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Notation: The length of any interval $I$ is denoted by $|I|.$

In the book Measure and Category by Oxtoby, in Chapter $1$ Measure and Category on real line, he states the following theorem and proof.

Theorem $1.5$ (Borel) If a finite or infinite sequence of intervals $I_n$ covers an interval $I,$ then $\sum|I_n| \geq |I|.$

Proof: Assume first that $I=[a,b]$ is closed and that all of the intervals $I_n$ are open. Let $(a_1,b_1)$ be the first interval that contains $a.$ If $b_1\leq b,$ let $(a_2,b_2)$ be the first interval of the sequence that contains $b_1.$ If $b_{n-1}\leq b,$ let $(a_n,b_n)$ ne the first interval that contains $b_{n-1}.$ This procedure must terminate with some $b_N>b.$ Otherwise the increasing sequence $\{b_n\}$ would converge to a limit $x\leq b,$ and $x$ would belong to $I_k$ for some $k.$ All but a finite number of the intervals $(a_n,b_n)$ would have to precede $I_k$ in the given sequence, namely, all those for which $b_{n-1}\in I_k.$ This is impossible, since no two of these intervals are equal.

I have two questions:

$(1)$ What is the meaning of 'first interval' in the proof?

For example, let $[a,b]=[0,1].$ The two intervals $(-1,0.5)$ and $(-0.1, 1)$ contain $a=0.$ Which interval should we choose for $(a_2,b_2)?$

$(2)$ I do not understand how author conclude the bold statement in the proof.

My understanding is that since $x$ is the limit of $\{b_n\},$ then for any open interval $I_k$ containing $x,$ by definition of limit, the 'tail' of the sequence $\{b_n\}$ is contained in $I_k.$ However, I fail to 'see' any contradiction in this situation.

Any help would be appreciated.

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Oxtoby is a lovely author, but I just learned he probably did not read Borel's proof because I have just read his actual proof, and Q2 has a horrible answer.

First Q. First means, from left to right. So in your example, $b_1=0.5$ and $b_2=1$. But remember you have to cover $[0,1]$. Your example is not a covering since $1\notin (-1,0.5)$ and $1\notin(-0.1,1)$.

Second Q. Let me just write a better proof for everyone who comes through here (only the compact case though).

Let $I=[a,b]$. Suppose we have a cover $\{I_n\}$ of $I$.

Simplest Route (uses Heine-Borel): Since $I$ is compact, $\{I_n\}$ has a finite sub-cover $\{I_{j_i}\}_{i=1}^n$. Since $I\subseteq \cup I_{j_i}$, this implies $$ |I|\leq \sum_{i=1}^n |I_{j_i}|\leq\sum_{i=1}^\infty |I_{i}| $$ so we are done.

Oxtoby/Borel (does not use Heine-Borel): Read this. The link shows the complete proof. Essentially, Oxtoby is trying to show a finite sub-cover of $I$, but the way he goes about it is not-at-all obvious. More precisely, it seems Oxtoby has made some kind of assumption using a priori knowledge.

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  • $\begingroup$ I see. So for Q2, what assumption has Oxtoby made in his proof? $\endgroup$ – Idonknow Oct 24 '17 at 4:46
  • $\begingroup$ By the way, in your proof of Q2, I think we need to show that why is length function monotone. Specifically, if $A\subseteq B,$ why is it that $|A| \leq |B|.$ $\endgroup$ – Idonknow Oct 24 '17 at 4:48
  • $\begingroup$ I frankly don't know what assumption he has made. That's why I said it seems. I edited the inequality. I forgot to write the middle step which is essential. If you don't believe it, then you can see this more explicitly, Oxtoby writes this explicitly in his proof after his contradiction assumption. $\endgroup$ – user494247 Oct 24 '17 at 5:24
  • $\begingroup$ By the way, don't get stuck on this. There are plenty of books that have measure theory written very well, such as Shakarchi's Real Analysis. $\endgroup$ – user494247 Oct 24 '17 at 5:33
  • $\begingroup$ Thanks. I will take a look on Shakarchi's real analysis. $\endgroup$ – Idonknow Oct 24 '17 at 5:57

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