I'm trying to prove that if $E \subset \mathbb{R}$ has finite measure and $\varepsilon \in \mathbb{R}$ such that $\varepsilon>0$, then $E$ is the disjoint union of a finite number of measurable sets, each of which has measure at most $\varepsilon$.
Here is what I've done:
Let $\varepsilon \in \mathbb{R}$ such that $\varepsilon>0$. Since $m^{*}(E) < +\infty$, there exists a countable collection $\{I_{k}\}_{k \in \mathbb{Z}^{+}}$ of open and bounded intervals (measurable sets) covering $E$ and such that $\sum \limits_{k=1}^{\infty } \ell ( I_{k})
If $x\in \mathbb{R}$, then we define $N( x,\varepsilon) := (x-\varepsilon/2,x+\varepsilon/2)$. For each $k\in \mathbb{Z}^{+}$ the collection $\{N( x,\varepsilon ) \colon x\in F_{k}\}$ is an open covering of $F_{k}$. The compactness of each $F_{k}$ implies that there exist $x_{1}^{k},\ldots ,x_{n_{k}}^{k}\in F_{k}$, with $n_{k}\in \mathbb{Z}^{+}$, such that $F_{k}\subseteq \bigcup _{i=1}^{n_{k}}N(x_{i}^{k},\varepsilon ) $. Thus $E\subseteq \cup _{k\in \mathbb{Z}^{+}}G_{k}\subseteq \cup _{k\in \mathbb{Z}^{+}}F_{k}\subseteq \cup _{k\in \mathbb{Z}^{+}}(\cup_{i=1}^{n_{k}}N( x_{i}^{k},\varepsilon ))$ ...
I can't figure out how to make this last union over $k$ finite. I suppose I've to use the fact that $m^*(G)
On the other hand, I think that if we remove the hypothesis of finite outer measure on $E$, then the conclusion is the same but with a countably infinite union. Am I right?
Thank you in advance.