# Lebesgue outer measure - Trouble understanding proof that outer Lebesgue measure of an interval is its length

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.

• How is $m^*$ defined? Mar 14 '21 at 18:29
• whoops, added the definition as an edit
– ecy
Mar 14 '21 at 19:12

$$m^*([a,b])$$ 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.

• 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?
– ecy
Mar 14 '21 at 19:37
• 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 $\inf X = 0$ Mar 14 '21 at 20:01
• 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.
– ecy
Mar 15 '21 at 1:14
• 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. Mar 15 '21 at 1:42
• 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 $m^*([a,b]) = b-a$
– ecy
Mar 15 '21 at 1:51

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 $$\inf \, \{\sum_{j\in e}l(j):e\in E\}\ge \inf \, \{\sum_{j\in f}l(j): f\in F\}.$$

• Hi! Thanks for the response, but I have the same follow-up question as I had for Benjamin.
– ecy
Mar 15 '21 at 1:18
• 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 $b-a\le\mu^*([a,b])\le b-a.$ Mar 16 '21 at 5:06