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.



Convergence of conditional measures for a convergent sequence of probabilities whose projection is constant
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](http://www.math.sns.it/media/doc/paper/1008/trasporto.pdf), 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).


