Without loss of generality, we model $(X_i)_{i\in\mathbb{N}}$ as the coordinate projections on the canonical probability space $(\mathsf{X}^{\mathbb{N}},\mathcal{X}^{\otimes\mathbb{N}},\mathbb{P})$. We proceed as follows:

1. Reverse filtration construction:

• For each $n\in\mathbb{N}$, let $\mathcal{F}_n$ be the $\sigma$-field generated by all measurable functions $f:\mathsf{X}^{\mathbb{N}}\to\mathbb{R}$ invariant under any permutation of the first $n$ coordinates, i.e. for any permutation $\pi$ of $\{1,\ldots,n\}$ and any $x\in\mathsf{X}^{\mathbb{N}}$,

$$ f(x_1,\ldots,x_n,x_{n+1},\ldots)=f(x_{\pi(1)},\ldots,x_{\pi(n)},x_{n+1},\ldots). $$

• $(\mathcal{F}_n)_{n\in\mathbb{N}}$ is a reverse filtration with $\mathcal{F}_n\supseteq\mathcal{F}_{n+1}$ and

$$ \mathcal{F}_\infty=\bigcap_{n\in\mathbb{N}}\mathcal{F}_n. $$

• $\mathcal{F}_\infty$ is generated by all functions invariant under any finite permutation of coordinates.

2. Conditional expectation and empirical averages:

• For any bounded measurable $h:\mathsf{X}\to\mathbb{R}$, the empirical average $\frac1n\sum_{i=1}^n h(X_i)$ is $\mathcal{F}_n$-measurable.
• For any $A\in\mathcal{F}_n$, exchangeability implies

$$ \mathbb{E}\!\left[\left(\frac1n\sum_{i=1}^n h(X_i)\right)\mathbf1_A\right] =\mathbb{E}[h(X_1)\mathbf1_A]. $$

• Hence

$$ \frac1n\sum_{i=1}^n h(X_i)=\mathbb{E}[h(X_1)\mid\mathcal{F}_n]. $$

• By the reverse martingale convergence theorem,

$$ \frac1n\sum_{i=1}^n h(X_i)\xrightarrow{\mathrm{a.s.}}\mathbb{E}[h(X_1)\mid\mathcal{F}_\infty]. $$

3. Multivariate functions:

• For bounded measurable $f:\mathsf{X}^k\to\mathbb{R}$,

$$ \mathbb{E}[f(X_1,\ldots,X_k)\mid\mathcal{F}_\infty] =\lim_{n\to\infty}\frac1{n(n-1)\cdots(n-k+1)} \sum_{\substack{1\le i_1,\ldots,i_k\le n\\ i_j\neq i_\ell}} f(X_{i_1},\ldots,X_{i_k}). $$

• But we have

$$ \lim_{n\to\infty}\frac1{n(n-1)\cdots(n-k+1)} \sum_{\substack{1\le i_1,\ldots,i_k\le n\\ i_j\neq i_\ell}} f(X_{i_1},\ldots,X_{i_k}) = \lim_{n\to\infty}\frac1{n^k}\sum_{i_1=1}^n\cdots\sum_{i_k=1}^n f(X_{i_1},\ldots,X_{i_k}). $$

• Hence, for product functions $f(x_1,\ldots,x_k)=f_1(x_1)\cdots f_k(x_k)$,

$$ \mathbb{E}[f_1(X_1)\cdots f_k(X_k)\mid\mathcal{F}_\infty] =\prod_{\ell=1}^k\mathbb{E}[f_\ell(X_1)\mid\mathcal{F}_\infty]. $$

• Thus, conditionally on $\mathcal{F}_\infty$, $(X_i)$ are independent.