Outer Measure is Countably Subadditive

[Bashyboy]

Homework Statement

If $\{E_k\}_{k=1}^\infty$ is any countable collection of sets, disjoint or not, then $m^* \left(\bigcup_{k=1}^{\infty} E_k \right) \le \sum_{k=1}^{\infty} m^*(E_k)$.

Homework Equations

$m^*(A) = \inf \{ \sum_{k=1}^{\infty} \ell (I_k) ~|~ A \subseteq \bigcup_{k=1}^{\infty} I_k \}$

Note: each $I_k$ is bounded, open interval.

The Attempt at a Solution

I am working through the proof of the above theorem in Royden and Fitzpatrick's Real Analysis and I am having trouble justifying the following claim they use in their proof: Let $\epsilon > 0$. For each natural number $k$, there is a countable collection $\{I_{k,i}\}_{i=1}^\infty$ of open, bounded intervals for which

$$E_k \subseteq \bigcup_{i=1}^\infty I_{k,i}$$

and

$$\sum_{i=1}^{\infty} \ell (I_{k,i}) < m^*(E_k) + \frac{\epsilon}{2^k}$$

I think they are just appealing to the following fact: Let $A \subseteq \Bbb{R}$ be nonempty; then $i = \inf(A)$ if and only if $\forall \delta > 0$, there exists an $a \in A$ such that $a < i + \delta$. Now since $m^*(E_k)$ is the infimum of certain sums, given $\frac{\epsilon}{2^k}$, there must exist $\{I_{k,i}\}_{i=1}^\infty$ satisfying the above two properties. Is this right?

 

[andrewkirk]

Yes. Let $S$ be the collection of all covers of $E_k$ by countable collections of bounded open intervals. For a cover $C\in S$, let $\phi(C)=\sum_{I\in C}\ell(I)$. Then the $A$ to which you refer is
$$A=\left\{\phi(C)\ :\ C\in S\right\}$$
and $m^*(E_k)=\inf A$. So for any $\epsilon>0$ there must exist $C\in S$ such that $\phi(C)<m^*(E_k)+\frac{\epsilon}{2^k}$, otherwise $\inf A\geq m^*(E_k)+\frac{\epsilon}{2^k}> m^*(E_k)=\inf A$, which is a contradiction.