Prove $E\left[E\left[Y|X_i,\,\cdots,\,X_n\right]| X_1,\,\cdots,\,X_{i-1},\,X_{i+1},\,\cdots,\,X_n\right]=E\left[Y| X_{i+1},\,\cdots,\,X_n\right]$. Suppose $X_1,\,\cdots,\,X_n$ are independent random variables on the probability space $\left(\Omega,\,\mathcal{A},\,P\right)$.
$Y$ is a nonnegative measurable function of $\left(X_1,\,\cdots,\,X_n\right)$.
Prove
$$
E\left[E\left[Y|X_i,\,\cdots,\,X_n\right]| X_1,\,\cdots,\,X_{i-1},\,X_{i+1},\,\cdots,\,X_n\right]=E\left[Y| X_{i+1},\,\cdots,\,X_n\right].
$$

I cannot prove this because neither $\sigma(X_i,\,\cdots,\,X_n)$ nor $\sigma\left(X_1,\,\cdots,\,X_{i-1},\,X_{i+1},\,\cdots,\,X_n\right)$ includes the other and I can't apply the standard formula.
 A: We can start showing that for any integrable random variable $Z$, and two sub-$\sigma$-algebras $\mathcal G_1$ and $\mathcal G_2$ such that $\mathcal G_2$ is independent of $\sigma(Z)\vee\mathcal G_1$, $$\mathbb E\left[Z \mid \mathcal G_1\vee \mathcal G_2\right]=\mathbb E\left[Z \mid \mathcal G_1\right]$$
(in order words, "adding" a conditioning by something independent of everything does not change the conditional expectation.
To this this, it suffices to show that for each $G_1\in\mathcal G_1$ and $
G_2\in\mathcal G_2$,
$$
\mathbb E\left[Z\mathbf{1}_{G_1}\mathbf{1}_{G_2}\right]=\mathbb E\left[\mathbb E\left[Z\mid\mathcal G_1\right]\mathbf{1}_{G_1}\mathbf{1}_{G_2}\right]$$
and conclude by $\pi$-$\lambda$-theorem (referred as Dynkin).
Now, in your context, apply it to $Z=\mathbb E\left[Y\mid X_k,i\leqslant k\leqslant n\right]$, $\mathcal G_1=\sigma\left(X_k,i+1\leqslant k\leqslant n\right)$ and $\mathcal G_2=\sigma\left(X_k,1\leqslant k\leqslant i-1\right)$.
A: Hint: Consider $\{A \in \sigma (\{X_1,X_2,\cdots, X_{i-1},X_{i+1},\cdots,X_n\}): \int_A LHS= \int_A RHS\}$. Using $\pi - \lambda$ Theorem it follows that if we prove that the equation here holds when $A$ is of the form $B \cap C$ where $A \in \sigma \{X_1,X_2,\cdots,X_{i-1})$ and $C \in \sigma(\{X_{i+1},X_{i+2},\cdots,X_n\})$ then we are done. But in this case each side is equal to $P(B)\int_C Y dP$.
