Exercise 1.6.21 (Besicovich covering lemma in one dimension)

Proof. Let denote the family of open intervals $I_1,\dots ,I_n$. Consider the subcollection of all intervals in that are not contained in the union of the other intervals, i.e.,

(1)

Let and enumerate the elements of by $I_1^{\prime },\dots ,I_m^{\prime }$. We argue that at $m$, it is not possible for a point to be contained in three of the intervals and that ${\bigcup <!--FUNCTION\; APPLICATION-->}_{i=1}^n{I}_n={\bigcup }_{j=1}^m{I}_m^{\prime }$ holds.

(i)

${\bigcup <!--FUNCTION\; APPLICATION-->}_{i=1}^n{I}_i={\bigcup }_{j=1}^m{I}_j^{\prime };$
To demonstrate that the former union is a subset of the latter, pick an arbitrary $x\in {\bigcup <!--FUNCTION\; APPLICATION-->}_{i=1}^n{I}_i$, i.e., $x\in I_i$ for $1\le i\le n$. If $I_i$ is not contained in the union of other intervals $I_1,\dots ,I_{i-1},I_{i+1},\dots ,I_n$, then $I_i=I_j^{\prime }$ for some $1\le j\le m$ by construction and we are done. If that is again not the case, then $I_i$ is contained in the union of some other intervals, i.e., $x\in I_{i_2}$ for some $i_2\ne i$. If $I_{i_2}$ satisfies the requirement of our refinement, then we are done. If not, we can find another set $I_{i_3}$ for some $i_3\ne i,i_2$. Continuing the argument in that manner, we must terminate at some $1\le i^{\prime }\le n$, latest at some $i_n\ne i,i_2,\dots ,i_{n-1}$. The resulting set $I_{i^{\prime }}$ cannot be contained in the union of the other intervals and must thus be equal to $I_j^{\prime }$ for some $1\le j\le m$ as desired.
The other direction follows directly from construction.

(ii)

Each point $x\in \mathbf{R}$ is contained in at most two of the $I_m^{\prime }$.
Suppose for the sake of contradiction that $x$ is contained in at least three refined intervals $I_{j_1}^{\prime }=(a_1,b_1)$, $I_{j_2}^{\prime }=(a_2,b_2)$ and $I_{j_3}^{\prime }=(a_3,b_3)$. Our strategy is to look for two intervals with the widest combined radius.

Let $a_i:=\min <!--FUNCTION\; APPLICATION-->\{a_1,a_2,a_3\}$ be the point residing furthest on the left, and let $b_j:=\max <!--FUNCTION\; APPLICATION-->\{b_1,b_2,b_3\}$ be the point residing furthest on the right. Then it is obvious that the associated intervals $(a_i,b_i)\cup (a_j,b_j)$ is going to contain the third interval, which has neither the point furthest to the left nor furthest to right. Thus, we arrive at a contradiction.

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