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On a measurable space, how is a measure being singular continuous relative to another defined? I searched on the internet and in some books to no avail and it mostly appears in a special case - the Lebesgue measure space $\mathbb{R}$.

  1. Do you know if singular continuous measures can be generalized to a more general measure space than Lebesgue measure space $\mathbb{R}$? In particular, can it be defined on any measure space, as hinted by the Wiki article I linked below?
  2. The purpose of knowing the answers to previous questions is that I would like to know to what extent the decomposition of a singular measure into a discrete measure and a singular continuous measure still exist, all wrt a refrence measure?

Thanks and regards!

PS: In case you may wonder, I encounter this concept from Wikipedia (feel it somehow sloppy though):

Given $μ$ and $ν$ two σ-finite signed measures on a measurable space $(Ω,Σ)$, there exist two $σ$-finite signed measures $ν_0$ and $ν_1$ such that:

  • $\nu=\nu_0+\nu_1\,$
  • $\nu_0\ll\mu$ (that is, $ν_0$ is absolutely continuous with respect to $μ$)
  • $\nu_1\perp\mu$ (that is, $ν_1$ and $μ$ are singular).

The decomposition of the singular part can refined: $$ \, \nu = \nu_{\mathrm{cont}} + \nu_{\mathrm{sing}} + \nu_{\mathrm{pp}} $$ where

  • $\nu_{\mathrm{cont}}$ is the absolutely continuous part
  • $\nu_{\mathrm{sing}}$ is the singular continuous part
  • $\nu_{\mathrm{pp}}$ is the pure point part (a discrete measure).
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It is a biiiigg mess-up. In fact you first decompose w.r.t. another measure then after having done that you can decompose into continuous and discrete or atomic and atomless or anything in between. It happens and that is where probably confusion comes from is that the Lebesgue measure itself is already purely continuous atomless and so on so the decomposition of absolutely continuous part will stay the same you might also call it somewhat regular continuous and regular discrete as well as singular continuous and sincular discrete. See the analogues? – Alexander Frei Nov 6 '14 at 20:04

I am not completely sure, and I cannot provide a publicly available reference, but I read in some lecture notes from our university that this decomposition can be generalized to $\mathbb{R}^n$. Let $\lambda$ be the Lebesgue measure on $\mathbb{R}^n$ and $\mu$ the measure under consideration. Then,

$$ \mu = \mu_a + \mu_s + \mu_d $$

where $\mu_d$ is discrete (i.e., supported on a countable set, with positive measure for every atom), $\mu_a$ is absolutely continuous w.r.t. $\lambda$ (i.e., it possesses a density), and $\mu_s$ is singularly continuous, i.e., it is supported on a Lebesgue null-set, and the atoms of this set have zero measure.

An example for $\mu_s$ in $\mathbb{R}^2$ would be a measure which is supported on a one-dimensional submanifold of $\mathbb{R}^2$, e.g., the uniform distribution on the unit circle.

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That's right and in fact it holds as a decomposition of the Banach space of signed Radon measures on an arbitrary locally compact group. The decomposition is unique and the total variations satisfy $|\mu| = |\mu_a|+|\mu_s|+|\mu_d|$ and similarly for the total variation norm. See e.g. Hewitt-Ross, Abstract harmonic analysis I, Theorem 19.20. – commenter Oct 29 '12 at 15:40

In the case of Borel measures on the real line, the continuous singular part $\nu_\mathrm{sing}$ can be characterized as follows: First let $$ F(x) = \nu_\mathrm{sing}((-\infty,x]). $$ (In the special case of probability measures, this is the cumulative probability distribution function.) Then $F$ is a continuous function, but $\nu_\mathrm{sing}$ and Lebesgue measure are mutually singular.

The Cantor function in the role of $F$ is an example. The Cantor distribution is a probability distribution no part of which has a density with respect to Lebesgue measure. But its cumulative distribution function is nonetheless continuous. I.e. there is no function $f$ such that for every Borel set $A$, $$ \nu(A) = \int_A f(x)\;dx + \nu_\mathrm{singular}(A) $$ for some other measure $\nu_\mathrm{singular}$ (except the trivial function $f=0)$.

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+1 Thanks! I was wondering if this concept is merely for Borel measure space $\mathbb{R}$? – Tim Jan 14 '12 at 1:41
Thanks! Do you know if singular continuous measures can be generalized to a more general measure space than Lebesgue measure space $\mathbb{R}$? In particular, can it be defined on any measure space, as hinted by the Wiki article I linked? – Tim Jan 14 '12 at 2:44
So you say rather study a generic Borel measure over the real line via its cumulative distribution function $F(x):=\nu((-\infty,x])$, is this right?? Then you can identify the purely atomic part as the discontinuities of the cumulative distribution function and the absolutely continuous part as those sections where it is differentiable. Is this right? Moreover, what if it is infinite at every point as for the Lebesgue measure for example? Next, how to proceed if it is not a Borel measure? – Alexander Frei Oct 14 '14 at 18:06

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