Monotone set function ($A\subseteq B\Rightarrow f(A)\le f(B)$)

A real-valued set function $f:\mathcal{P}(V)\to \mathbb{R}$ is monotone in the nondecreasing convention whenFor a finite ground set $V$, this is equivalent to every marginal increment being nonnegative. Necessity follows by comparing $S$ and $S\cup \{x\}$. Conversely, enumerate the finitely many elements of $B\setminus A$ and add them one at a time; summing their nonnegative increments gives $f(B)-f(A)\ge 0$. A normalised set function with this property is a nonnegative function.

The finiteness condition in that converse matters. On $V=\mathbb{N}$, define $f(S)=1$ for finite $S$ and $f(S)=0$ for infinite $S$. Adding one element never changes whether $S$ is finite, so all marginal increments are zero. Yet $f(\varnothing )=1>0=f(\mathbb{N})$, so the set function is not monotone. Single-element tests alone cannot compare sets differing in infinitely many elements.