# decomposing a measure into relative atom and atomless parts

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 $A\in\mathcal{X}$?

• Notice that the family $(\mu_y)_{y \in \mathcal Y}$ is not defined pointwise, but only in an $L^\infty$ sense. Therefore, it doesn't really mean anything to talk about "each $\mu_y$". May 29, 2018 at 15:23

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.