Prob. 15, Chap. 2, in Royden's REAL ANALYSIS: For any set of positive measure and $\epsilon > 0$, there are finitely many disjoint measurable sets ... Here is Prob. 15, Chap. 2, in the book Real Analysis by H.L. Royden and P.M. Fitzpatrick, 4th edition:

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 $\epsilon$.

Here $E \subset \mathbb{R}$ of course, and $m^*(E) < \infty$, where $m^*(E)$ is the infimum of the the set of all the sums of the form $\sum_{k = 1}^\infty l \left( I_k \right)$, where $\left\{ I_k \right\}_{k=1}^\infty$ is a countable collection of non-empty, bounded open intervals which cover $E$ and, for each $k$, $l \left( I_k \right)$ denotes the length of the interval $I_k$.
How to proceed from here?
 A: For each $n\in \Bbb Z$, let
$$ S_n := E\cap \big[ n\epsilon, (n+1)\epsilon \big) .$$
Then $$+\infty > m^*(E)=\sum_{n\in \Bbb Z} m^* \left( S_n \right)= \lim_{p\to \infty} \sum_{n \in \mathbb{Z}, \lvert n \rvert \leq p} m^* \left( S_n \right). $$
So take $p\in \Bbb N$ large enough that
$$ m^*(E)-\epsilon \leq \sum_{ n \in \mathbb{Z}, \lvert n \rvert \leq p} m^* \left( S_n \right) \leq m^*(E). $$
Let
$$ \mathscr{B} := \left\{ S_n \colon n \in \mathbb{Z}, \lvert n \rvert \leq  p \right\}, $$ and let
$$ \mathscr{A} :=  \mathscr{B} \bigcup \left\{ E\setminus \cup_{B \in \mathscr{B} } B  \right\}.$$
Then $\mathscr{A}$ is a finite pairwise-disjoint measurable family with
$$ \bigcup_{A \in \mathscr{A}} A =E ,$$ and $$ A \in \mathscr{A} \implies m(A) \le \epsilon.$$
A: The statement is trivial if $E$ is an interval. For example, if $E = (a, b),$ then we may write $$E = \bigcup_{0 \leq k \leq \lfloor \frac{b - a}{\varepsilon} - 1 \rfloor} (a + k \varepsilon, a + (k + 1) \varepsilon] \cup (a + \lfloor \frac{b - a}{\varepsilon}\rfloor \varepsilon, b).$$ Thus the statement is true for intervals. The same can be done for any finite union of intervals, since this can be written as a finite union of disjoint intervals.
Assume now that $E$ is a general measurable set with finite measure and let $\varepsilon > 0$. Then let $\bigcup_{1 \leq n \leq N} I_n$ such that $m(E \Delta \bigcup_{1 \leq n \leq N} I_n) < \varepsilon.$ Cover the union of intervals as discussed in the previous paragraph. Note that $E \Delta \bigcup_{1 \leq n \leq N} I_n = (E \setminus \bigcup_{1 \leq n \leq N} I_n) \cup(\bigcup_{1 \leq n \leq N} I_n \setminus E)$ and this has measure smaller than $\varepsilon.$ Then so does $E \setminus \bigcup_{1 \leq n \leq N} I_n.$ Then a covering such as the one you desire is given by the sets covering $\bigcup_{1 \leq n \leq N} I_n$ intersected with $E$ afterwards (of course, the measure is still less than $\varepsilon$) and $E \setminus \bigcup_{1 \leq n \leq N} I_n.$ I hope this helps. :)
