We have a measure space $(\Omega, \mathcal F, \mu)$. Suppose that $\rho: \Omega \to \Bbb R$ is a measurable function, where $\rho$ is non-negative, and $\int \rho d\mu >0$. Define $\nu: \mathcal F \to \Bbb R$ by $\nu(A)=\int \chi_A \rho d\mu$, where $\chi_A$ denotes the characteristic function of set A.
I am trying to show that $\nu$ is a well defined measure. I know that i have to prove the following three things:
i. $μ(∅)=0$
ii. $μ(A)≥0$ for all $A∈ \mathcal F$
iii. If ${A_i}∈ \mathcal F$ is a sequence of pairwise disjoint sets, then $μ(⋃A_i)= ∑ μ(Ai)$.
However I dont not know how to apply these steps when the integral is present, as is the case here. Any help appreciated!
I'll write $f$ instead of $\rho$, mainly for cosmetics.
(i) $\chi_\emptyset f$ is the constantly $0$ function, so $\int \chi_\emptyset f=0$.
(ii) Integrals of non-negative functions are non-negative by definition.
(iii). Fix a sequence of measurable, pairwise disjoint sets $A_i$. Put
$$ h_n:=\sum_{i=1}^n \chi_{A_i} f$$
$$h:=\sum_{i=1}^\infty \chi_{A_i} f$$
Both $h$ and every $h_n$ is measurable. Also, $h_n\to h$ pointwise and $h_n\le h_{n+1}$. By the Monotone Convergence Theorem $$ \lim_{n\to \infty}\int h_n d\mu =\int hd\mu $$ Since the sets are pairwise disjoint, we have that $\sum_{i=1}^\infty \chi_{A_i} = \chi_{\bigcup_{i=1}^\infty A_i}$. Therefore, $$ \int hd\mu=\int \chi_{\bigcup_{i=1}^\infty A_i} fd\mu =\nu \left(\bigcup_{i=1}^\infty A_i\right). $$ On the other hand, by linearity, $$ \int h_n d\mu = \sum_{i=1}^n \int \chi_{A_i}fd\mu=\sum_{i=1}^n \nu(A_i) $$