Let $(S,\Sigma,\mu)$ be a measure space. Show that if $(A_n)$ is a pairwise disjoint sequence of sets in $\Sigma$ and $A = \bigcup_n A_n$ then $$1_A = \sum_n 1_{A_n}.$$
My try:
To show that $1_A = \sum_n 1_{A_n}$, we need to show that they are equal on every point $x\in S$.
Fix $x\in S$. We have two cases to consider: either $x\in A$ or $x\not\in A$.
If $x\in A$, then by definition of $A$, there exists $n$ such that $x\in A_n$. Therefore, $1_A(x) = 1$ and $1_{A_n}(x) = 1$ for that $n$, and $1_{A_k}(x) = 0$ for all $k\neq n$. Thus, we have $$\sum_n 1_{A_n}(x) = 1_{A_n}(x) = 1,$$ and so $1_A(x) = \sum_n 1_{A_n}(x)$.
If $x\not\in A$, then $x\not\in A_n$ for all $n$, and so $1_A(x) = 0$ and $1_{A_n}(x) = 0$ for all $n$. Thus, we have $$\sum_n 1_{A_n}(x) = 0,$$ and so $1_A(x) = \sum_n 1_{A_n}(x)$.
Since we have shown that $1_A(x) = \sum_n 1_{A_n}(x)$ for every $x\in S$, we conclude that $1_A = \sum_n 1_{A_n}$.
Question
Does my logic make sense? Should I only discuss a point in $x \in A$ or I can use the union as well?