Let $(X, d)$ be a metric space and $\mathcal{B}(X)$ the Borel $\sigma$-algebra of X. Let $\mu, \nu$ be two real-valued signed measures defined on $(X, \mathcal{B}(X))$ and $f : X \to \mathbb{R}$ Borel measurable, $\alpha, \beta \in \mathbb{R}$.
Can anyone explain to me if the following takes place $$\int f(x) d(\alpha \mu + \beta \nu)(x) = \alpha \int f(x) d\mu(x) + \beta \int f(x) d\nu(x)?$$
Thank you!
Yes. By definition if $E$ is measurable and $\mu(E)$, $\nu(E)$ are finite then $$ (\alpha \mu + \beta \nu)(E) = \alpha \mu(E) + \beta \nu(E).$$
In integral notation this becomes $$\int \chi_E d(\alpha \mu + \beta \nu) = \alpha \int \chi_E \, d\mu + \beta \int \chi_E \, d\nu.$$ From here you can prove the result for simple $f$ and from there for (sufficiently integrable) measurable $f$ via the usual limiting processes.