How is a singular continuous measure defined?

Question

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!

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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).

Answer 1

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 $f=0)$.

Answer 2

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.