I'm taking a course in Real Analysis covering Measure Theory. We are using Real Analysis by Royden/Fitzpatrick as the text. I'm going through the book and I'm struggling to understand the approach of Proposition 1 which states: The outer measure of an interval is it's length. This makes me think I'm misunderstanding the definition of the outer measure.
I understand the first part which tells us that if we have an interval $[a,b]$, let $\epsilon > 0$ be given. Then $ [a,b] \subseteq (a- \epsilon, b+ \epsilon)$ so $m^*([a,b]) \leq \ell((a- \epsilon, b+ \epsilon) = b - a + 2\epsilon$. Since this holds for any $\epsilon > 0$ then $m^*([a,b]) \leq b-a$.
My question is: Why can't we use the same trick to show the other inequality? We have that $(a - \epsilon, b + \epsilon)$ is a cover of $[a,b]$. Then $\ell(a-\epsilon ,b+\epsilon) = b - a + 2\epsilon \geq b-a$. Since the outer measure, in this case, is defined as $$m^*([a,b]) = \inf \left\{ \sum_{k=1}^{\infty} \ell (I_k) : [a,b] \subseteq \cup_{k=1}^{\infty} I_k \right\},$$ then why can't we say $b-a + 2\epsilon \geq m^*([a,b]) \geq b-a$, by virtue of $m^*([a,b])$ being defined as the infimum?
We have an open cover that is bounded below by $b-a$, so can't we say that the the infimum over all open covers is bounded below by $b-a$? The textbook takes a different approach using the compactness of the interval. I understand the proof of that. I just don't understand we we have to take that approach. This makes me think I'm misunderstanding something subtle about the definition of the outer measure. If someone could help clarify the definition that would be greatly appreciated!