Let $\mu_1: \mathcal{M} \to [0,\infty)$ be a finite measure and $\mu_2: \mathcal{M} \to [0,\infty]$ be a measure (can take infinite value).
Now define $\mu: \mathcal{M} \to [-\infty,\infty]$ by $\mu = \mu_1 - \mu_2$. I want to show $\mu$ is a signed measure.
The conditions "$\mu(\emptyset)=0$" and "$\mu$ can only take either $\infty$ or $-\infty$" is clear (in this case $\mu$ takes $-\infty$).
I need to show for any {$\{E_k\}$} is a countable collection of disjoint measurable sets, we have $$\mu \Big( \bigcup_k^{\infty} E_k \Big) = \sum_k^{\infty} \mu(E_k).$$
I decide to split into two cases:
$\mu \Big( \bigcup_k^{\infty} E_k \Big)$ is finite. I managed to prove this case.
$\mu \Big( \bigcup_k^{\infty} E_k \Big) = -\infty$. If there exists some $k_0$ such that $\mu(E_{k_0}) = - \infty$, then we have the equality. But the problem is when every $\mu(E_k)$ is finite. In this case, $\mu_1(E_k)$ and $\mu_2(E_k)$ are also finite for every $k$. How to show $\sum_k^{\infty} \mu(E_k) = -\infty$?
Any help is highly appreciated. Thank you.
Actually I read Royden and Fitzpatrick's textbook:
So I just want to take an example.
We have
$$\mu\left(\bigcup_{k}E_k\right)=\mu_1\left(\bigcup_{k}E_k\right)-\mu_2\left(\bigcup_{k}E_k\right).$$
Using disjointness, this equals
$$\sum_k\mu_1(E_k)-\sum_k\mu_2(E_k)=\sum_k(\mu_1(E_k)-\mu_2(E_k))=\sum_k\mu(E_k).$$