Constant bound for the 1 dimensional Besicovitch covering theorem on real line I recently looked through the proof of the Gagliardo–Nirenberg Interpolation Inequality, see proof and it says that for real line $R$, there exists a sequence of open intervals $\{I_k\}$, which covers the compact support domain with
$$
\sum_k \chi_{I_k}\le 4
$$
I have read the proof of the Besicovitch covering theorem but if set $N=1$, I cannot get the constant bound $4$ here. Could anyone tell me that why the bound here is $4$?
Thanks !!
 A: Given a  covering by open intervals of a compact set $K \subset {\mathbb R}$, there is a finite subcover  $S$, and we will use intervals from this subcover. Let $S_1$ be the  set of intervals in $S$ that cover $x_1=\min K$. Choose $I_1$ as the interval in $S_1$ with the largest right endpoint, (breaking ties arbitrarily, e.g., in favor of the longest interval).
If $m \ge 1$ and $I_1,\ldots,I_m$ have been chosen and do not cover $K$, let $S_{m+1}$ be the set of intervals in $S$ that cover
$$x_{m+1}=\min\{ K\setminus \cup_{j=1}^m I_j\} \,,$$ and choose  $I_{m+1}$ as the interval in $S_{m+1}$ with the largest right endpoint  (breaking ties as before).
Since $S$ is finite, this process must stop, yielding a finite subcover $\{I_j\}_{j=1}^n$ of $K$. Now for each $m \in [2,n-1]$, the interval $I_{m+1}$ cannot cover $x_m$ (otherwise it would be chosen instead of $I_m$), so $I_{m+1}$ is disjoint from
$\cup_{j=1}^{m-1} I_j$. Thus the intervals  $\{I_{2k} : 2k \le n\}$ are pairwise disjoint, and so are the intervals in  $\{I_{2k+1} : 2k+1 \le n\}$. Thus
$$
\sum_k \chi_{I_k}\le 2 \,.
$$
