I am stuck understanding Royden's version of the proof that an interval's outer Lebesgue measure is equal to its length, specifically just the first part where we consider a closed, bounded interval $[a,b]$, and want to prove that $m^*([a,b]) = b-a$. (full proof is here)
I understand the first step, which we just consider an arbitrary open interval $(a-\varepsilon, b+\varepsilon)$ and arrive at $m^*([a,b]) \leq b-a$. From here, I agree that if we can show $m^*([a,b]) \geq b-a$, then we have $m^*([a,b]) = b-a$, which is what we want for closed bounded intervals.
For the converse, I understand the application of Heine-Borel on an arbitrary open cover of $[a,b]$ and obtaining the subsequent sum of lengths of the intervals in a finite subcover. We get from these calculations that for any arbitrary finite subcover $\{I_k\}_{k=1}^N$, we have $\sum_{k=1}^N l(I_k) > b-a$.
The connection I'm not making here is why $\sum_{k=1}^N l(I_k) > b-a$ implies $m^*([a,b]) \geq b-a = l([a,b])$. The former is a statement about the sum of lengths over a finite subcover of $[a,b]$, so it seems like if anything, I want to say that $m^*([a,b]) \leq b-a$ since $m^*(A) = \inf\{\sum_k l(I_k)|A\subseteq I_k\}$.
EDIT: for reference, Royden defines Lebesgue outer measure of a set $A$ of real numbers as $m^*(A) = \inf\{\sum_{k=1}^\infty l(I_k) | A \subseteq \bigcup_{k=1}^\infty I_k\}$, where $l$ of an interval is defined to be the difference of its endpoints if the interval is bounded, and $\infty$ otherwise.
Best Answer
$m^*([a,b])$ is defined as the infimum over all covers of $[a,b]$. If for any cover finite cover of $A$ $\sum_{k=1}^N l(I_k)>b-a$. Then: $$m^*([a,b])=\inf \{\sum_{k=1}^\infty l(I_k) | [a,b] \subset \bigcup_k I_k \} \geq \inf \{\sum_{k=1}^N l(I_k)| [a,b] \subset \bigcup_k I_k, N \in \mathbb{N} \} \geq b-a$$ because the infimum is the biggest lower bound.