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This comes from Royden and Fitzpatrick's Real Analysis:

Proposition 3 If $\{E_k\}_{k=1}^\infty$ is any countable collection of sets, disjoint or not, then

$m^*(\bigcup_{k=1}^\infty E_k) \le \sum_{k=1}^\infty m^*(E_k)$

Note, $m^*(A)$ denotes the outer measure of $A$. There are a few parts of the proof I am having trouble with. Here is the first

For each natural number $k$, there is a countable collection $\{I_{k,i} \}_{i=1}^\infty$ of open, bounded intervals for which $E_k \subseteq \bigcup_{i=1}^\infty I_{k,i}$.

This seems to rely on there being a open cover for any set of real numbers. Is this true because $\mathbb{R}$ can be written as $\bigcup_{n=1}^\infty (-n,n)$, a countable union of open, bounded sets?

Here is another part of the proof giving me trouble:

Now, $\{I_{k,i} \}_{k,i=1}^\infty$ is a countable collection of open, bounded intervals that cover $\bigcup_{k=1}^\infty E_k$: the collection is countable since it is a countable collection of countable collections. Thus, by definition of the outer measure,

$m^* \left( \bigcup_{k=1}^\infty E_k \right) \le \sum_{k,i=1}^\infty \ell (I_{k,i}) =...$

How did they obtain this inequality? The most I can get is

$m^* \left( \bigcup_{k=1}^\infty E_k \right) \le m^* \left( \bigcup_{k,i=1}^\infty I_{k,i} \right)$,

by the monotone property. Perhaps we need to show that $m^* \left( \bigcup_{k,i=1}^\infty I_{k,i} \right) \le \sum_{k,i=1}^\infty \ell(I_{k,i})$, which would be a special case of the theorem we are trying to prove.

EDIT: This is in response to Dunham's answer, but everyone else is at liberty to chime in:

Regarding part 1, yes it might be only one example of a collection, but it suffices to prove that every set of real numbers has an open cover, right? Regarding part 2, I know that the definition of the outer measure involves the infimum, that's precisely what is confusing me. Why wouldn't we have $\sum_{k,i=1}^\infty \ell (I_{k,i}) \le m^* \left( \bigcup_{k=1}^\infty E_k \right)$ instead?

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    $\begingroup$ $m^{\ast}(\bigcup \dotsc) = \inf S$, and $\sum \ell(I_{k,i})$ is an element of $S$. Hence $m^{\ast}(\bigcup\dotsc) \leqslant \sum \ell(I_{k,i})$. $\endgroup$ – Daniel Fischer Oct 25 '16 at 21:23
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Part 1: You are essentially correct. You have shown one example of a collection of intervals covering $E_k$.

Part 1-Edit: If you look at the proof in the book, they actually ask for an open cover satisfying an additional property:

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

The existence comes from the definition

$m^*(E_k)= \inf \left\{ \sum_{k=1}^\infty \ell(\bar{I}_{k,i}) | E_k \subset \cup_{i=1}^{\infty} \bar{I}_{k,i} \right\}$.

Recall that if the infimum of a set of numbers is $a$, then something between $a$ and $a+\epsilon$ must be in the set.

Part 2: Look back to the definition of an outer measure. It is the infimum, over all coverings by intervals, of the sum of the lengths of those intervals. $\{I_{k,i}\}$ is one example of a covering, hence the inequality.

Part 2-Edit Every element of a set is at least as large as the infimum. $\{I_{k,i}\}_{k,i}$ is a cover of $\cup_{k} E_k$, so $\sum_{k,i} \ell(I_{i,k})$ is an element of the set in the definition of $m^*(\cup_{k} E_k)$.

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  • $\begingroup$ Regarding part 1, yes it might be only one example of a collection, but it suffices to prove that every set of real numbers has an open cover, right? Regarding part 2, I know that the definition of the outer measure involves the infimum, that's precisely what is confusing me. Why wouldn't we have $\sum_{k,i=1}^\infty \ell (I_{k,i}) \le m^* \left( \bigcup_{k=1}^\infty E_k \right)$ instead? $\endgroup$ – user193319 Oct 25 '16 at 19:56
  • $\begingroup$ I added edits to my answer. I can try to explain further if something is unclear. $\endgroup$ – Dunham Oct 25 '16 at 22:30

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