Signed measure: verification Let $f\colon X\to[-\infty,+\infty]$ a measurable function such that $$\int_Xf^+\;d\mu<\infty\quad\text{or}\quad \int_Xf^-\;d\mu<\infty,$$ then $$\nu(E)=\int_Ef\;d\mu$$ is a signed measure. 
I have to prove the countable additivity of $\nu$.
Let $\{E_k\}$ a sequence of disjoint measurable set and let $E=\cup_n E_k.$
By definition $$\nu(E)=\nu^+(E)-\nu^-(E)=\int_Ef^+d\mu-\int_Ef^-d\mu$$
since $\nu^+$ and $\nu^-$ are measures we have $$\nu(E)=\sum_{k}\nu^+(E_k)-\sum_{k}\nu^-(E_k)$$
I can conclude as $$\nu(E)=\sum_k[\nu^+(E_k)-\nu^-(E_k)]=\sum_k\nu(E_k)$$

Question. iIt's true or false? if it is fake how can i repair it?

 A: To conclude, you want to claim that
$$\sum_{k=1}^\infty \nu^+(E_k)-\sum_{k=1}^\infty \nu^-(E_k) = \sum_{k=1}^\infty \left( \nu^+(E_k) + \nu^+(E_k)\right)$$
right?
You just need to make sure that you don't get undefined sums of extended reals, that is, sums of type $\infty - \infty$. To do that you need to show that under these assumptions the following is always true
\begin{align}
\sum_{k=1}^\infty \nu^+(E_k) < \infty && \text{ or } && \sum_{k=1}^\infty \nu^-(E_k) <\infty
\end{align}
Which is done by splitting the assumption that
\begin{align}
\int_Xf^+ d\mu < \infty && \text{ or } && \int_Xf^- d\mu < \infty
\end{align}
in two cases. First, assume that $\int_Xf^+ d\mu < \infty$. Then
$$\sum_{k=1}^\infty \nu^+(E_k) = \nu^+(E) \leq \nu^+(X) = \int_Xf^+ d\mu < \infty$$
The second case, when $\int_Xf^- d\mu < \infty$, is done analogously.
This means that your argument follows.
A: Assume that $\int_{X}fd\mu=+\infty$ in this case we have $\nu^-<\infty$
If $\nu^+(E)=+\infty$, then $\nu^+(E)=\sum_{k}\nu^+(E_k)$. We have two cases to consider
i) $\nu^+(E_k)<\infty$ for every k. We have
$$\sum_{k\le n} \nu(E_k)=\sum_{k\le n}\nu^+(E_k)- \nu^-(E_k)=\sum_{k\le n} \nu^+(E_k)- \sum_{k\le n}\nu^-(E_k)\to+\infty $$
Because $\sum_{k\le n}\nu^-(E_k)$ converges.
ii) Exists $k\geq 0$ such that $\nu^+(E_{k})=+\infty$.
Define $k_{0}=\min_k \{k\geq 0\mbox{ such that }\nu^+(E_{k})=+\infty\}$
$$\sum_{k\le k_0} \nu(E_k)=\sum_{k\le k_0}\nu^+(E_k)- \nu^-(E_k)=+\infty$$
This last imply $\sum_{k}\nu(E_k)=+\infty.$ And in both cases $\nu(E)=\sum_{k}\nu(E_k).$ (For the case $\nu(E)<\infty$ we can use you argue)
Analogue for the case $\int_{X}fd\mu=-\infty$ and if $\int_X fd\mu<\infty$ you argue is correct
