 Solution

Let $T$ swap coordinate blocks $0,\dots ,n$ and $n+1,\dots ,2n+1$, fixing all other coordinates. The symmetric law makes $T$ measure preserving, and invariance of the event $A$ gives $F\circ T=F$ pointwise. Put ${\mathcal{B}}_n=\sigma (X_{n+1},\dots ,X_{2n+1})$.

Conditional expectation transforms covariantly under this block swap:To check it, the left side is ${\mathcal{B}}_n$-measurable, and for $C\in {\mathcal{B}}_n$ the change of variables by $T$ sends its defining integral to the conditional-expectation identity on $T(C)\in {\mathcal{E}}_n$, with $F\circ T=F$. Thus the two versions agree. Measure preservation now givesThis is the first requested estimate.

The L1 contraction of conditional expectation follows from the pointwise bound $|\mathbb{E}[Z\mid \mathcal{H}]|\le \mathbb{E}[|Z|\mid \mathcal{H}]$: take expectations to get $\Vert \mathbb{E}[Z\mid \mathcal{H}]{\Vert }_1\le \Vert Z{\Vert }_1$. Applied to a difference, it makes conditional expectation a contraction between any two integrable inputs.

Let $U_n=\mathbb{E}[F\mid {\mathcal{B}}_n]$. Since ${\mathcal{B}}_n\subseteq {\mathcal{G}}_n$, $U_n$ is ${\mathcal{G}}_n$-measurable. HenceCombining this with the other given approximation bound yieldsThe left side does not depend on $n$, and $\epsilon >0$ was arbitrary. It is therefore zero. For a $\mathcal{G}$-measurable version $Y=\mathbb{E}[F\mid \mathcal{G}]$, the event $C=\{Y>1/2\}$ belongs to $\mathcal{G}$ and satisfiesThis proves that exchangeable invariant events are tail events modulo null sets. It does not assert literal pointwise equality of the two events or sigma-algebras before completing them with null sets.