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A nonnegative function that is a uniformly continuous function on the interval $[0,\infty )$ and has a finite integral tends to zero. Otherwise uniform continuity gives disjoint intervals of a fixed width on which the function exceeds a fixed positive value, contradicting integrability. This supplies a useful decay argument for Lyapunov functions with merely continuous vector fields.

To make the fixed width explicit, suppose some $\eta >0$ occurs as a lower bound $f(t_n)\ge \eta$ along a sequence with $t_n\to \infty$. Choose $\delta >0$ from uniform continuity for output tolerance $\eta /2$. On every interval $(t_n-\delta /2,t_n+\delta /2)$ lying in $[0,\infty )$, the function is greater than $\eta /2$. Extract a subsequence whose centers are more than $\delta$ apart. Each disjoint interval contributes at least $\eta \delta /2$ to the integral, so the first $N$ contribute at least $N\eta \delta /2$. This contradicts finiteness of the integral. The same input width at every center is the essential step; continuity alone would allow progressively narrower peaks.

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.

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 $f:\mathcal{P}(V)\to \mathbb{R}$ 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.

A real-valued function $f:X\to \mathbb{R}$ is nonnegative when $f(x)\ge 0$ for every input $x\in X$. For example, $f(x)=x^2$ and $g(x)=|x|$ are nonnegative functions on the real line.

Products and nonnegative linear combinations preserve the property: if $f,g\ge 0$ and $\alpha ,\beta \ge 0$, then $f(x)g(x)\ge 0$ and $\alpha f(x)+\beta g(x)\ge 0$ at every point. A partition of unity uses nonnegative functions whose values sum to one, so its weighted combinations lie in the convex hull of the values being combined.

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 tensor-product surface basis is $b_{jk}(u,v)={\varphi }_j(u){\psi }_k(v)$. If both univariate bases consist of nonnegative functions, thenIf both factors are strictly positive at the chosen parameters, their product is strictly positive there as well. Its support is the Cartesian product of the two factor supports. With partition of unity, this nonnegativity additionally places the parametric surface point in the convex hull of the active control points.

The monotone convergence theorem says that if $0\le f_n\uparrow f$ almost everywhere and the functions are measurable on a measure space, thenwith infinite integrals permitted. The limit is measurable after the usual modification on any exceptional null set.

The dominated convergence theorem says that if measurable functions $f_n$ converge almost everywhere to $f$ and satisfy $|f_n|\le g$ almost everywhere for an integrable nonnegative function $g$, then $f$ is integrable andThe theorem applies equally to real or complex functions. In particular domination must be by one fixed Lebesgue integrable function, rather than by a different bound for each $n$.

Let $S_N={\sum }_{k=0}^N{a}_k$. These are nonnegative measurable functions increasing to the extended-valued sum $S$. Monotone convergence givesA nonnegative function with finite integral cannot be infinite on a set of positive measure. Hence $S(x)<\infty$ almost everywhere, which is exactly almost-everywhere convergence of the nonnegative series. One may redefine the sum arbitrarily on the exceptional null set to obtain an ordinary integrable representative.

The positive part of a real-valued function $f$ is the nonnegative function $f_{+}(x)=\max \{0,f(x)\}$. It retains positive values and replaces negative values by zero: if $f(x)=-3$, then $f_{+}(x)=0$, while $f(x)=2$ gives $f_{+}(x)=2$.

Together with the negative part $f_{-}(x)=\max \{0,-f(x)\}$, it satisfies $f=f_{+}-f_{-}$ and $|f|=f_{+}+f_{-}$. These identities reduce Lebesgue integration of a signed function to integration of nonnegative functions. The resulting difference of integrals is defined only when it does not subtract two infinite quantities; integrability requires both integrals to be finite.