Countable subadditivity of the Lebesgue measure 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.

This link gives a proof without using monotonicity mentioned above. It is derived from the sense of "giving a room of $\epsilon$".
http://mathonline.wikidot.com/countable-subadditivity-of-the-lebesgue-outer-measure#toc0
