# 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.