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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.

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  • $\begingroup$ How is $m^*$ defined? $\endgroup$
    – Ben S.
    Mar 14 '21 at 18:29
  • $\begingroup$ whoops, added the definition as an edit $\endgroup$
    – ecy
    Mar 14 '21 at 19:12
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$m^*([a,b])$ 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.

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  • $\begingroup$ Thanks for the response! I think I'm almost there, but I'm still not following the last inequality - the linked proof (equivalent to Royden's proof) gets us to $\sum_{k=1}^N l(I_k) > b-a$ for any finite subcover, but how is this strict inequality turned into $\geq$ as you stated? $\endgroup$
    – ecy
    Mar 14 '21 at 19:37
  • $\begingroup$ Because if you have a set $X$ and you know that for all $x$ in a set $X$ you have a lower bound $c$ such that $x>c$, then the infimum $\inf X$ must not be strictly larger than $c$ it can be equal to it. For example consider $X:=\{\frac{1}{n} | n \in \mathbb{N} \}$, then for all $x \in X$, $x>0$ but $\inf X = 0 $ $\endgroup$
    – Ben S.
    Mar 14 '21 at 20:01
  • $\begingroup$ Your example makes sense to me in the following way: for the set $\{ \frac{1}{n} | n \in N \}$, we know the inf is 0. For this case, if it were any other positive $\varepsilon$, I know that by the archimedean property, I can find an $n$ such that $\frac{1}{n} < \varepsilon$, which is a contraditiction. It is this argument that nets us the fact that inf(X) = 0. However, absent an argument for why we can make $\sum_k l(I_k)$ for some finite subcover $\{I_k\}$ arbitrarily close to $b-a$, it seems I can't use the same argument here. $\endgroup$
    – ecy
    Mar 15 '21 at 1:14
  • $\begingroup$ I'm not sure if I understand your question correctly. Your original question was asking why the lower bound $b-a$ on the sum of lengths of a finite covers implies a lower bound on $m^*([a,b])$. Now your asking why $b-a$ is a lower bound at all. $\endgroup$
    – Ben S.
    Mar 15 '21 at 1:42
  • $\begingroup$ Sorry, maybe I'm not expressing myself well. I followed your answer halfway - I'm convinced that $m^*([a,b]) > b-a$, but still not that $m^*([a,b]) \geq b-a$, which is what I believe we need, combined with $m^*([a,b]) \leq b-a$ from earlier, to deduce $m^*([a,b]) = b-a$ $\endgroup$
    – ecy
    Mar 15 '21 at 1:51
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Let $E$ be the set of families of open intervals of open intervals that cover $[a,b].$ Let $F$ be the set of finite families of open intervals of open intervals that cover $[a,b].$

For any $e\in E$ there is an $f\in F$ with $f\subseteq e.$ So for any $e\in E$ there is an $f\in F$ with $$\sum_{j\in e}l(j)\ge \sum_{j\in f}l(j).$$ So we have $$\inf \, \{\sum_{j\in e}l(j):e\in E\}\ge \inf \, \{\sum_{j\in f}l(j): f\in F\}.$$

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  • $\begingroup$ Hi! Thanks for the response, but I have the same follow-up question as I had for Benjamin. $\endgroup$
    – ecy
    Mar 15 '21 at 1:18
  • $\begingroup$ Let $M=\{\sum_{f\in F}l(f):f\in F\}.$ Then $\emptyset\ne M\subseteq (b-a,\infty),$ so $\mu^*([a,b])\ge\inf M\ge b-a.$ On the other hand $\mu^*([a,b])\le \inf \{l((a-r,b+r)):r>0\}=\inf \{b-a+2r: r>0\}=b-a.$ So we have $b-a\le\mu^*([a,b])\le b-a.$ $\endgroup$ Mar 16 '21 at 5:06

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