Show that $f(X_{1},...,X_{n})\in\mathcal G$ 
Suppose  $X_{1},\cdots,X_{n}$ are independent random variables from $(\Omega,\mathcal{F})$ to $(\mathbb{R},\mathcal{B}(\mathbb{R}))$; $f:\mathbb R^{n}\mapsto \mathbb R$ and $f\in \mathcal{B}(\mathbb{R}^{n})/\mathcal{B}(\mathbb{R})$. Define $\mathcal{F_{i}}:=\sigma(X_{i})$ where $\sigma(X_{i})=X_{i}^{-1}(\mathcal{B}(\mathbb{R})),i=1,...,n$ and $\mathcal{G}:=\sigma(\cup_{i=1}^{n}\mathcal{F_{i}}).$
Show that $f(X_{1},...,X_{n})\in\mathcal G$.

For any $S\in\mathcal B(\mathbb R)$, since $f$ is a Borel measurable function,  $f^{-1}(S)\in\mathcal B(\mathbb R^n)$ obviously .But I don't understand that  why we have $\{\omega\in\Omega:(X_{1}(\omega),...,X_{n}(\omega))\in f^{-1}(S)\} \in \mathcal G ?$
 A: For every $i\in\{1,\dots,n\}$ the function $X_i:\Omega\to\mathbb R$ is $\mathcal F_i/\mathcal B(\mathbb R)$ measurable.
Then consequently for every $i\in\{1,\dots,n\}$ the function $X_i:\Omega\to\mathbb R$ is $\mathcal G/\mathcal B(\mathbb R)$ measurable.
Then consequently the function $X=(X_1,\dots, X_n):\Omega\to\mathbb R^n$ is $\mathcal G/\mathcal B(\mathbb R)^n$ measurable.
Here $(\mathbb R^n,\mathcal B(\mathbb R)^n)$ is the product of the $n$ copies of measurable space $(\mathbb R,\mathcal B(\mathbb R))$ and a function $X$ that has that space as codomain is measurable iff the projections $\pi_i\circ X$ are measurable for $i=1,\dots,n$.
Fortunately it can be proved that $\mathcal B(\mathbb R)^n=\mathcal B(\mathbb R^n)$ so that we can also state that $X:\Omega\to\mathbb R^n$ is $\mathcal G/\mathcal B(\mathbb R^n)$ measurable.
(If you are not familiar with this then have a look here)
Next to that we have that $f:\mathbb R^n\to\mathbb R$ is $\mathcal B(\mathbb R^n)/\mathcal B(\mathbb R)$ measurable.
Then based on the general rule that a composition of measurable functions is again measurable we conclude that $f\circ X:\Omega\to\mathbb R$ is $\mathcal G/\mathcal B(\mathbb R)$ measurable.
