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.

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If this is homework please tag it as such. –  Hans Parshall Oct 7 '11 at 18:47
This is not homework... –  Jack Oct 7 '11 at 18:51
Sorry Jack; many questions phrased similarly are homework. –  Hans Parshall Oct 7 '11 at 18:52
I know this question is old, but it is the first result when searching for this topic. Why can we assume in this proof that $m\left(\bigcup G_n\right) = \sum m\left(G_n\right)$, just because the $G_i$ are pairwise disjoint? –  mb7744 Dec 11 '14 at 15:56

Given a union of sets $\bigcup_{n = 1}^\infty F_n$, you can create a disjoint union of sets as follows.
Set $G_1 = F_1$, $G_2 = F_2 \setminus F_1$, $G_3 = F_3 \setminus (F_1 \cup F_2)$, and so on. Can you see what $G_n$ needs to be?
Using $m(\bigcup_{n = 1}^\infty G_n)$ and monotonicity, you can prove $m(\bigcup_{n = 1}^\infty F_n) \leq \sum_{n = 1}^\infty m(F_n)$.
I can instead see that $G_{n+1}=F_{n+1}$ \ $\bigcup F_k$. I'm not sure about $G_n$. –  Jack Oct 7 '11 at 19:19
What does $k$ range over? If you know a general form for $G_{n + 1}$, what is different about $G_n$? –  Hans Parshall Oct 7 '11 at 19:20
$k$ goes from $1$ to $n$. Then I suppose $G_n = F_n$\ $\bigcup F_k$ where $k$ goes from $1$ to $n-1$. right? –  Jack Oct 7 '11 at 19:24