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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.

Michael Hardy
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Subhasish Mukherjee
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    And how would you argue that the outer measure of an open interval $(a, b)$ is (at least) $b-a$? – Martin R Jan 31 '21 at 17:51
  • IT is taken as definition (see 1st paragraph of the OP) – Tito Eliatron Jan 31 '21 at 17:52
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    @TitoEliatron: No, the definition is the infimum of $\sum_{k=1}^\infty \ell(I_k)$ over all coverings with open intervals $I_k$. – Martin R Jan 31 '21 at 17:54
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    @MartinR I've been ruminating on this for more than a day, and your comment just made me realize it! I was missing something obvious. It wasn't explicitly proved that the outer measure of $(a,b)$ is at least $b-a$, and I was conflating it with length. When I thought of how to prove it, I realized I couldn't do it directly and would need some more machinery (such as Heine-Borel). Thank you very much! – Subhasish Mukherjee Jan 31 '21 at 17:55
  • @MartinR "where ℓ(I) for an open interval (a,b) is just b−a" – Tito Eliatron Jan 31 '21 at 18:00
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    @TitoEliatron: Yes, but it is not immediately obvious that $|I| = \ell(I)$ for an open interval. – Martin R Jan 31 '21 at 18:02
  • Ok! now I have my epiphany! – Tito Eliatron Jan 31 '21 at 18:03
  • Some mathematician in the late 19th century published a "proof" that the outer measure of [0,1] is 0. – DanielWainfleet Jan 31 '21 at 22:10

1 Answers1

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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.

Martin R
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  • I have one confusion in this. Could you please help me understand why the following doesn't work?: I claim that $|(a,b)|=b-a$. Proof: Given any $\epsilon>0$, choose $A_1= (a-\frac{\epsilon}4, b+\frac{\epsilon}4), A_k=\emptyset , \forall k\ge 2$. Then, clearly $I\subset \cup_k A_k$. Moreover, $b-a\le \sum_k l(A_k)=b-a+\frac{\epsilon}2<\color{red}{b-a}+\epsilon$. This satisfies the infimum definition hence the result follows. – Koro May 16 '23 at 12:34
  • @Koro: That is the argument in the second paragraph of the question. It shows that $|[a,b]| \le b-a$. – Martin R May 16 '23 at 12:41
  • I don't understand. The second paragraph takes an open cover of [a,b] and then proceeds. I'm taking open cover of (a,b). I am using the definition of infimum and then using the fact that infimum is unique. $|A|=\inf{\sum_k l(I_k): A\subset \cup I_k}, I_k$'s are open intervals. I claim that |(a,b)|= b-a. If I can show that for every $\epsilon>0$, there are $I_k$'s such that $(a,b)\subset \cup I_k$ with the property that $\sum_k l(I_k)<b-a+\epsilon$, then the proof is complete. Isn't it? – Koro May 16 '23 at 12:44
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    @Koro: That does not show that there is not some other cover of $(a, b)$ such that $\sum \ell(I_k)$ is strictly less than $b-a$. So you still have to show that $|(a, b)| \ge b-a$. For that part one needs the Heine-Borel theorem, i.e. the completeness of the real numbers. – Martin R May 16 '23 at 12:52
  • I am starting to understand it now. Thanks a lot. – Koro May 16 '23 at 13:27
  • @Koro: Here is a demonstration that this fails for intervals of rational numbers, so the completeness is really needed: https://math.stackexchange.com/a/948585/42969 – Martin R May 16 '23 at 13:31