Absolutely Continuous, Atomic, and Singular Spectra

Three Ways Spectral Mass Can Be Distributed

A spectral measure need not have a density. Its mass may be spread across frequencies in a way described by a density, concentrated at individual frequencies, or concentrated on a set of Lebesgue measure zero without any point masses. Distinguishing these possibilities matters because a density describes only the first kind of mass; it cannot encode atoms or singular-continuous mass.

Let \((X_n)_{n\ge0}\) be a real-valued weakly stationary sequence on a probability space \((\Omega,\mathcal{F},\mathbb{P})\), with finite second moments. Write \(\gamma(h)=\operatorname{Cov}(X_{n+h},X_n)\), for integer lag \(h\), and use the spectral convention from the earlier material: a finite positive measure \(F\) on the frequency circle \([-\pi,\pi)\) satisfies \(\gamma(h)=\int e^{ih\lambda}F(d\lambda)\). For a real-valued sequence, \(F\) is symmetric. In particular, \(F([-\pi,\pi))=\gamma(0)=\operatorname{Var}(X_n)\). The endpoints of this interval are identified when treating it as a circle.

The Lebesgue decomposition theorem says that every finite positive measure has a unique decomposition into an absolutely continuous part and a singular part. This is a standard measure-theoretic result; its proof is beyond this tutorial. The singular part itself splits uniquely into its atoms and a singular-continuous remainder. Thus the spectral measure has the form

where \(f\ge0\) is integrable, the frequencies \(\lambda_j\) are distinct, \(a_j>0\), and \(F_{\mathrm{sc}}\) is singular continuous. The sum may be finite, countably infinite, or empty. In the absolutely continuous part, the Radon–Nikodym theorem supplies the density \(f\); the density is defined up to changes on a Lebesgue-null set. Neither an atom nor a nonzero singular-continuous measure can be represented by an integrable density with respect to \(d\lambda\).

Proof. For uniqueness of the absolutely continuous/singular split, suppose \(F=\alpha+\sigma\), where \(\alpha\ll m\) and \(\sigma\) is concentrated on a Lebesgue-null set \(N\). On any measurable set \(A\subseteq N^c\), \(\sigma(A)=0\), so \(\alpha(A)=F(A)\). On \(N\), absolute continuity gives \(\alpha(N)=0\), so \(\sigma(A)=F(A)\) for measurable \(A\subseteq N\). These identities determine both measures: \(\alpha(A)=F(A\cap N^c)\) and \(\sigma(A)=F(A\cap N)\) for every measurable \(A\). Hence the two measures are unique. For any finite measure \(\sigma\), its set of atoms \(D=\{\lambda:\sigma(\{\lambda\})>0\}\) is countable: for each positive integer \(r\), there can be only finitely many points with mass at least \(1/r\), since \(\sigma\) has finite total mass, and every positive mass is at least \(1/r\) for some \(r\). Consequently \(\sigma_{\mathrm{at}}=\sum_{\lambda\in D}\sigma(\{\lambda\})\delta_\lambda\) is well-defined, and \(\sigma-\sigma_{\mathrm{at}}\) is positive and has no atoms. If \(\sigma\) is singular, both measures are concentrated on its null supporting set, so the remainder is singular continuous. The atomic part is fixed by its singleton masses, which proves uniqueness of this further split. Finally, if \(F\) is symmetric, reflection of any of these components is another decomposition with the same defining properties. Uniqueness therefore makes each component symmetric. \(\square\)

The three pieces contribute separately to the covariance integral. The density contributes \(\int e^{ih\lambda}f(\lambda)\,d\lambda\); the atoms contribute \(\sum_j a_je^{ih\lambda_j}\); and the singular-continuous part contributes \(\int e^{ih\lambda}F_{\mathrm{sc}}(d\lambda)\). When \(F\) is symmetric, these integrals are real and may be written with \(\cos(h\lambda)\). A density exists for the absolutely continuous component, but not for the atomic or nonzero singular-continuous components.

Reading a Mixed Spectrum Correctly

A single spectral measure may contain all three components. The following example also shows why each covariance term must carry the mass of its own component. It uses the same singular-continuous probability measure \(\nu\) just constructed.

Common Pitfall: Confusing Singular with Atomic

“Singular” does not mean “made of atoms.” Both atomic and singular-continuous measures are singular with respect to Lebesgue measure. The distinction is whether any singleton has positive mass: an atomic measure has at least one such point, while a singular-continuous measure has none. A density describes the absolutely continuous part only; writing a point mass or a singular-continuous measure as an ordinary density loses its actual distribution of mass.

A second common error is to omit a component’s weight when computing its covariance contribution. In the mixed example, a wrong calculation would replace \(\frac12\int\cos(h\lambda)\nu(d\lambda)\) by \(\int\cos(h\lambda)\nu(d\lambda)\). At \(h=0\), that gives \(1+\frac12+1=\frac52\), while the displayed measure has total mass \(2\). Substituting the actual component \(\frac12\nu\) gives \(1+\frac12+\frac12=2\), as required. Checking \(h=0\) is a quick way to detect a missing or duplicated mass.

In practice, classify each part by asking three questions: Is it absolutely continuous with respect to Lebesgue measure, and therefore expressible through a density? Does it assign positive mass to any single frequency? If neither description applies, is it singular and atomless? These tests distinguish the components without treating every non-density as an atom or assuming that every spectral measure has a density.