Let $\lbrace F_n \rbrace$ be a sequence of sets in a $\sigma$-algebra $\mathcal{A}$. I want to show that $m\left(\bigcup F_n\right)\leq \sum m\left(F_n\right)$ where $m$ is a countable additive measure defined for all sets in a $\sigma$ algebra $\mathcal{A}$.
I think I have to use the monotonicity property somewhere in the proof, but I don't how to start it. I'd appreciate a little help.
Thanks.
Added: From Hans' answer I make the following additions. From the construction given in Hans' answer, it is clear the $\bigcup F_n = \bigcup G_n$ and $G_n \cap G_m = \emptyset$ for all $m\neq n$. So $m\left(\bigcup F_n\right)=m\left(\bigcup G_n\right) = \sum m\left(G_n\right).$ Also from the construction, we have $G_n \subset F_n$ for all $n$ and so by monotonicity, we have $m\left(G_n\right) \leq m\left(F_n\right)$. Finally we would have $\sum m(G_n) \leq \sum m(F_n).$ and the result follows.