TITLE: If $m^*(E)=\infty$, then $E=\bigcup_{k=1}^{\infty}E_k$, $E_k$ measurable and $m^*(E_k)<+\infty$
QUESTION [9 upvotes]: 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?

REPLY [7 votes]: 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).