Suppose $X, Y \subset \mathbb{R}^n$ are compact, and fix $\mu_X$ a probability measure on $X$, and $\mu_Y$ a probability measure on $Y$.

Consider a sequence of probability measures $\gamma_n$ on $X\times Y$ with the property that $\mu_X(A) = \gamma_n(A \times Y)$ and $\mu_Y(B) = \gamma_n(X \times B)$.

Consider also the disintegration $d \gamma_n(x,y) = d \mu_X(x) d \gamma_n^x(y)$ of $\gamma_n$.

What I am wondering is the following:

If $\gamma_n$ weak-$*$ converges to a probability measure $\gamma_{\infty}$, is it necessarily true that the disintegration $\gamma_n^x$ weak-$*$ converges to $\gamma_\infty^x$ (for $\mu_X$-a.e. $x$) ?

Playing around with the definitions, my intuition says this won't hold, because we are asking for some kind of pointwise control from weak-$*$ convergence. On the other hand, it doesn't seem clear how to construct a counterexample.


1 Answer 1


Let me describe my example in probability language.

Let $U \sim U(-1,1)$ be a uniformly distributed random variable on $[-1,1]$. Let $\xi = \operatorname{sgn}(U)$; note that $|U|$ and $\xi$ are independent.

Let $A_n \subset [0,1]$ be a sequence of "typewriter" sets, such that $m(A_n) \to 0$ yet $1_{A_n}$ diverges everywhere. Let $$\xi_n = \begin{cases} -\xi, & |U| \in A_n \\ \xi, & \text{otherwise}.\end{cases}$$ Then it is clear that $\xi_n \overset{d}{=} \xi$, i.e. a fair coin flip, for every $n$. (For instance, $\xi_n$ obviously takes only the values $\pm 1$, and if we write $\xi_n = (-1)^{1_{A_n}(|U|)} \xi$, independence gives $E [\xi_n] = E[(-1)^{1_{A_n}(|U|)}] E[\xi] = 0$.)

Note that $P((U, \xi_n) \ne (U,\xi)) = P(|U| \in A_n) = m(A_n) \to 0$, so $(U, \xi_n) \to (U,\xi)$ in probability and thus also weakly.

To translate, let $X = [-1,1] \setminus \{0\}$ (just to avoid annoyance), $Y = \{-1,1\}$, $\mu_X,\mu_Y$ the respective uniform measures, and $\gamma_n$ the joint law of $(U, \xi_n)$. Likewise $\gamma_\infty$ is the joint law of $(U,\xi)$. We have shown $\gamma_n \to \gamma_\infty$ weakly, and that the marginals of $\gamma_n, \gamma_\infty$ are $(\mu_X,\mu_Y)$.

Now $$\gamma_n^x = \begin{cases} \delta_{-\operatorname{sgn}(x)}, & |x| \in A_n \\ \delta_{\operatorname{sgn}(x)}, & |x| \notin A_n. \end{cases}$$ In particular, for every $x$ we have $\gamma_n^x = \delta_{1}$ for infinitely many $n$ and $\gamma_n^x = \delta_{-1}$ for infinitely many $n$. This fails to converge weakly.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .