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.