I recently looked through the proof of the Gagliardo–Nirenberg Interpolation Inequality, see [proof](https://arxiv.org/pdf/1812.04281.pdf) 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](http://aurora.asc.tuwien.ac.at/%7Efunkana/downloads_general/sem_talebi.pdf) 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 !!



Constant bound for the 1 dimensional Besicovitch covering theorem on real line
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
$$
\sum\_k \chi\_{I\_k}\le 2 \,.
$$


