 Maximum of element weights

Let $V$ be a nonempty finite set with real weights $w(v)$. For nonempty subsets put $f(S)={\max }_{v\in S}w(v)$ and choose $f(\varnothing )=c\le {\min }_{v\in V}w(v)$. This defines a monotone set function and a submodular set function on the full power set.

To check diminishing returns, for $x\notin S$ the marginal increment is ${\Delta }_{x}f(S)=\max \{0,w(x)-f(S)\}$. If $A\subseteq B$, then $f(A)\le f(B)$, so ${\Delta }_{x}f(A)\ge {\Delta }_{x}f(B)$. Choosing the value of the empty set no larger than every weight makes this calculation valid even for $A=\varnothing$.

For weights $2,5,9$ and $c=0$, adding the weight-$5$ element to a weight-$2$ selection increases $f$ by $3$; adding it to a selection already containing weight $9$ increases $f$ by zero.