The following example with $\nn=\{1,2,3\}$ illustrates the proof. \begin{equation*} \ss={\large\{}\{1\},\{2\},\{3\},\{1,2\},\{1,3\},\{2,3\}{\large\}} \end{equation*} $\ss$ contains two subcollections that are balanced: the singleton partition and the three pairs. Two weighting schemes for $\ss$ are \begin{equation*} \delta=\{1/2,1/2,1/2,1/4,1/4,1/4\}\quad \text{ and }\quad\lambda=\{1/3,1/3,1/3,1/3,1/3,1/3\}. \end{equation*} Then \begin{equation*} \delta-\lambda=\left\{\frac{2}{12},\frac{2}{12},\frac{2}{12}, \frac{-1}{12},\frac{-1}{12},\frac{-1}{12}\right\} \end{equation*} Continuing, \begin{equation*} \alpha\delta+(1-\alpha)\lambda=\left\{\frac{4+2\alpha}{12},\frac{4+2\alpha}{12}, \frac{4+2\alpha}{12},\frac{4-\alpha}{12},\frac{4-\alpha}{12},\frac{4-\alpha}{12} \right\} \end{equation*} Both $\delta$ and $\lambda$ are interior to the set of balanced weighting schemes, so the range of $\alpha$ over which their affine combinations are balanced weighting schemes is larger than $[0,1]$. In fact it is $[-2,4]$. At both endpoints are balanced weighting schemes with 0s, corresponding to the two balanced subcollections.