Royden 4th ed., Chapter 2, #15

Question

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 $E$.

Answer 1

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 $$ \begin{align*} E\setminus E_0&=\left[\left(E\cap\bigcup_{k=1}^N I_k\right)\cup E_0\right]\cap E_0^C\\ &=\left(E\cap\left(\bigcup_{k=1}^N I_k\right)\cap E_0^C\right)\cup \left(E_0\cap E_0^C\right)\\ &=\left(E\cap\left(\bigcup_{k=1}^N I_k\right)\cap E_0^C\right)\cup\emptyset\\ &=\left(E\cap\left(\bigcup_{k=1}^N I_k\right)\cap E_0^C\right)\\ &\subseteq\bigcup_{k=1}^N I_k. \end{align*} $$

Answer 2

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 $< \epsilon$: $$ E = ((-\infty, A) \cap E) \cup \bigcup_j (I_j \cap E) \cup (E \cap [B, +\infty)). \square $$