Consider a measure space $(\Omega,\mathcal{F},\mu)$.
Let $f: (\Omega,\mathcal{F}) \rightarrow (\mathbb{R}, \mathcal{B}_{\mathbb{R}})$ be a non-negative and mesurable. Then, $$\nu({A})=\int_{{\Omega}} 1_{A}(x) f(x) d\mu$$ where $A \in \mathcal{F}$ defines a measure on $(\mathcal{F}, [0,\infty])$
show that if $\int_{\Omega} f(x) d\mu=1$, then $\nu(\Omega)=1$.
POST: So, from the definition of $\nu$, we have $$\nu({\Omega})=\int_{{\Omega}} 1_{\Omega}(x) f(x) d\mu=\int_{{\Omega}} f(x) d\mu=1$$
$\int_\Omega f(x)dx=\int_\Omega 1_\Omega(x)f(x)dx$, because $1_\Omega(x)$ is the same as multiplying $f(x)$ by one on the entire domain of the integral.