Where does empirical distribution of an exchangeable random sequence converge to? Let $(X_n)_{n \ge 1}$ be an exchangeable random sequence and let $\alpha$ be the corresponding de-Finetti random measure. I want to show that the empirical distribution of the sequence converges to $\alpha$ almost surely.
Note: For an iid-P random sequence $(X_n)_{n \ge 1}$, Glivenko Cantelli lemma states that the empirical distribution converges to $P$ almost surely.
 A: I wonder why this question has gone unanswered this long.
By de Finetti's theorem any random sequence $(\xi_n)_{n\geq1}$ in a Borel space $S$ is conditionally i.i.d., say, given $\mathcal{F}$.
Fix $f\geq0$ measurable on $S$. Then
\begin{eqnarray*}
P\left(\frac{1}{n}\sum_{k\leq n}f(\xi_k)\rightarrow\int fd\alpha \right)&=&EE\left(1\left\{\frac{1}{n}\sum_{k\leq n}f(\xi_k)\rightarrow\int fd\alpha \right\}\bigg|\mathcal{F}\right)\\
&=&E \int1\left\{\frac{1}{n}\sum_{k\leq n}f(x_k)\rightarrow\int fd\alpha \right\}d\alpha^\infty(x)=E1=1,
\end{eqnarray*}
where the law of large numbers was used (conditionally) in the next to last equality. An approximation (since $S$ is Borel) now implies
\begin{gather}\label{empirical}
\eta_n\equiv n^{-1}\sum_{k\leq n}1_B(\xi_k)\rightarrow \alpha B\:\:\: a.s., \:\:\:\:\: B\in\mathcal{S}.
\end{gather}
For an alternate proof one can also exploit the fact that the $(\eta_n)$ form a reverse martingale. For a proof of this fact, a good reference is  $\textit{Probabilistic Symmetries and Invariance Principles}$ by Olav Kallenberg, where also the above result can be found.
