Convergence of conditional distributions implies convergence of unconditional distributions Let $X,X_1,X_2,\dots$ be real random variables defined on a probability space $(\Omega,\mathcal A ,P)$. Given a sub-$\sigma$-algebra $\mathcal F\subset\mathcal A$, let $P_{X}$ denote the distribution of $X$ and let $P_{X|\mathcal F}$ denote the conditional distribution of $X$ given $\mathcal F$, i.e. a Markov kernel from $(\Omega,\mathcal F)$ to $(\mathbb R ,\mathcal B(\mathbb R))$ such that $\omega\mapsto P_{X|\mathcal F}(\omega,B)$ is a version of $P(X\in B|\mathcal F)$ for each $B\in \mathcal B(\mathbb R)$. Suppose that
$$P_{X_n|\mathcal F}(\omega,\cdot)\Rightarrow P_{X|\mathcal F}(\omega,\cdot) \quad P\text{-almost surely},$$
where $\Rightarrow$ denotes convergence in distribution (weak convergence). (The above definition makes sense since conditional distributions are almost surely unique). Is it then also the case that $P_{X_n}\Rightarrow P_X$ ?
I tried the following:
Let $f\in C_b(\mathbb R)$. From the definition of weak convergence we have
$$\int f dP_{X_n|\mathcal F}(\omega,\cdot)\to \int f dP_{X|\mathcal F}(\omega,\cdot)\quad \text{as } n\to \infty, \quad P\text{-almost surely.}  $$ Moreover, from the properties of conditional distributions we have
$$E[f\circ X|\mathcal F](\omega)=\int f  dP_{X|\mathcal F}(\omega,\cdot)  \quad P\text{-almost surely, } $$ for each $n$. Therefore $E[f\circ X_n|\mathcal F]\to E[f\circ X|\mathcal F]$ $P$-almost surely. From the dominated convergence theorem we get
$$E[f\circ X_n]=\int f dP_{X_n}\to \int f dP_X \quad \text{as } n\to \infty.$$
As $f\in C_b(\mathbb R)$ was arbitrary, we indeed have $P_{X_n}\Rightarrow P_X$.
Is this correct? Thanks a lot for your help.
 A: This is a long comment that first better in the answer section:
Some attention is required to what versions of conditional probabilities are taken, for in general conditional probabilities are not representable as stochastic kernels unless some regularity assumptions are satisfied (which $\mathbb{R}$ satisfies).
Let $X_\infty$ denote $X$.
Since $\mathbb{R}$ (with the usual Euclidean structure) is a nice space (Polish), for each $n\in\mathbb{N}\cup\{\infty\}=\overline{\mathbb{N}}$,  the conditional probability $P[X_n\in\cdot|\mathcal{F}]$ is regular, that is there is a stochastic kernel $\mu_n$ from $(\Omega,\mathcal{F}$ to $(\mathbb{R},\mathscr{B}(\mathbb{R})$, and a null set $N_n\in\mathcal{A}$ such that

*

*$P[X_n\in B|\mathcal{F}](\omega)=\mu_n(\omega, B)$ for all Borel set $B$ and $\omega\in N_n$

*For any Borel set $B$, $\omega\mapsto\mu_n(\omega,B)$ is $\mathcal{F}$-measurable

A simple monotone class arguments shows that for any $f\in\mathcal{C}_b(\mathbb{R})$,
$$\begin{align}E[f(X_n)|\mathcal{F}](\omega)=\int_\mathbb{R} f(x)\mu_n(\omega,dx),\qquad\omega\in N_n\tag{1}\label{one}\end{align}$$
For all $\omega$ outside $N=\bigcup_{n\in\overline{\mathbb{N}}}N_n$ (which is a negligible set) condition \eqref{one} holds.
If for any $\omega\in\Omega\setminus N$ we have that $\mu_n(\omega,\cdot)$ converges weakly to $\mu_\infty(\omega,\cdot)$, then by dominated convergence
$$\begin{align}
E[f(X_n)]&=\int_\Omega E[f(X_n)|\mathcal{F}](\omega)\,P(d\omega)\\
&=\int_{\Omega\setminus N}\Big(\int_\mathbb{R} f(x)\mu_n(\omega,dx)\Big)\,P(d\omega)\xrightarrow{n\rightarrow\infty}\int_{\Omega\setminus N}\Big(\int_\mathbb{R} f(x)\mu_\infty(\omega,dx)\Big)\,P(d\omega)\\
&=\int_\Omega E[f(X_\infty)|\mathcal{F}](\omega)\,P(d\omega)=E[f(X_\infty)]
\end{align}$$
Which is what you sketched in your posting.
The latter condition (on the kernels $\mu_n)$ however, seems to me that would be much harder to verify than to directly show that $X_n$ converges weakly in law.
