I want to show that for finite signed measures $\mu, \nu$ defined on $\sigma$-algebra $\mathcal{A}$ the following triangle inequality holds
$ | \mu +\nu |(A) \leq |\mu|(A) +|\nu|(A) $
for all $A\in \mathcal{A}$. If $E, F$ form a Hahn decomposition where $E$ is negative and $F$ is positive, then by definition $\mu^+(A) = \mu(A \cap F)$ and $\mu^-(A) = -\mu(A \cap E)$. Further by definition $|\mu|(A) = \mu^+(A) + \mu^-(A)$ and it can be shown that $\mu = \mu^+-\mu^-$ where at least one if the measures $\mu^+$ and $\mu^-$ is finite.
Hence
$ | \mu +\nu |(A) = (\mu +\nu)^+(A) + (\mu +\nu)^-(A) $
and
$ (\mu +\nu)^+(A) = (\mu^++v^+-\mu^--v^-)(A\cap F) = \mu(A\cap F) + v(A\cap F) + \mu(A\cap F \cap E) + v(A\cap F \cap E) = \mu(A\cap F) + v(A\cap F) + \mu(A\cap \emptyset) + v(A\cap \emptyset) = \mu(A\cap F) + v(A\cap F). $
Similar holds for $(\mu +\nu)^-(A)$. I don't understand how we would ever get a strict inequality not just an equality. How to show this? This answer does not seem to adress why there's not only an equality.