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Let $f$ be a normalised set function that is a submodular set function, and order a finite subset $T=\{x_1,\dots ,x_n\}$. Set $P_j=\{x_1,\dots ,x_j\}$ and $P_0=\varnothing$. Then the displayed inequality holds for every $U\subseteq T$, with equality for $U=T$. The right side is a modular set function lying below $f$ on the power set of $T$ and touching it at the top.

Proof. Reveal the elements of $U$ in the chosen order. Immediately before adding $x_j\in U$, the conditioning set is $U\cap P_{j-1}\subseteq P_{j-1}$. By diminishing returns, its marginal increment is at least $f(P_j)-f(P_{j-1})$. Sum and use $f(\varnothing )=0$. The same telescoping sum gives equality at $T$. The individual increments need not be nonnegative.

Suppose $f$ is a normalised set function and a submodular set function, and $T_1,\dots ,T_t\subseteq T$ cover every element of $T$ exactly $k$ times, counting repetitions in the indexed collection. Then the displayed inequality holds. Neither the monotone set function property nor the nonnegative function property is required.

Proof. Fix an ordering of $T$ and write $a_j=f(P_j)-f(P_{j-1})$ for its successive marginal increments. The chain lower bound for a submodular set function gives $f(T_i)\ge {\sum }_{x_j\in T_i}{a}_j$. Summing counts each $a_j$ exactly $k$ times, and ${\sum }_j{a}_j=f(T)$. Exact coverage matters because the $a_j$ may be negative. For a monotone set function, the same argument also allows each element to occur at least $k$ times.

For a weighted hypergraph, the hypergraph cut function totals the weights of hyperedges meeting both a subset and its complement. It is a normalised set function and a symmetric set function. With nonnegative weights it is a nonnegative function and a submodular set function. It is generally not a monotone set function: a single two-element hyperedge of weight one has cut value one on either singleton, but zero on the whole ground set.

Proof of submodularity for nonnegative weights. For one hyperedge $e$ of size at least two, let $q_e$ indicate whether it is split. Adding $x\notin e$ has zero marginal increment. For $x\in e$, with $x\notin S$, the marginal increment is $1$ if $S\cap (e\setminus \{x\})$ is empty, $-1$ if it equals $e\setminus \{x\}$, and zero otherwise. These values can only decrease under inclusion of $S$. Thus diminishing returns holds for $q_e$, and taking a nonnegative weighted sum preserves it. For a singleton or empty hyperedge, $q_e$ is identically zero. A negative-weight two-element hyperedge gives a counterexample to submodularity: its two singleton values sum to a negative number, while the values at their union and intersection are zero.

On a finite ground set, a modular set function is a set function satisfying the displayed equality for every pair of subsets. Equivalently,To prove this representation, apply the equality to disjoint $S$ and $\{x\}$: $h(S+x)-h(S)=h(\{x\})-h(\varnothing )$. Add these fixed marginal increments along any ordering of $S$. Conversely, the contributions of each element and of the constant cancel in the displayed identity. Thus a modular set function need not be a normalised set function; its normalised instances are precisely the real-valued set functions with finite additivity on a finite power set. Adding or subtracting a modular set function preserves submodularity.

Every real-valued submodular set function on a finite ground set $V$ is the sum of a normalised set function $g$ that is nonnegative, monotone and submodular, and a modular set function $h$. A useful explicit choice isProof. The formula makes $h$ a modular set function, so $g$ is a submodular set function. It is a normalised set function because $h(\varnothing )=f(\varnothing )$. For $x\notin S$, diminishing returns gives ${\Delta }_{x}f(S)\ge {\Delta }_{x}f(V\setminus \{x\})=d_x$, hence ${\Delta }_{x}g(S)\ge 0$. Thus $g$ is a monotone set function, and starting at zero makes it a nonnegative function. The decomposition separates all potentially negative terminal marginal increments into the simple additive term. A constant term in $h$ is necessary unless $f$ is already a normalised set function.

A real-valued set function is monotone, in the nondecreasing convention, when enlarging its input subset cannot decrease its value. Equivalently, each marginal increment ${\Delta }_{x}f(S)$ is nonnegative. This follows by adding elements individually; the converse follows by considering $S\subseteq S\cup \{x\}$. A normalised set function with this property is a nonnegative function.

For a set function $f:2^V\to \mathbb{R}$, the four properties are:

Here monotone set function means nondecreasing under inclusion; it does not require strict increase.

Use the chain lower bound for a submodular set function. Order $T=\{x_1,\dots ,x_n\}$, set $P_j=\{x_1,\dots ,x_j\}$ with $P_0=\varnothing$, and put $a_j=f(P_j)-f(P_{j-1})$. For each $U\subseteq T$, reveal its elements in this same order. Since $U\cap P_{j-1}\subseteq P_{j-1}$, diminishing returns and the normalised set function property giveApply this to each $T_i$ and interchange the finite sums. Every element occurs exactly $k$ times, soThis is the exact-cover inequality for submodular set functions. Some $a_j$ can be negative: exact multiplicity, rather than merely at least $k$ occurrences, is what makes the argument valid without monotonicity. In fact the nonnegative function hypothesis is unnecessary here. If $T=\varnothing$, every $T_i$ is empty and both sides vanish.

The monotone decomposition of a submodular set function subtracts each element's smallest possible marginal increment. For $x\in V$, defineThe set function $h$ is a modular set function: the contributions of each element, and the constant $f(\varnothing )$, cancel inSubtracting this equality from the submodularity inequality for $f$ proves that $g$ is a submodular set function. Also $g(\varnothing )=0$, so it is a normalised set function.

For $x\notin S$, the inclusion $S\subseteq V\setminus \{x\}$ and diminishing returns implyAdding elements one at a time now proves that $g$ is a monotone set function. Since it starts at zero, it is also a nonnegative function. Thus $f=g+h$ has all the required properties.

Here modular set function means equality in the intersection–union identity; a constant term is allowed. If one imposed the additional convention $h(\varnothing )=0$, the claimed decomposition of an arbitrary $f$ would require $f(\varnothing )=0$. The displayed constant is essential for the problem as stated. For $V=\varnothing$, take $g=0$ and $h=f$.

The rank of a matroid is a normalised set function, a monotone set function, and a submodular set function. Adding one element increases it by at most one.

Proof of submodularity. Choose a maximal matroid independent set $I_0$ in $A\cap B$, extend it to a maximal matroid independent set $I_A$ in $A$, then extend it to one $I_U$ in $A\cup B$. The matroid independence axioms make their sizes $r(A\cap B)$, $r(A)$ and $r(A\cup B)$. Every element of $I_U\setminus I_A$ lies outside $A$, since otherwise it could be added to $I_A$ inside $A$. Hence $I_0\cup (I_U\setminus I_A)$ is an independent subset of $B$, so its size $r(A\cap B)+r(A\cup B)-r(A)$ is at most $r(B)$.

The remaining assertions follow from the definition of the rank of a matroid. The empty set has maximum independent size zero, increasing the available subset cannot reduce that maximum, and deleting a newly added element from an independent subset decreases its size by at most one.