Finite measure divided into small measures 
Show that if $E$ has finite measure and $\epsilon>0$, then $E$ is the disjoint union of a finite number of measurable sets, each of which has measure at most $\epsilon$.

First, I'm quite confused with the problem statement. If it's true, it means any set $E$ with finite measure is the disjoint union of a finite number of measurable sets, and hence must be measurable. But any set with finite measure might not be measurable, or is it?

Let $n$ be such that $1/n<\epsilon/2$. For $z\in\mathbb{Z}$, let $I_z=[z/n,(z+1)/n)$. Then $E=\bigcup_{z\in\mathbb{Z}}E\cap I_z$ and hence $$\mu(E)=\sum_{z\in\mathbb{Z}}\mu(I_z\cap E)=\sum_{n=0}^\infty\mu(I_n\cap E)+\sum_{n=0}^\infty\mu(I_{-n-1}\cap E)$$ $$=\sum_{n=0}^\infty\Big(\mu(I_n\cap E)+\mu(I_{-n-1}\cap E)\Big).$$ Since this series converges, $\lim_{n\to\infty}\mu(I_n\cap E)+\mu(I_{-n-1}\cap E)=0$. So for $N$ large enough,
$$\sum_{n=N}^\infty\Big(\mu(I_n\cap E)+\mu(I_{-n-1}\cap E)\Big)<\epsilon.$$
It follows that you can partition $E$ into the sets $(I_n\cup I_{-n-1})\cap E$ for $n<N$ and $$\bigcup_{n=N}^\infty \Big((I_n\cup I_{-n-1})\cap E\Big).$$
