TITLE: Is always true that: $ v(\cup_{n=1}^{\infty} A_n) \leq \sum\limits_{n=1}^{\infty} v(A_n)$?
QUESTION [11 upvotes]: Let $v : \mathcal {P}(\mathbb R) \to [0,\infty]$ a set-function such that $\displaystyle{ v(F \cup E) \leq v(E) + v(F)  \quad \forall E,F \subset \mathbb R}$.
Is always true that: $ \displaystyle{ v\left(\bigcup_{n=1}^{\infty} A_n\right) \leq \sum_{n=1}^{\infty} v(A_n)}$ ?
I think this is not but I can't find a counterexample. Any help?
Thank's in advance!

REPLY [3 votes]: The purpose of that question is to show that if $\nu$ subadditive it isn't implies that $\nu$ $\sigma$-additive. The answers above shows that fact.
It's well known that, if $\nu$ is a measure, then:
$$ \nu  \ \sigma\text{-additive} \ {\Longrightarrow} \  \nu  \ \sigma\text{-subadditive} \ {\Longrightarrow} \ \nu  \ \text{subadditive}  $$