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?

  • $\begingroup$ 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

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).

  • $\begingroup$ 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

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