# Setting

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 $$\int f(x,y) d\mu(x,y) = \int_0^1 \int_0^1 f(x,y)d\mu_{x}(y) dx.$$

# Question

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.

# Proof

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.

• What does it mean to approximate a measurable set by a continuous function? – Michael Greinecker yesterday
• @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. – Pablo Lessa yesterday

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).

• 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. – Pablo Lessa yesterday
• 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. – Pablo Lessa yesterday

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 $$\mu_x=\lambda$$.

• 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 $f_2$. – Pablo Lessa 4 hours ago