In Sheldon Axler's book, Measure Integration, and Real Analysis, he defines outer measure of a set as $|A| = \inf\big\{\sum_{k=1}^\infty \ell(I_k): I_1, I_2, \dots \text{are open intervals such that} A\subset \bigcup_{k=1}^\infty I_k\big\}$, where $\ell(I)$ for an open interval $(a,b)$ is just $b-a$. He later proves that outer measure preserves order, i.e. $A\subset B \Rightarrow |A| \le |B|$.
Later, we are trying to prove that the outer measure of the closed interval $[a,b]$ is $b-a$. We bound it from above by saying for $\varepsilon > 0$, $(a-\varepsilon, b+\varepsilon), \varnothing, \varnothing,\dots$ is a sequence of open intervals whose union contains $[a,b]$, so $|[a,b]|\le b-a+2\varepsilon$ which with the definition of outer measure implies $|[a,b]| \le b-a$. The next section is confusing to me:
Is the inequality in the other direction obviously true to you? If so,
think again, because a proof of the inequality in the other direction
requires that the completeness of $\mathbf{R}$ is used in some form…Thus
something deeper than you might suspect is going on with the
ingredients needed to prove that $|[a, b]| ≥ b − a$.
He then goes onto prove it using the Heine-Borel theorem. However, because outer measure preserves order and $(a,b)$ is a subset of $[a,b]$, couldn't we easily bound it from below with that? Is the open interval not thought of as a subset? I don't quite understand the reasoning and feel I'm missing something obvious. Any help would be appreciated.
Best Answer
You are arguing that $$ |[a, b]| \ge |(a, b)| \ge b-a \, , $$ but the right inequality needs to be justified. We know that $\ell((a, b)) = b-a$, but it not obvious from the definition that $|(a, b)| = \ell((a, b))$.
It is in fact easier to prove $$ |[a, b]| \ge b-a $$ first, because any open covering $\bigcup_{k=1}^\infty I_k$ of the compact interval $[a, b]$ contains a finite sub-covering, that is where the Heine-Borel theorem comes into play.