decomposing a measure into relative atom and atomless parts

Question

Let $(X,\mathcal{X},\mu)$ be a standard probability space. A measure $\mu$ is atomic if $\mu$ is supported on at most countable many atoms, and is atomless if $\mu$ has no atom (a point x is an atom if $\mu(\{x\})>0$).

Now let $(X,\mathcal{X},\mu)$ be an extension of $(Y,\mathcal{Y},\nu)$ and $\mu=\int_{Y}\mu_{y}d\nu(y)$ be the disintegration of $\mu$ with respect of $\nu$. Then for each $\mu_{y}$, there is a unique way to write $\mu_{y}=\mu_{y,c}+\mu_{y,d}$, where $\mu_{y,c}$ is an atomless and $\mu_{y,d}$ is an atomic measure on $(X,\mathcal{X})$.

Question: let $\mu_{c}=\int_{Y}\mu_{y,c}d\nu(y)$ and $\mu_{d}=\int_{Y}\mu_{y,d}d\nu(y)$. Is the decomposition $\mu=\mu_{c}+\mu_{d}$ well defined? i.e. Is $A$ measurable for $\mu_{c}$ and $\mu_{d}$ for all $A\in\mathcal{X}$?

Answer 1

It is generally true that if $(X,\mathcal{X})$ is a measurable space, $(Y,\mathcal{Y})$ a standard Borel space, and $\kappa:X\to\mathcal{P}(X)$ a transition probability, then the measure-valued function that maps $x$ to the atomic part of $\kappa(x)$ is measurable. The rest is a red herring.

With the usual $\sigma$-algebra on $\mathcal{P}(X)$ (the space of probability measures on $X$) generated by evaluation of measurable sets, the function $j:\mathcal{P}(X)\times X\to [0,1]$ given by $j(\mu,x)=\mu\big(\{x\}\big)$ is measurable. You can find the details in Appendix A of "Fundamentals of Nonparametric Bayesian Inference" (2017) by Ghosal and van der Vaart. Let $C\subseteq\mathcal{P}(X)\times X$ be the set on which $j$ is strictly positive. Let $g_\mu$ be the $\mu$-section of $C$. By a standard generalization of Fubini's theorem for transition probabilities, the function $\mu\mapsto \int g_\mu~\mathrm d\mu$ is measurable and so is the function $\mu\mapsto \int_A g_\mu~\mathrm d\mu$, which gives you the desired measurability.

For another approach, look at Theorem 2.12 of

Dubins, Lester, and David Freedman. "Measurable sets of measures." Pacific Journal of Mathematics 14.4 (1964): 1211-1222.

Answer 2

This fact is one of the consequences of Rokhlin's theory of Lebesgue probability spaces (also known as standard probability spaces, or Lebesgue–Rokhlin probability spaces), see the exceptionally well written Wikipedia article Standard probability space and the references therein or the 1952 English translation of Rokhlin's original paper (MR22591 and MR30584). Section 4.1 of Rokhlin's paper (pp.40-43 of the English tranlation) explicitly contains the statement you are asking about. Alas, this theory is virtually unknown in the probabilistic community.