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The usual statement of the Lebesgue Decomposition Theorem says that given two $\sigma$-finite measures $\mu$ and $\nu$ on a measure space, we can decompose $\nu = \nu_1 + \nu_2$, where $\nu_1$ is absolutely continuous with respect to $\mu$ and $\nu_2$ and $\mu$ are mutually singular.

Wikipedia (link text) says that there is a "refinement" of this result, where the singular part $\nu_2$ can be further decomposed into a discrete measure and a singular continuous measure.

I understand what a discrete measure is, but what exactly is the definition of a singular continuous measure? I was also wondering if anyone knew of a reference for this refined result, since I haven't been able to find it anywhere.

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@user1736: I believe "$\mu$ is singular continuous measure" means that the measure of any point is $0$, and there is a Borel set of Lebesgue measure $0$ whose $\mu$-measure equals the $\mu$ measure of the entire space. – Arturo Magidin Dec 3 '10 at 6:18
up vote 5 down vote accepted

That decomposition is commonly encountered in probability theory. In a more general setting, suppose that $\rho$ is a $\sigma$-finite measure on $\mathcal{B}(\mathbb{R}^n)$. Then there is a unique decomposition $\rho = \rho_{ac} + \rho_d + \rho_{cs}$, such that: 1) $\rho_{ac}$ is absolutely continuous, that is, $\rho_{ac}$ is zero on sets of Lebesgue measure zero; 2) $\rho_d$ is discrete, that is, $\rho_d$ is zero on the complement of some countable set $C$; 3) $\rho_{cs}$ is continuous singular, that is, $\rho_{cs}$ is zero at every point $x \in \mathbb{R}^n$ (= continuous measure), and is zero on the complement of some set $B$ of Lebesgue measure zero (= singular measure).

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See also (the section "Examples on $\mathbb{R}^n$"). – Shai Covo Dec 3 '10 at 9:11

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