let $(X, \Sigma, \mu)$ be a measure space, $f : X \to [0, +\infty] $ be a measurable function and $$\nu : \Sigma \to [0, +\infty] \;\;\;\;\;\; A \mapsto \nu_f(A) = \int_A f d\mu $$
1st question is to show that $\nu_f$ is a measure.
well $\nu_f(\emptyset) = 0$ since $\mu(\emptyset)$
let $\{ A_i\}_{i \in \mathbb{N}} \subset \Sigma$ be a sequence of disjoint sets
$\nu(\bigcup_{i \geq 1} A_i) = \int_{\bigcup_{i \geq 1} A_i} f d\mu = \int_{X} f \chi_{\bigcup_{i \geq 1} A_i} d\mu = \int_{X} \sum_{i \geq 1} f \chi_{ A_i} d\mu = \int_{X} \lim_{n \to \infty} \sum_{i = 1}^{n} f \chi_{ A_i} d\mu$
now if $f$ was non-negative I would use the monotone convergence theorem to conculde the $\sigma$-additivity but in this case I'm a bit lost.
EDIT : I'm dumb because $f$ is in fact positive according to the given data.
but what if $f$ was any measurable function ?