Independence system/ graph-theory $ (E,S)$ is an independence system with $ w: E \rightarrow \mathbb{R_+}$.
$E= \{ e_1,...,e_m\} $ is the set of edges with $ w(e_1) \geq....\geq w(e_m), w(e_{m+1}):=0$. Define the set: $ E_i:= \{ e_1,...,e_i\}$ and $ T \subset E$ with $ w'(T) = \sum_{ e \in T} w(e)$.
I will show: $$ w'(T)= \sum_{ i=1}^m |T \cap E_i|\big(w(e_i)-w(e_{i+1})\big)$$
Can somebody give me a hint to do that?
 A: It is enough to think what is $w'(T_j)$ where $T_j =\{e_j\}$.
Let $$w''(T_j)=\sum_{ i=1}^m |T_j \cap E_i|\big(w(e_i)-w(e_{i+1})\big)$$
then 
\begin{eqnarray}w''(T_j) &=&\underbrace{|T_j \cap E_1|}_{=0}\big(w(e_1)-w(e_{2})\big)+...+\underbrace{|T_j \cap E_{j-1}|}_{=0}\big(w(e_{j-1})-w(e_{j})\big) +\\
&+&  \underbrace{|T_j \cap E_{j}|}_{=1}\big(w(e_{j})-w(e_{j+1})\big)+...+\underbrace{|T_j \cap E_{m}|}_{=1}\big(w(e_m)-w(e_{m+1})\big) \\
&=& \underbrace{0+0+...+0}_{j-1}+\big(w(e_{j})-w(e_{j+1})\big)+...+\big(w(e_m)-w(e_{m+1})\big) \\
&=&w(j)-w(e_{m+1})\\
&=&w(j)-0\\
&= & w'(T_j)
\end{eqnarray}
So $w'$ and $w''$ are the same functions on singeltons and now should not be difficult to see that they are the same functions:
For every $T$ and every $i$ we have: $$|T\cap E_i| = \sum _{j\in T}|T_j\cap E_i|$$ so: 
\begin{eqnarray}w''(T) &=& \sum _{i=1}^m\color{red}{\Big(}\sum _{j\in T}|T_j\cap E_i|\color{red}{\Big)}\big(w(e_i)-w(e_{i+1})\big)\\
&=& \sum _{j\in T}\color{red}{\Big(}\sum _{i=1}^m|T_j\cap E_i|\big(w(e_i)-w(e_{i+1})\big)\color{red}{\Big)}\\
&= &\sum _{j\in T}w'(T_j)\\
&= & w'(T)
\end{eqnarray}
