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I am currently trying to grasp some ideas on Lebesgue-Rokhlin spaces from Bogachev, "Measure Theory", vol. 2. Such spaces are also known as standard probability spaces but the definitions are not unique in the literature.

Let us restrict to probability measures.

Theorem 9.4.7: If $(M, \mathcal{M}, \mu)$ is Lebesgue-Rokhlin then it is isomorphic mod0 to $([0,1], \mathcal{B}[0,1], \nu)$ where $\nu = c \lambda + \sum_{n=1}^\infty \alpha_n \delta_{1/n}$ where $\alpha_n = \mu(a_n)$ and $a_n$ is the countable family of atoms of $\mu$.

Now every Borel probability measure on $[0,1]$ can be decomposed into an absolutely continuous part, a singular continuous part and an atomic part. In the above theorem I am somehow missing the singular continuous part. Is it not explicitely mentioned or what is the intuition that one only considers the absolutely continuous part and the atomic part? Is it "hidden" behind the Lebesgue measure?

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  • $\begingroup$ It looks like the singular continuous part is intentionally omitted. That is, part of what the theorem asserts is that you can find an appropriate $\nu$ which has no singular continuous part. So the theorem is already nontrivial even if $M = [0,1]$. $\endgroup$
    – Nate Eldredge
    Commented Nov 20, 2015 at 18:53
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    $\begingroup$ For example, the Cantor measure on [0,1] (which is singular continuous) is isomorphic mod0 to Lebesgue measure (which has no singular continuous part), where the isomorphism $T$ is the Cantor staircase function. $\endgroup$
    – Nate Eldredge
    Commented Nov 20, 2015 at 18:56
  • $\begingroup$ Thanks ... of course (see Example 3.6.2, where he also references to Chapter 9 for generalization). $\endgroup$
    – yada
    Commented Nov 20, 2015 at 19:51

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Forget about the atomic part. The gist of the theory is that (up to isomorphism) there is only one purely non-atomic Lebesgue-Rolhlin space, and it is isomorphic to the unit interval endowed with the Lebesgue measure.

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