C \] has all three parts: Gaussian density \(\frac12\phi(x)\), a jump of size \(1/4\) at \(0\), and one quarter of the atomless Cantor measure. A distribution can therefore be continuous without possessing a density, and can possess a density without being purely continuous.
Cantor singular measure
Singularity need not mean discreteness.
Example. The Cantor distribution is continuous and singular with respect to Lebesgue measure. It assigns full mass to the Cantor set, which has Lebesgue measure zero, but has no atoms.
Detecting singularity
Singular and absolutely continuous parts never overlap.
Corollary. If \(\nu\ll\mu\) and \(\nu\perp\mu\), then \(\nu=0\).
Proof. Singularity gives a set \(N\) with \(\mu(N)=0\) and \(\nu(X\setminus N)=0\). Absolute continuity gives \(\nu(N)=0\). Hence \(\nu(X)=0\).