So I am going through this paper. Could someone please explain how they proved this lemma:
Lemma 1.2 Let $I_1,\dots,I_n\subset\mathbb R$ be segments of lengths $\ell_1,\dots,\ell_n$ with midpoints $c_1,\dots,c_n$. Assume the union $\bigcup I_i$ is a segment (i.e. the family of segments is non-separable). Then the segment $I$ of length $\sum \ell_i$ with midpoint at the center of mass $c=\frac{\sum \ell_i c_i}{\sum \ell_i}$ covers $\bigcup I_i$.
P.S. I tried to access the research paper where this is proved but it seems it is not accessible to the public domain. Thank you!!
I read Goodman & Goodman proof and this is a short summery in OP's notations.
Let $\bigcup_{i=1}^n I_i=[a,b]$. We may assume that $I_i=[a_i,b_i]$ with $a=a_1\leq a_2\leq a_3\leq\dots\leq a_n$. Then for any $1\leq i\leq n$, $$c_i-a=a_i+\frac{l_i}{2}-a\leq l_1+l_2+\dots+l_{i-1}+\frac{l_i}{2}$$ which implies that $$ \begin{align}c-a&=\frac{\sum_{i=1}^n l_i c_i}{\sum_{i=1}^n l_i}-a=\frac{\sum_{i=1}^n l_i (c_i-a)}{\sum_{i=1}^n l_i}\\ &\leq \frac{\sum_{i=1}^n l_i \left(l_1+l_2+\dots+l_{i-1}+\frac{l_i}{2}\right)}{\sum_{i=1}^n l_i}\\ &=\frac{\sum_{1\leq j<i\leq n} l_i l_j+\sum_{i=1}^n\frac{l^2_i}{2}}{\sum_{i=1}^n l_i}=\frac{(\sum_{i=1}^n l_i)^2}{2\sum_{i=1}^n l_i}=\frac{\sum_{i=1}^n l_i}{2} \end{align}$$ which means that $c-\frac{\sum_{i=1}^n l_i}{2}\leq a.$ In a similar way, we show that $b\leq c+\frac{\sum_{i=1}^n l_i}{2}$ and we are done, $$\bigcup_{i=1}^n I_i=[a,b]\subset \left[c-\frac{\sum_{i=1}^n l_i}{2},c+\frac{\sum_{i=1}^n l_i}{2}\right].$$