[proofplan] Independence of random variables is defined via their generated $\sigma$-algebras: $X_1, \ldots, X_n$ are independent if for all Borel sets $B_1, \ldots, B_n$, $\mathbb{P}(X_1 \in B_1, \ldots, X_n \in B_n) = \prod_{i=1}^n \mathbb{P}(X_i \in B_i)$. We show the same factorisation holds for $f_1(X_1), \ldots, f_n(X_n)$ by expressing each event $\{f_i(X_i) \in C_i\}$ as $\{X_i \in f_i^{-1}(C_i)\}$ and applying the independence of the $X_i$. [/proofplan] [step:Rewrite $\{f_i(X_i) \in C_i\}$ as a preimage event for $X_i$] Let $C_1, \ldots, C_n \subset \mathbb{R}$ be arbitrary (Borel) sets. For each $i$, the event $\{f_i(X_i) \in C_i\}$ consists of all $\omega \in \Omega$ such that $f_i(X_i(\omega)) \in C_i$, which holds if and only if $X_i(\omega) \in f_i^{-1}(C_i)$. Defining $B_i = f_i^{-1}(C_i) \subset \mathbb{R}$, \begin{align*} \{f_i(X_i) \in C_i\} = \{X_i \in B_i\}. \end{align*} [/step] [step:Apply independence of $X_1, \ldots, X_n$ to the preimage events] Since $X_1, \ldots, X_n$ are independent, the product formula holds for any choice of sets $B_1, \ldots, B_n \subset \mathbb{R}$: \begin{align*} \mathbb{P}(f_1(X_1) \in C_1, \ldots, f_n(X_n) \in C_n) &= \mathbb{P}(X_1 \in B_1, \ldots, X_n \in B_n) \\ &= \prod_{i=1}^n \mathbb{P}(X_i \in B_i) \\ &= \prod_{i=1}^n \mathbb{P}(f_i(X_i) \in C_i). \end{align*} Since this holds for all choices of $C_1, \ldots, C_n$, the random variables $f_1(X_1), \ldots, f_n(X_n)$ are independent. [/step]