I want to verify my proof for the following problem:
Suppose $\mu$ is a measure on a measurable space $(X, \mathcal{S})$ and $f: X \rightarrow[0, \infty]$ is an $\mathcal{S}$ -measurable function. Define $v: \mathcal{S} \rightarrow[0, \infty]$ by $$ v(A)=\int \chi_{A} f d \mu $$ for $A \in \mathcal{S}$. Prove that $\nu$ is a measure on $(X, \mathcal{S})$.
Attempt The part that $v(\varnothing)=0$ is trivial. Given a countable union of disjoint sets $A=A_1\cup \ldots \cup A_n$, we know that any $x\in X$ can only belong to one $A_i$. Hence $\chi_A(x) = \chi_{A_1\cup \ldots \cup A_n}(x) = \chi_{A_1}(x) +\ldots +\chi_{A_n}(x)$. Plugging this into v:
$v(\bigcup_{i=1}^n A_i)=v(A) = \int \chi_A f d\mu =\sum_{i=1}^n \int \chi_{A_i}fd\mu = \sum_{i=1}^n v(A_i) $