Given a measure space $(X,\mathcal{M},\mu)$ and a measurable function $f:X\rightarrow \overline{\mathbb{R}}$ such that at least one of $f^+$ or $f^-$ is integrable, show that $\nu:\mathcal{M}\rightarrow \overline{\mathbb{R}}$ defined by $\nu(E)=\int_E f d\mu$ is a signed measure over $X$.
Now I've already proved that $\nu(\varnothing)=0$ and that $\nu$ assumes only one of the values $+\infty$ and $-\infty$, it remains to prove that for every disjoint sequence $(E_i)_i$ of measurable sets, one has $\nu(\cup_i^{+\infty}E_i)=\sum_{i=1}^{+\infty}\nu(E_i)$, where the series either converges absolutely or diverges to $+\infty$ or to $-\infty$. First of all, suppose that the series converges, so we have to prove that it converges absolutely, that is $\sum_{i=1}^{+\infty}|\nu(E_i)|< +\infty$.
$\sum_{i=1}^{+\infty}|\nu(E_i)|\leq\sum_{i=1}^{+\infty}\int_{E_i}|f| d\mu=\int_{\cup_i E_i}|f| d\mu=\int_{\cup_i E_i}(f^++f^-) d\mu$ and this last term is finite, because at least one of $f^+$ or $f^-$ is integrable and also we are assuming that the series (without modulus) converges, that is $-\infty<\sum_{i=1}^{+\infty}\nu(E_i)=\int_{\cup_i E_i}(f^+-f^-) d\mu<+\infty$.
Is it correct? And now, how can I prove that if it diverges, it diverges to $+\infty$ or to $-\infty$? I think that the idea is: 1) $\cup_i E_i$ is measurable; 2) since $f$ is measurable and at least one of $f^+$ or $f^-$ is integrable, $\sum_{i=1}^{+\infty}\nu(E_i)=\sum_{i=1}^{+\infty}\int_{E_i}f d\mu=\int_{\cup E_i} f d\mu$ exists, so, if it is not finite, it should be $-\infty$ or $+\infty$.