Royden 4th ed., Chapter 2, #15 Prepping for a comprehensive test in August and I am working on a problem from Royden 4th ed. (Chapter 2, #15): Show that if $E$ has finite measure and $\varepsilon>0$, 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 have so far:
Let $\varepsilon>0$ and let the Lebesgue outer measure of a set $E$, $m^*(E)$, be positive and finite. Let $\{I_k\}_{k=1}^\infty$ be a countable collection of bounded open intervals that cover $E$ such that
$$
\sum_{k=1}^\infty\ell(I_k)\le m^*(E)+\varepsilon.
$$
Since our sum is convergent, there exists $N\in\mathbb{N}$ such that 
$$
\sum_{k=N+1}^\infty\ell(I_k)\le\varepsilon.
$$
Then define 
$$
E_0=E\cap\bigcup_{k=N+1}^\infty I_k.
$$
Since $E_0\subseteq\bigcup_{k=N+1}^\infty I_k$, we have
$$
m^*(E_0)=m^*\left(E\cap\bigcup_{k=N+1}^\infty I_k\right)\le m^*\left(\bigcup_{k=N+1}^\infty I_k\right)\le\sum_{k=N+1}^\infty\ell(I_k)<\varepsilon.
$$
Here is where I am stuck:
It seems intuitive that if $E_0$ is covered by $\bigcup_{k=N+1}^\infty I_k$, then $E\setminus E_0$ is covered by $\bigcup_{k=1}^N I_k$. But I cannot figure out the proof.
If this is true, then I can finish the proof: $E\setminus E_0$ is covered by a finite number of bounded intervals, so there exists some interval $[a,b)$ such that $\bigcup_{k=1}^N I_k\subseteq [a,b)$. Then all I do is choose $M$ large enough so that $M\varepsilon>b-a$ and then subdivide $[a,b)$ into intervals of width $(b-a)/M<\varepsilon$. Then I intersect $E\setminus E_0$ with each of these intervals and union them all together (along with $E_0$) to get a finite union of disjoint intervals with width less than $\varepsilon$ that is equal to $E$.
 A: I have another solution, which seems [at least] easier for me. Of course the knowledge I used exceeds the scope of that specific section. Any comments and criticism are welcome. 
Consider a function
$$
\newcommand{\abs}[1]{\left\vert #1 \right\vert}
\newcommand\rme{\mathrm e}
\newcommand\imu{\mathrm i}
\newcommand\diff{\, \mathrm d}
\DeclareMathOperator\sgn{sgn}
\renewcommand \epsilon \varepsilon
\newcommand\trans{^{\mathsf T}}
\newcommand\F {\mathbb F}
\newcommand\Z{\mathbb Z}
\newcommand\R{\Bbb R}
\newcommand \N {\Bbb N}
\newcommand \k {\Bbbk}
\renewcommand\geq\geqslant
\renewcommand\leq\leqslant
\newcommand\bm\boldsymbol
\newcommand\stpf\blacktriangleleft
\newcommand\qed\blacktriangleright
\newcommand\upint{\mspace{9mu}\overline {\vphantom \int\mspace{9mu}}\mspace{-18mu}{\int}}
\newcommand\lowint{\mspace{1mu} \underline{
\vphantom \int \mspace {9mu}} \mspace {-10mu}
\int}
f(x) = m(E \cap (-\infty, x)), x \in \R. 
$$
Then $f (-\infty) = 0, f(+\infty ) = m(E)$. We claim that $f \in \mathcal C \R$. For $h > 0$, note that for each $x \in \R$, 
$$
f(x+h) - f(x) = m((-\infty, x+h) \cap E) - m((-\infty, x) \cap E), 
$$
and note that
$$
(-\infty, x+h )\cap E = ((-\infty, x) \cap E) \cup ([x,x+h) \cap E) \subseteq ((-\infty, x) \cap E) \cup [x,x+h) 
$$
thus by sub-additivity and monotonicity, 
$$
f(x) \leq f(x + h) \leq f(x) + m([x,x+h)\cap E) \leq f(x) + h. 
$$
Thus $\lim_{h \to 0^+} f(x+h) - f(x) = 0$. Similarly $\lim_{h \to 0^-} f(x+h) - f(x) = 0$. Therefore $f$ is continuous at $x$. Now we can apply the Intermediate Value Theorem to $f$, and conclude that there is some $A \in \R$ s.t. $f(A) = \min (\epsilon /2, m(E)/4)$. 
Using the similar method, we could also find some $B > A$ that $m( E \cap [B , +\infty)) = \min (\epsilon /2, m(E)/4)$. The rest is to partition the interval $[A, B]$ into subintervals $I_j$ [left close right open] with the same length $\epsilon / 2$, and clearly $I_j \cap E $ has measure $\leq m(I_j) = \epsilon /2 < \epsilon$. Obviously the decomposition is finite. Putting all these together we obtain a finite disjoint union of measurable sets, each of which has measure $< \epsilon$:
$$
E = ((-\infty, A) \cap E) \cup \bigcup_j (I_j \cap E) \cup (E \cap [B, +\infty)).  \square
$$
A: I figured it out:
Since $E\subseteq\bigcup_{k=1}^\infty I_k$, then 
$$
\begin{align*}
E&=E\cap\bigcup_{k=1}^\infty I_k\\
&=E\cap\left(\bigcup_{k=1}^N I_k\cup\bigcup_{k=N+1}^\infty I_k\right)\\
&=\left(E\cap\bigcup_{k=1}^N I_k\right)\cup \left(E\cap\bigcup_{k=N+1}^\infty I_k\right)\\
&=\left(E\cap\bigcup_{k=1}^N I_k\right)\cup E_0.
\end{align*}
$$
Thus
$$
\begin{align*}
E\setminus E_0&=\left[\left(E\cap\bigcup_{k=1}^N I_k\right)\cup E_0\right]\cap E_0^C\\
&=\left(E\cap\left(\bigcup_{k=1}^N I_k\right)\cap E_0^C\right)\cup \left(E_0\cap E_0^C\right)\\
&=\left(E\cap\left(\bigcup_{k=1}^N I_k\right)\cap E_0^C\right)\cup\emptyset\\
&=\left(E\cap\left(\bigcup_{k=1}^N I_k\right)\cap E_0^C\right)\\
&\subseteq\bigcup_{k=1}^N I_k.
\end{align*}
$$
