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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.

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    $\begingroup$ And how would you argue that the outer measure of an open interval $(a, b)$ is (at least) $b-a$? $\endgroup$
    – Martin R
    Commented Jan 31, 2021 at 17:51
  • $\begingroup$ IT is taken as definition (see 1st paragraph of the OP) $\endgroup$
    – Tito Eliatron
    Commented Jan 31, 2021 at 17:52
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    $\begingroup$ @TitoEliatron: No, the definition is the infimum of $\sum_{k=1}^\infty \ell(I_k)$ over all coverings with open intervals $I_k$. $\endgroup$
    – Martin R
    Commented Jan 31, 2021 at 17:54
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    $\begingroup$ @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! $\endgroup$
    – Subhasish Mukherjee
    Commented Jan 31, 2021 at 17:55
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    $\begingroup$ @TitoEliatron: Yes, but it is not immediately obvious that $|I| = \ell(I)$ for an open interval. $\endgroup$
    – Martin R
    Commented Jan 31, 2021 at 18:02

1 Answer 1

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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.

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  • $\begingroup$ 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. $\endgroup$
    – Koro
    Commented May 16, 2023 at 12:34
  • $\begingroup$ @Koro: That is the argument in the second paragraph of the question. It shows that $|[a,b]| \le b-a$. $\endgroup$
    – Martin R
    Commented May 16, 2023 at 12:41
  • $\begingroup$ 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? $\endgroup$
    – Koro
    Commented May 16, 2023 at 12:44
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    $\begingroup$ @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. $\endgroup$
    – Martin R
    Commented May 16, 2023 at 12:52
  • $\begingroup$ I am starting to understand it now. Thanks a lot. $\endgroup$
    – Koro
    Commented May 16, 2023 at 13:27

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