$1_{\bigcup_{i=1}^{\infty}A_n}=\sum_{n=1}^{\infty}1_{A_n}$
Solution 1:
Let $A:=\cup_{n=1}^\infty A_n$.
If $x\in A$, then $1_A(x)=1$. Since the union is disjoint, $x$ must be in $A_k$ for exactly one $k$. Thus $$ \sum_{n=1}^\infty 1_{A_n}(x)=1_{A_k}(x)=1 $$
If $x\not\in A$, then $1_A(x)=0$. But since $x$ is not in $A$, $x$ cannot be in $A_n$ for any $n$. Hence, the $1_{A_n}(x)=0$ for each $n$ and $\sum_{n=1}^\infty 1_{A_n}(x)=0$.