Let $X_1, X_2, X_3$ be mutually independent random variables from $(\Omega,\mathcal{F},P)$ to $(\mathbb{R},\mathcal{B})$. Show that $X_3$ is independent of $\sigma(X_1,X_2)$, i.e., for all $B \in \mathcal{B}$ and $A \in \sigma(X_1,X_2)$, $P(X_3^{-1}(B) \cap A) = P[X_3^{-1}(B)]P[A]$.
My attempt:
We know from the mutual independence of $X_1,X_2,X_3$ that for all $B_1,B_2,B_3 \in \mathcal{B}$,
$$ P[X_1^{-1}(B_1) \cap X_2^{-1}(B_2) \cap X_3^{-1}(B_3)] = P[X_1^{-1}(B_1)]P[X_2^{-1}(B_2)]P[X_3^{-1}(B_3)] $$
Take an arbitrary $B_3 \in \mathcal{B}$ and $A \in \sigma(X_1,X_2)$. Now, $\sigma(X_1,X_2) = \sigma(\sigma(X_1) \cup \sigma(X_2))$.
I have worked out three cases.
First case: $A \in \sigma(X_1) \cup \sigma(X_2)$.
Since $X_3$ and $X_1$ (or $X_2$) are independent, $P[X_3^{-1}(B_3) \cap A] = P[X_3^{-1}(B_3)]P[A]$, and we are done.
Second case: $A =C_1 \cap C_2$ for some $C_1 \in \sigma(X_1)$ and $C_2 \in \sigma(X_2)$.
$C_1=X_1^{-1}(B_1)$ for some $B_1 \in \mathcal{B}$ and $C_1=X_1^{-1}(B_2)$ for some $B_2 \in \mathcal{B}$. Hence, by independence of $X_1,X_2,X_3$, $$P[X_3^{-1}(B_3) \cap (C_1 \cap C_2)] = P[X_3^{-1}(B_3)]Pr[C_1]Pr[C_2] = P[X_3^{-1}(B_3)]Pr[C_1 \cap C_2]$$
and we are done.
Third case: $A =C_1 \cup C_2$ for some $C_1 \in \sigma(X_1)$ and $C_2 \in \sigma(X_2)$.
We can show that $A^c$ and $X_3^{-1}(B_3)$ are independent by case 2. Hence, $A$ and $X_3^{-1}(B_3)$ are independent.
$\tag*{$\blacksquare$}$
How do we proceed for a general set $A$ in $\sigma(X_1,X_2)$?