Suppose $X_1,..,X_n$ are independent random variables and $X_i'$ is an independent copy of $X_i$, then how does one show that $$E[f(X_1,..,X_n)|X_1,..,X_{i-1},X_{i+1},..,X_n]$$ and $$E[f(X_1,..,X_i',..,X_n)|X_1,..,X_{i-1},X_{i+1},..,X_n]$$ are equal almost everywhere and why is $f(X_1,..,X_n)$ and $f(X_1,..,X_i',..,X_n)$ conditionally independent over $G:=\sigma(X_1,..,X_{i-1},X_{i+1},..,X_n)$?
I am trying to argue using measure theoretic arguments but I can't seem to be able to show this. I am stuck trying to understand the $\sigma$-algebras generated by both functions which are measurable on $\sigma(X_1,..,X_{i-1},X_{i+1},..,X_n)$. Any ideas?
i.e. Possible proof of conditional independence:
Note: $\sigma(f(X_1,..,X_n))\subset \sigma(X_1,..,X_n)$ and $\sigma(f(X_1,..,X_i',..,X_n))\subset \sigma(X_1,..,X_i',..,X_n)$. Let $A\in \sigma(X_1)\cup...\cup\sigma(X_n) $ and $B\in \sigma(X_1)\cup..\cup\sigma(X_i')..\cup\sigma(X_n)$ then $$E[1_A1_B|G]=E[1_A|G]E[1_B|G]$$ This implies that it holds for all $A\in \sigma(X_1,..,X_n)$ and $B\in \sigma(X_1,..,X_i',..,X_n)$.
Let $Y=[X_1,\ldots,X_{n-1}]$. Then $$ \mathsf{E}[f(X_1,\ldots,X_{n-1},X_n)\mid Y]=\varphi(Y) \quad\text{a.s.} $$ and $$ \mathsf{E}[f(X_1,\ldots, X_{n-1},X_n')\mid Y]=\varphi'(Y) \quad\text{a.s.}, $$ where $\varphi(y)=\mathsf{E}f(y_1,\ldots, y_{n-1}, X_n)$ and $\varphi'(y)=\mathsf{E}f(y_1,\ldots, y_{n-1}, X_n')$. (See, e.g., Lemma 6.2.1 on page 236 here.) But $\varphi(y)=\varphi'(y)$ because $X_n\overset{d}{=}X_n'$.