Define by
$$
m^*(A) := \inf\left\{ \sum_i |I| : A \subseteq \bigcup_i I_i \right\}
$$
the outer measure of some set $A \subseteq \mathbb R$. Then we have $m^*(I) = |I|$
for each interval (open, half-open and closed). The following proof is from J.C.Oxtoby: Measure and Category, and I do not understand why it is neccessary that we select a finite cover, i.e. utilise Heine-Borel. First I present his proof, and then a variant which does not uses the selection of a finite subcover.
Proof from the book: The inequality $m^*(I) \le |I|$ is clear, since $I$ covers itself. To prove the inverse inequality, let $\varepsilon$ be an arbitrary positive number and let $\{ I_i \}$ be an open covering of $I$ such that $\sum|I_i| < m^*(I) + \varepsilon$. Let $J$ be a closed subinterval of $I$ such that $|J| > |I| – \varepsilon$. By the Heine-Borel theorem, $J \subseteq \bigcup_{i=1}^k I_i$ for some $k$. Let $K_1, \ldots, K_n$ be an enumeration of the closed intervals into which $\overline I_1, \ldots, \overline I_k$ are divided by all the $(r-1)$-dimensional hyperplanes that contain an $(r-1)$-dimensional face of one of the intervals $I_1, \ldots, I_k$ or $J$, and let $J_1,\ldots, J_m$ be the closed intervals into which $J$ is divided by these same hyperplanes. Then each interval $J_i$ is equal to at least one of the intervals $K_j$. Consequently,
$$
|J| = \sum_{i=1}^m |J_i| \le \sum_{j=1}^n |K_j| \le \sum_{i=1}^k |I_k|
< m^*(I) + \varepsilon
$$
[Remark: In the book is written $\ldots \sum_{j=1}^n |K_j| = \sum_{i=1}^k |I_k| \ldots$, but I think it must be an less than equal, as the other sum might count some intersection areas multiple times, whereas the $K_i$ are disjoint]. Therefore $|I| \le m^*(I) + 2\varepsilon$. The desired inequality follows by letting $\varepsilon \to 0$. $\square$
Now I guess by the usage of Heine-Borel the proof gets complicated, I do not see why it should be necessary? If we just use this subdivision on $\{ I_i \}$ with those hyperplanes, we get a countable number of disjoint intervals too, as each intervall is hit by at most countable many hyperplanes, and therefore is subdivided in at most countable many intervals, and taken all of them (countable many) we got again countable many. So here is my presumably simpler proof (but of course I might have overlooked something, for which I ask you…)
(Simple Proof): Let $\varepsilon > 0$ be given and let $\{ I_i \}$ be a cover by intervals such $\sum_i |I_i| \le m^*(I) + \varepsilon$. Now let $K_1, K_2, K_3, \ldots$ be the intervals into which $\overline I_1, \overline I_2, \ldots$ are divided by the $(r-1)$-dimensinal hyperplanes that contain an $(r-1)$-dimensional face of one of the intervals $I_i$. Then we have
$$
|I| = \sum_i |K_i| \le \sum_i |I_i| \le m^*(I) + \varepsilon
$$
and hence $|I| \le m^*(I) + \varepsilon$, which gives $|I| \le m^*(I)$ for $\varepsilon \to 0$. $\square$
So is anything wrong with this proof?
Best Answer
You are dealing with an infinite number of hyperplanes. Cutting the original intervals by all of these planes may leave you with a bunch of sets that are no longer intervals with positive volume.