How is a singular continuous measure defined? 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 $\mathbb{R}$. 


*

*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?

*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).
  

 A: 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 $f=0)$.
A: A singular (say, probability) measures $\mu$ with respect to the Lebesgue measure $\lambda$ on $\mathbb R^d$ satisfies by definition: there exists a Borel set $S$ such that $\mu(S)=1$ and $\lambda(S)=0$. To obtain a continuous singular measure, that is satisfying $\mu(\{x\})=0$ for any $x\in\mathbb R^d$, the idea is to find a measure supported on a set $S$ having positive dimension but strictly less than $d$. When $d=1$, you can use sets of fractional dimension, like the cantor set, but in higher dimension you can find more easy examples, e.g. taking for $\mu$ a measure on the unit circle (resp. sphere) in $\mathbb R^2$ (resp. $\mathbb R^d$).  
A: 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.
