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 $E$.

• Perhaps if you read more carefully, starting with "Here is where I am stuck:" and then the sentence that follows it? But never mind, I solved it earlier today. Thanks for the comment though. Jeez. – Laars Helenius May 9 '14 at 2:53
• OK. Thanks for the clarification. Your comment came off as very snarky. Apologies. I have edited the question to reflection the original statement. The only thing I need to clarify is making the arbitrary intervals disjoint intervals, which is always possible in a $\sigma$-algebra. – Laars Helenius May 9 '14 at 3:19

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*}