Let $A_1,A_2,\ldots,A_n$ be independent subsets of probability space $(\Omega, \Sigma, P)$ (For every $I\subseteq \{1,2,\ldots,n\}$, $P(\bigcap_{j\in J}A_j)=\prod_{j\in J}P(A_j) )$. Prove that $1_{A_1},1_{A_2},\ldots,1_{A_n}$ is independent, i.e. for every Borel set $B_1,B_2,\ldots,B_n$ of $\mathbb{R}$ we have $P(\cap_{i=1}^n \{w\in \Omega: 1_{A_i}(w)\in B_i\} )= \prod_{i=1}^n P(\{w\in \Omega: 1_{A_1}(w)\in B_i\})$.
My partial answer:
Since $1_{A_i}(w) \in \{0,1\}$ for every $w\in \Omega$, then we have
$$ \{w\in \Omega: 1_{A_i}(w)\in B_i\}=\begin{cases} \emptyset & \text{if } 1\notin B_i \text{ and } 0\notin B_i ,\\ A_i & \text{if } 1\in B_i \text{ and } 0\notin B_i, \\ A_i^c & \text{if } 1\notin B_i \text{ and } 0\in B_i, \\ \mathbb{R} & \text{if } 1\in B_i \text{ and } 0\in B_i. \\ \end{cases} $$ First, we prove that if $A_1,A_2,\ldots,A_n$ is independent then $A_1^c,A_2^c,\ldots,A_n^c$ is independent. If there is $i_0$ such that $1\notin B_{i_0} \text{ and } 0\notin B_{i_0}$ then we have $$P\left(\bigcap_{i=1}^n \{w\in \Omega: 1_{A_i}(w)\in B_i\} \right)= 0=\prod_{i=1}^n P(\{w\in \Omega: 1_{A_1}(w)\in B_i\}).$$ Hence, we can assume for the case $1\in B_i \text{ or } 0\in B_i$ for every i.
Thanks.