Q:

Details of why measurable functions of independent random variables independent.

I'm looking through this proof from Durrett's Probability book

I don't understand why 

holds.  Can someone explain this to me?

A:

$f_i$ is measurable which means that if $A$ is any Borel subset of $\mathbf{R}$, the preimage $f_i^{-1}(A)$ is a Borel subset of $\mathbf{R}^{m(i)}$. Likewise, $(X_{i,1},\dots,X_{i,m(i)})$ is $\sigma((X_{i,1},\dots,X_{i,m(i)}))$ measurable which means that because $f_i^{-1}(A)$ is a Borel subset of $\mathbf{R}^{m(i)}$,
$$ (X_{i,1},\dots,X_{i,m(i)})^{-1}(f_i^{-1}(A)) \in \sigma((X_{i,1},\dots,X_{i,m(i)})). \tag{1} $$
The left hand side of $(1)$ is exactly $f_i(X_{i,1},\dots,X_{i,m(i)})^{-1}(A)$. For the right hand side, one can show that

$$ \sigma((Y_1,\dots,Y_m)) = \sigma(\sigma(Y_1) \cup \dots \cup \sigma(Y_m)) $$

from which it follows that
$$ \sigma((X_{i,1},\dots,X_{i,m(i)})) = \sigma\left( \bigcup_{j = 1}^{m(i)} \sigma(X_{i,j}) \right) = \mathcal{G}_i. $$
Hence $(1)$ says that the preimage of every Borel subset of $\mathbf{R}$ under $f_i(X_{i,1},\dots,X_{i,m(i)})$ is in $\mathcal{G}_i$. Therefore $f_i(X_{i,1},\dots,X_{i,m(i)}) \in \mathcal{G}_i$.

