8
$\begingroup$

Reading Royden's fourth edition of Real Analysis. I'm working with outer measure defined as

$$m^*(E)=\inf\left\{\sum_{n=1}^\infty l(I_n):\,E\subset \bigcup_{n=1}^\infty I_n\right\},$$

where each $I_n$ is a bounded, open interval. Also, $E$ is measurable if and only if

$$m^*(A)=m^*(A\cap E)+m^*(A\cap E^C),$$

for every set $A$.

In reading the proof of Theorem 11 on page 40, I start with $E$ a measurable set. Then I suddenly read the statement: "Consider the case where $m^*(E)=\infty$. Then $E$ may be expressed as the disjoint union of a countable collection $\{E_k\}_{k=1}^\infty$ of measurable sets, each of which has finite outer measure.

I am stuck on this last sentence. How come this is true?

$\endgroup$
  • $\begingroup$ Is $E$ assumed to be measurable from the beginning? $\endgroup$ – Giuseppe Negro Jun 26 '13 at 21:01
  • $\begingroup$ Yes, E is measurable. $\endgroup$ – David Jun 26 '13 at 21:06
  • $\begingroup$ I've added this fact (E is measurable) to the question above. $\endgroup$ – David Jun 26 '13 at 21:15
6
$\begingroup$

This is false unless we assume that $E$ is measurable. We can construct nonmeasurable sets which have outer measure $\infty$ which contain no measurable set of positive measure (take a Bernstein set, where both the set $B$ and its complement have nonempty intersection with every uncountable closed set). Let $B$ be such a set.

Suppose $B = \bigcup_{i=1}^\infty E_i$ is the disjoint union of countably many measurable sets. Then the only possibility for $E_i$ are sets of measure $0$ (because $E_i \subset B$), which means that $B$ has measure $0$, which is clearly a contradiction.


If $E$ is measurable, then we can just take $E_i = E \cap ( i,i+1 ]$, which is the intersection of two measurable sets. $E = \bigsqcup_{i \in \mathbb Z} E_i$ ($\sqcup$ denotes disjoint union).

$\endgroup$
  • $\begingroup$ I know that this was answered a long time ago, but I'm also stuck at this. Why are we sure that if $x\in E$, then there exists an $i \in \mathbb{Z}$ such that $x \in (i, i+1]$? $\endgroup$ – Kurome Mar 5 '16 at 9:47
  • $\begingroup$ @Kurome The ceiling function $\lceil x \rceil$ has the property that $\lceil x \rceil - 1 < x \le \lceil x \rceil$. $\endgroup$ – A.S Mar 5 '16 at 19:07
  • $\begingroup$ This seems like general result. Isn't this also true even if $m(E)< +\infty$? $\endgroup$ – Kurome Mar 6 '16 at 2:55
  • $\begingroup$ @Kurome yes it's a general result, but it's trivial that if $E$ has finite measure, then $E$ can be decomposed into finitely many sets of finite measure. The only interesting case is the case where $m(E) = +\infty$. Perhaps I misunderstand what you're getting at. $\endgroup$ – A.S Mar 6 '16 at 3:00
  • $\begingroup$ No that's enlightening. So if $m(E)<+\infty$, we can rewrite $E$ only for finite number of $E_i = E \cap (i,i+1]$? Because of we let $i$ run to all $\mathbb{Z}$ we'll get a contradiction where $E= \infty$, is that right? $\endgroup$ – Kurome Mar 6 '16 at 3:07

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.