# Continuity of disintegrations

Suppose that $\pi:Y\to X$ is a continuous surjection from one compact metric space to another. Given a regular probability measure $\mu$ on $Y$ with pushforward measure $\nu:=\pi^*\mu$, it is known that there is a family $(\mu_x)_{x\in X}$ of probability measures on $Y$ such that $x\mapsto \mu_x(B)$ is Borel measurable for each Borel subset $B$ of $Y$, such that almost every $\mu_x$ lives on the fiber above $x$, and such that $\int_Y f(y)d\mu(y)=\int_{x\in X}\int_{\pi^{-1}(x)}f(y)d\mu_x(y)d\nu(x)$.

My question is: suppose in advance that $\mathcal{O}$ is some fixed nonempty open subset of $Y$. Is it true that there is a probability measure $\mu$ as above for which $\mu(\mathcal{O})>0$ and such that there is a disintegration as in the previous paragraph for which the map $x\mapsto \mu_x(\mathcal{O})$ is upper semi-continuous? (A previous version of this question asked for continuity.) This is motivated by the fact that any Radon measure on a compact metric space is upper semi-continuous with respect to the Hausdorff metric.

• How about $\mu := \delta_y$ for some fixed $y\in\mathcal{O}$? Then you can take the disintegration to be $\mu_x = \delta_y$ for every $x$, which trivially makes $x\mapsto \mu_x$ continuous and satisfies your requirements. But perhaps you want $\mu$ to have full support? May 5, 2018 at 6:44
• Thanks to @TobiasFritz I understand that I do not understand what you're asking. What about the identity map $[0,1]\to [0,1]$? then the disintegration is $\mu_x=\delta_x$ for every $\mu$ and for every $\mathcal{O}\subsetneq [0,1]$, $x\mapsto \mu_x(\mathcal{O})$ is not continuous in any boundary point of $\mathcal{O}$. May 5, 2018 at 7:36

Let me start by addressing the question hinted on in the title, which in my view is more interesting than the question eventually asked in the body.

Given $\pi:X\to Y$ as above, we may endow the space $\text{Prob}(Y)$ with the weak* topology and consider its closed subspace $Z=\cup_x \text{Prob}(\pi^{-1}(x))$ consisting of all probability measures which images in $X$ are delta measures. Identifying $\delta_x$ with $x$ we consider the map $\pi_*:Z\to X$. The disintegration of a probability measure $\mu\in\text{prob}(Y)$ is a $\pi_*(\mu)$-a.e defined measurable map $X\to Z$, $x\mapsto \mu_x$, as described in the question. One may ask whether such a map could be represented as a continuous map (and that was my initial interpenetration of the question above).

The answer to this question is that in case $Y=X=[0,1]$ and $\pi:Y\to X$ is the Cantor-Lebesgue function (see https://en.wikipedia.org/wiki/Cantor_function) there are no local continuous sections $X\to Z$. Indeed, any such section is easily seen to be discontinuous at every diadic point of $X=[0,1]$. In particular, the disintegration function of any probability measure on $Y$ could not be represented by a continuous function.

Below is an edit answering the new version in the question's body.

Consider $X=Y=[-1,1]$ and $\mathcal{O}=(0,1]\subset Y$.

Let $\pi_1:Y\to X$ be the identity map. Then for any $\mu\in \text{Prob}(Y)$ we have that $\mu_x=\delta_x$, and the function $x\mapsto \mu_x(\mathcal{O})$ equals $\chi_{\mathcal{O}}$ which is lower semicontinuous, but not upper semi-continuous.

Let $\pi_2:Y\to X$ be given by defined by $\pi_2(x)=\min\{x,0\}$. Then for every $\mu\in \text{Prob}(Y)$ the function $x\mapsto \mu_x(\mathcal{O})$ equals $\mu(\mathcal{O})\cdot\delta_0$ and if $\mu(\mathcal{O})>0$ this function is upper semi-continuous but not lower semi-continuous.

We conclude that no semi-continuity is guaranteed in general.

• I see how this shows that Lebesgue measure doesn’t work. But why couldn’t some other measure still satisfy the conclusion for some given open set? Apr 30, 2018 at 18:50
• This is not about disintegration of a measure - it is about the fact that every map $X\to Z$, where $Z=\cup \text{Prob}(\pi^{-1}(x))$ endowed with the natural topology, is discontinuous at every diadic point of $X=[0,1]$. Apr 30, 2018 at 19:13
• the above example works for every open set in $X$. if you want an example that continuity fails at just one point, you can use $Y$ being the graph of $\sin 1/x$ on $(0,1]$ union the segment $[-1,1]$ in the y-axis and $X$ being the segment $[0,1]$ in the x-axis. Apr 30, 2018 at 19:37
• Is it possible that if I rephrased my question to merely ask for upper semi-continuity that it would then have a positive answer? Or does your example show that even this is not true? May 2, 2018 at 3:07
• Not for any reasonable pre-order on the target space that I can think of right now (leaving aside too week ones like the pre-order given by inclusions of supports). And of course, you are most welcome, I am happy to help. May 4, 2018 at 6:08

I know one situation when this happens namely when $X,Y$ are smooth manifolds $\pi$ is a smooth map and $\mu$ is the volume measure defined by a Riemann metric $g_Y$ on $Y$, $\mu=dV_{g_Y}$.

Fix a Riemann metric $g_X$ on $X$ and assume that $\dim Y=N\geq n=\dim X$. Then the coarea formula shows the pushforward $\nu$ has the form $\newcommand{\eH}{\mathscr{H}}$ $$\nu=\rho dV_{g_X},\;\;\rho(x)=\int_{\pi^{-1}(x)} \frac{1}{J_\pi(y)} d\eH_{N-n}(y),$$ where $J_\pi(y)$ is the Jacobian of the differential $d\pi(y): T_yY\to T_{\pi(y)} X$ and $\eH_k$ denotes the $k$-th dimensional Hausdorff measure on $Y$ defined by the metric $g_Y$

You can then define $\newcommand{\eO}{\mathscr{O}}$ $$\mu_x (V):= \frac{1}{\rho(x)}\int_{\pi^{-1}(x)\cap V} \frac{1}{J_\pi(y)} d\eH_{N-n}(y),\;\;V\subset Y.$$ Let $U\subset X$ be the set of regular values of $\pi$. This is an open subset since $Y$ is compact. It is also dense. As set $\eO$ you can take $\eO=\pi^{-1}(U)$. If $x\in U$, then $\pi^{-1}(x)$ is a compact smooth submanifold of $X$ and $J_\pi\neq 0$ on $\eO$. The map

$$U\ni x\mapsto \mu_x(\eO)$$ is continuous. For $x\in X\setminus U$ we have $\mu_x(\eO)=0$. If $\pi$ is a submersion then $U=X$.

• That’s a nice example but unfortunately in my context I cannot assume that my spaces are manifolds. Apr 30, 2018 at 17:35