Question

_{This is a sequel to another question I have asked.}

The notion of disintegration is a refinement of conditional probability to spaces which have more structure than abstract probability spaces; sometimes this is called regular conditional probability. Let $Y$ and $X$ be two nice metric spaces, let $\mathbb P$ be a probability measure on $Y$, and let $\pi : Y \to X$ be a measurable function. Let $\mathbb P_X(B) = \mathbb P(\pi^{-1} B)$ denote the push-forward measure of $\mathbb P$ on $X$. The disintegration theorem says that for $\mathbb P_X$-almost every $x \in X$, there exists a nice measure $\mathbb P^x$ on $Y$ such that $\mathbb P$ "disintegrates":$$\int_Y f(y) ~d\mathbb P(y) = \int_X \int_{\pi^{-1}(x)} f(y) ~d\mathbb P^x(y) d\mathbb P_X(x)$$ for every measurable $f$ on $Y$.

This is a beautiful theorem, but it's not strong enough for my needs. Fix a Borel set $B \subseteq X$, and let $p(x) = \mathbb P^x(B)$. Part of the theorem is that $p$ is a measurable function of $x$. Suppose that the map $\pi : Y \to X$ is continuous instead of simply measurable. My question: What is a general sufficient condition for $p(x)$ to be continuous?

To me, this is an obvious question to ask, since if $x$ and $x'$ are two close realizations of a random $x \in X$, then the measures $\mathbb P^x$ and $\mathbb P^{x'}$ should be close too, at least in many natural situations. However, in my combing through the literature, I haven't been able to find an answer to this question. My guess is that most people are content to integrate over $x$ when they use the theorem. For my purposes, I need some estimates which I get by continuity.

At this point, I've managed to prove and write down a pretty good sufficient condition for the case I care about (Banach spaces), using an abstract Wiener space-type construction. However, I am hoping that an expert can point me toward a good reference that does this in wider generality.

Answer 1

I think you need some hypothesis on the measure to be pushed, at least in the very common case where $\pi$ is the projection to a factor in a product.

Take any family $\mathbb{P}^x$ of measures in a space $X'$, where $x$ runs over $X$, and let $Y=X'\times X$, $\pi$ be the projection on $X$, and $\mathbb{P}=\int_X \mathbb{P}^x dx$ where $dx$ is any measure on $X$. Then $\pi$ is very regular (smooth if $X'$ and $X$ are smooth manifolds for example) but yet, any kind of lack of regularity can appear in $\mathbb{P}^x$ (which are by construction the disintegration measures, since they are unique up to a negligible set).

I guess that in this setting, assuming $\mathbb{P}$ to be absolutly continuous with continuous density would be sufficient.

Answer 2

Probably is not general as you want, but if you don't think before about that can be a begining...

Proposition: If $\pi:Y\to X$ is bijective function such that $\pi^{-1}$ is continuous then $\mathbb{P}^{x_n}\to\mathbb{P}^{x}$ (weak topology) whenever $x_n\to x$.

Proof: it follows from the Disintegration Theorem that for all $B\in\mathcal{B}(Y)$ we have

$$ \begin{array}{rcl} \mathbb{P}(B)&=&\displaystyle\int_X\int_{\pi^{-1}(x)}\chi_B(y)\ d\mathbb{P}^x(y)\ d\mathbb{P}_X(x) &=&\displaystyle\int_X \mathbb{P}^x(B\cap\pi^{-1}(x)) \ d\mathbb{P}_X(x) \end{array} $$ For the other hand we have that $$ \mathbb{P}(B)=\displaystyle\int_X \chi_B(\pi^{-1}(x))\ d\mathbb{P}_X(x) =\displaystyle\int_X \chi_B(\pi^{-1}(x))\delta_{\pi^{-1}(x)}(\pi^{-1}(x)) \ d\mathbb{P}_X(x) $$ so $$ \displaystyle\int_X \mathbb{P}^x(B\cap\pi^{-1}(x)) \ d\mathbb{P}_X(x)=\mathbb{P}(B)=\displaystyle\int_X \delta_{\pi^{-1}(x)}(B\cap\pi^{-1}(x)) \ d\mathbb{P}_X(x) $$

Since $\mathbb{P}^x$ is a probability measure and $B\cap\pi^{-1}(x)$ is a singleton or empty set, we have $$ \delta_{\pi^{-1}(x)}(B\cap\pi^{-1}(x))\geq \mathbb{P}^x(B\cap\pi^{-1}(x)) $$ and from previous integral equality almost surely we have $$ \delta_{\pi^{-1}(x)}(B\cap\pi^{-1}(x))=\mathbb{P}^x(B\cap\pi^{-1}(x)). $$ Fix $x$ and take $B=\pi^{-1}(x)$, then from the above equality, follows that $\mathbb{P}^x=\delta_{\pi^{-1}(x)}$.

If $x_n\to x$ then $p(x_n)=\mathbb{P}^{x_n}$ converge to $p(x)$ in the weak topology. In fact, by the continuity of $\pi^{-1}$ we get that $\int f\ p(x_n) \to\int f\ p(x)$ for all bounded uniformly continuous functions $f$.