Let $X,Y$ be Polish spaces and suppose that $X$ is compact. Denote by $\mathcal{Mes}(X,\mathcal{P}(X\times Y))$ the set of (Borel) measurable functions from $X$ to the set of Borel probability measures $\mathcal{P}(X\times Y)$ on $X\times Y$. We equip this space with the uniform metric given on any two $f,g\in \mathcal{Mes}(X,\mathcal{P}(X\times Y))$ by
$$
D(f,g) := \sup\_{x \in X} \max\{1,d\_{LP:Y}(f(x),g(x))\}
,
$$
where $d\_{LP:Y}$ denotes the Lévy-Prokhorov metric on $\mathcal{P}(Y)$. Equip $\mathscr{P}(X)$ with the Lévy-Prokhorov metrics denoted by $d\_{LP:X}$. Let $\pi:X\times Y \mapsto X$ be the natural projection $(x,y)\mapsto x$ and fix a measure $\mu \in \mathcal{P}(X\times Y)$.




---


By the Disintegration Theorem, we know that the [*disintegration*](https://en.wikipedia.org/wiki/Disintegration_theorem#Statement_of_the_theorem) of $\mu$ by $\pi$ associates to $\mu$ the ($\pi\_{\#}\mu$-a.s.-unique) function $\operatorname{Dis}\_{\pi}:x\mapsto \mu\_x \in \mathcal{Mes}(X,\mathcal{P}(X\times Y))$. Therefore:
$$
\operatorname{Dis}\_{\pi}: \mathcal{P}(X\times Y)\mapsto \mathcal{Mes}(X,\mathcal{P}(X\times Y)).
$$
Is this map ever continuous? I.e.:
$$
\mbox{Does: } d\_{LP:X}(\mu\_n,\mu)\mapsto 0 \mbox{ imply that: }
\sup\_{x \in X} d\_{LP:X\times Y}\left(
\operatorname{Dis}\_{\pi}(\mu\_n)\_x,\operatorname{Dis}\_{\pi}(\mu)\_x
\right)\mapsto 0?
$$


I expect this is false in general, but are there some references describing when it holds?



Is disintegration continuous?
No, at least if $X$ is uncountable and $Y$ has at least two points. Let $K$ be a compact subset of $Y$ with at least two points (any finite such subset will do). Under this assumption, there exists an atomless Borel probability measure $\nu$ on $X$. Let $\mathcal{P}\_{\nu,K}(X\times Y)$ be the closed set of Borel probability measures $\mu$ on $X\times Y$ such that $\pi\_{\#}\mu=\nu$ and $\mu(X\times K)=1$. In this space, the set of probability measures supported on the graph of a measurable function is dense. This is an old result from optimal control theory, where it represents the denseness of relaxed controls within the space of controls. So your mapping will be discontinuous at any point in$ \mathcal{P}\_{\nu,K}(X\times Y)$. The disintegrations of probability measures supported on the graph of measurable functions (which are basically measurable functions) are bounded away from the corresponding disintegration under your sup-metric, but not under the Lévy-Prokhorov metric on $\mathcal{P}(X\times Y)$.


There is also the issue of whether your mapping is really well-defined. Since disintegrations are only $\pi\_{\#}\mu=\nu$ unique, which selection one chooses matters, especially for your fine topology on the space of measurable functions.


The mentioned denseness results holds in much greater generality, but a version sufficient for this argument is given by Theorem 3 of [Milgrom, Paul R., and Robert J. Weber. "[Distributional strategies for games with incomplete information.](https://web.stanford.edu/%7Emilgrom/publishedarticles/DistributionalStrategies.pdf)" Mathematics of operations research 10.4 (1985): 619-632.]


