Let $X$ be a standard Borel measurable space (for example, $\Bbb{R}$).
Let $p$ be an exchangeable probability measure on the countably infinite product $X^\Bbb{N}$.
Call a measurable set $A\subseteq X^\Bbb{N}$
- Strictly exchangeable if for any finite permutation $\sigma:X^\Bbb{N}\to X^\Bbb{N}$, $\sigma^{-1}(A)=A$;
- Almost surely exchangeable if for any finite permutation $\sigma:X^\Bbb{N}\to X^\Bbb{N}$, $p(\sigma^{-1}(A)\,\Delta\,A)=0$, where $\Delta$ denotes the symmetric difference.
Is it true that if $A$ is almost surely exchangeable, then there exists a strictly exchangeable $B$ such that $p(A\,\Delta\,B)=0$?
(I think I've found a proof, but it seems very specific to permutations, and I would like to see other proofs that might generalize to arbitrary actions.)
A reference would also be welcome.
Best Answer
The key observation is that strict exchangeability means $1_A(\sigma(X)) = 1_A(X)$ for all $X$ while almost sure exchangeability means the equality only holds almost surely.
Suppose $A$ is almost surely exchangeable. Let $f_n(X) = \frac{1}{n!}\sum_{\sigma \in S_n}1_A(\sigma(X))$. Then $f_n$ is strictly exchangeable over $S_n$ and $f_n = 1_A$ almost surely. Let $f(X) = \limsup_n f_n(X)$. Then $f(X)$ is strictly exchangeable and $f(X) = 1_A(X)$ almost surely. Let $B = \{f(X) = 1\}$. Then $1_B(X) = I(f(X) = 1)$, from which it is clear that $1_B(X) = 1_A(X)$ almost surely and that $1_B$ is strictly exchangeable.