1
$\begingroup$

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 $E$.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Laars Helenius May 9 '14 at 2:53
  • $\begingroup$ 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. $\endgroup$ – Laars Helenius May 9 '14 at 3:19
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.