Suppose $\mu_n$ is a sequence of probability measures on $[0,1]\times [0,1]$ converging to a limit probability $\mu$ meaning that $$ \lim_{n\to+\infty}\int f(x,y)d\mu_n(x,y) = \int f(x,y)d\mu(x,y)$$ for all continuous $f:[0,1]\times [0,1] \to \mathbb{R}$.

Suppose furthermore that all these probabilities project to the uniform measure on the first coordinate. This implies there are Borel mappings (conditional probabilities) $x \mapsto \mu_{n,x}$ and $x \mapsto \mu_{x}$ from $[0,1]$ to the space of probabilities on $[0,1]$ satisfying $$\int f(x,y) d\mu_n(x,y) = \int_0^1 \int_0^1 f(x,y)d\mu_{n,x}(y) dx,$$ and $$\int f(x,y) d\mu(x,y) = \int_0^1 \int_0^1 f(x,y)d\mu_{x}(y) dx.$$


I'm looking for a reference for the fact that $\lim_{n\to+\infty}\mu_{n,x} = \mu_x$ for almost every $x \in [0,1]$.

More generally, I'm looking for some reference covering the situation when $\mu_n$ are probabilities on some compact space with constant push-forward under some continuous mapping of that space.


Here's a proof of the claim (I would still love to have a reference).

Take $f(x,y) = h(x)g(y)$ with $h$ and $g$ continuous and notice that $$\lim_{n \to +\infty}\int_0^1 h(x)(\mu_{n,x}(g) - \mu_x(g)) dx = 0,$$ where we use $m(g)$ for the integral of $g$ with respect to the measure $m$.

Using $h$ to approximate the set $A_{\epsilon} = \lbrace x \in [0,1]: \liminf_{n \to +\infty} \mu_{n,x}(g) - \mu_x(g) > \epsilon\rbrace$ and Fatou's lemma (all functions are bounded) $$\frac{\epsilon}{2}|A_{\epsilon}| \le \int_0^1 h(x)\liminf_{n \to +\infty}(\mu_{n,x}(g)-\mu_x(g)) dx \le \liminf_{n \to +\infty} \int_0^1 h(x)(\mu_{n,x}(g)-\mu_x(g)) dx = 0,$$ where $|A|$ denotes the Lebesgue measure of $A$. This shows that $A_\epsilon$ has measure $0$.

Since this holds for all $\epsilon > 0$ and also for the function $-g$ we get $$\lim_{n \to +\infty}\mu_{n,x}(g) = \mu_x(g),$$ for almost every $x$.

Intersecting the full measure sets where this holds, over all $g$ in a countable dense set of continuous functions on $[0,1]$, the claim follows.

  • $\begingroup$ What does it mean to approximate a measurable set by a continuous function? $\endgroup$ Jul 1, 2020 at 14:08
  • $\begingroup$ @MichaelGreinecker I was thinking: 1- Take a compact subset of your set which has almost the same measure. 2- The characteristic function of the compact set is a monotone decreasing limit of continuous functions. $\endgroup$ Jul 1, 2020 at 14:29

2 Answers 2


This is false. Generally, disintegration behaves poorly with respect to weak convergence. I believe the error in your proof is the first inequality, which I don't see how to justify.

Many counterexamples arise from a well known phenomenon in optimal transport. For any probability measure $\mu$ on $[0,1] \times [0,1]$ with uniform first marginal, there exists a sequence $\mu_n$ of probability measures on $[0,1] \times [0,1]$ with uniform first marginal such that (1) $\mu_n \to \mu$ weakly and (2) each $\mu_n$ is supported on the graph of a continuous function. That is, each $\mu_n$ is of the form $\mu_n(dx,dy)=dx\delta_{f_n(x)}(dy)$ for some continuous $f_n$. See Theorem 9.3 of Ambrosio's lecture notes, for example, and approximate the Borel maps therein in $L^1$ by continuous ones.

Now, for example, if $\mu$ is Lebesgue measure (or more generally if the disintegration $\mu_x$ is nonatomic for a.e. $x$), and $\mu_n$ is supported on the graph of a measurable function for each $n$, then there is no way we can have $\mu_{n,x} \to \mu_x$ weakly, because $\mu_{n,x}$ is a delta for each $n$ whereas $\mu_x$ is not (and the set of delta measures is weakly closed).

  • $\begingroup$ Thank you very much! This is a really nice answer. I still don't see how the proof is wrong, but I'll look into it. $\endgroup$ Jul 1, 2020 at 14:14
  • $\begingroup$ I think I found the mistake. Repeating the argument for $-g$ only proves that $\liminf \mu_{n,x}(g) \le \mu_x(g) \le \limsup \mu_{n,x}(g)$ almost surely. $\endgroup$ Jul 1, 2020 at 15:23

A simple special case of Dan's answer above: Define $f_n:[0,1] \to [0,1]$ by $f_n(x)= nx \mod 1$ and define $g_n:[0,1] \to [0,1]^2$ by $g_n(x)=(x,f_n(x))$. The pushforward $\mu_n=\lambda g_n^{-1}$ of Lebesgue measure $\lambda$ on $[0,1]$ is the uniform measure on the graph of $f_n$. The sequence $\mu_n$ converges weakly to Lebesgue measure $\mu$ on $[0,1]^2$ but for each $x$ in the unit interval, $\mu_{n,x}$ are Dirac measures that cannot converge weakly to $\mu_x=\lambda$.

  • $\begingroup$ That is really simple. Thanks! I noticed that restricting $n$ to powers of $2$ this is the mixing property for the iterates of the mapping $f_2$. $\endgroup$ Jul 2, 2020 at 15:02

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