Integrating with respect to a linear combination of two signed measures 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!
 A: 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.
A: Ok. Thanks for the reminder, Leucippus.
Dealing with this argument, as i said, you have to take care, because you don't know yet, if the monotone convergence theorem is still valid for signed measure.
Despite of this, we are able to prove this using only the definition of integral.
By definition,
$\int f d\mu := \sup \{\sum_{i=1}^N \inf_{E_i} \{f(x)\} \mu(E_i)\} \text{ ,}$
where the sup is taken over all partitions
But we already now that if $A,B \subset \mathbb{R}$, then $\sup A+B = \sup A + \sup B$
So, applying the equallity above to $\int f(x) d(\alpha \mu + \beta \nu)(x) = \alpha \int f(x) d\mu(x) + \beta \int f(x) d\nu(x)$, we obtain the required identity.
Now, using the Hahn Decomposition Theorem you can prove the fact that te monotone convergence theorem also works for signed measures
