Properties of the Open-Set Content

From a Positive Functional to Open-Set Content

Let $X$ be a locally compact Hausdorff space. The open sets of $X$ can be assigned values using a positive linear functional: test an open set by all continuous functions that stay between zero and one and have compact support inside it, then take the largest functional value those tests can achieve. The resulting set function is called open-set content here. Its basic properties follow from the order-preserving nature of the functional and from the ability to split a compactly supported function across a finite open cover.

Write $C_c(X)$ for the real-valued continuous functions on $X$ with compact support, where $\operatorname{supp}f$ is the closure of $\{x:f(x)\ne0\}$. Let $I:C_c(X)\to\mathbb R$ be linear and positive, meaning $I(f)\ge0$ whenever $f\ge0$. Positivity implies order preservation: if $f\le g$, then $g-f\ge0$, so $I(f)\le I(g)$.

Definition: For an open set $U\subseteq X$, define its content by $$ c(U)=\sup\{I(f): f\in C_c(X),\ 0\le f\le1,\ \operatorname{supp}f\subseteq U\}. $$ The supremum is taken in $[0,\infty]$. Such an $f$ is called admissible for $U$. The zero function is admissible for every open set, so $c(U)\ge0$, and $c(\varnothing)=0$.

No measure is assumed in this definition: $c$ is a set function on the open subsets of $X$, and it may take the value $+\infty$. We use the standard cutoff fact for locally compact Hausdorff spaces: if $K$ is compact, $U$ is open, and $K\subseteq U$, there is $q\in C_c(X)$ with $0\le q\le1$, $q=1$ on $K$, and $\operatorname{supp}q\subseteq U$. This is the Urysohn cutoff property.

Monotonicity and a Finite Cutoff Partition

Proposition: If $U$ and $V$ are open subsets of $X$ and $U\subseteq V$, then $c(U)\le c(V)$.

Proof. Every function admissible for $U$ is admissible for $V$, because its support is contained in $U\subseteq V$. Thus the set of values whose supremum defines $c(U)$ is contained in the corresponding set for $c(V)$. Taking suprema gives $c(U)\le c(V)$, including when either supremum is infinite. $\square$

For subadditivity, the essential step is to split a test function over a finite cover of its compact support. The following construction gives the precise partition needed. Suppose a compact set $K$ is covered by finitely many open sets $U_{i_1},\ldots,U_{i_r}$. By the cutoff property and compactness, choose finitely many compact sets $K_j$ covering $K$, each contained in one of these open sets, and choose cutoffs $q_j$ such that $q_j=1$ on $K_j$ and $\operatorname{supp}q_j\subseteq U_{i(j)}$. Repetitions among the indices $i(j)$ are allowed. Put $S=\sum_{j=1}^r q_j$, and define

\psi_j=\frac{q_j}{\max(1,S)}.

The denominator is a positive continuous function, so each $\psi_j$ is continuous. Also $0\le\psi_j\le1$, $\operatorname{supp}\psi_j\subseteq\operatorname{supp}q_j$, and $\sum_j\psi_j=S/\max(1,S)\le1$ on $X$. Since $S\ge1$ on $K$, $\sum_j\psi_j=1$ there. These functions form a finite cutoff partition on $K$, subordinate to the specified open sets.

Countable Subadditivity

Theorem: For any sequence of open sets $U_1,U_2,\ldots$ in $X$, $$ c\left(\bigcup_{n=1}^{\infty}U_n\right)\le\sum_{n=1}^{\infty}c(U_n), $$ where the series of nonnegative extended values is the limit of its partial sums.

Proof. Set $U=\bigcup_{n=1}^{\infty}U_n$, and let $f$ be admissible for $U$. If $f=0$, then $I(f)=0$ and $I(f)\le\sum_n c(U_n)$. Otherwise $K=\operatorname{supp}f$ is compact and is covered by the $U_n$. A finite subcollection $U_{n_1},\ldots,U_{n_r}$ covers $K$. Construct the cutoff partition $\psi_1,\ldots,\psi_r$ on $K$ just described, subordinate to these open sets. Define $f_j=f\psi_j$. Each $f_j$ is continuous; its support is a closed subset of the compact set $K$, hence compact, and is contained in $U_{n_j}$. Moreover $0\le f_j\le1$, so $f_j$ is admissible for $U_{n_j}$. Since the partition sums to one on $K$, and $f=0$ outside $K$, we have $f=\sum_{j=1}^r f_j$ on all of $X$. Linearity gives $I(f)=\sum_j I(f_j)$.

It remains important not to charge the same open set once for each piece assigned to it. Let $J=\{n_1,\ldots,n_r\}$ be the set of distinct indices in the finite subcover, and, for each $i\in J$, define $h_i=\sum_{j:n_j=i}f_j$. This function is continuous with compact support contained in $U_i$. Also, $$ 0\le h_i=f\sum_{j:n_j=i}\psi_j\le f\sum_{j=1}^r\psi_j\le1. $$ Thus $h_i$ is admissible for $U_i$. Regrouping the finite sum by distinct indices now gives $$ I(f)=\sum_{i\in J}I(h_i)\le\sum_{i\in J}c(U_i)\le\sum_{n=1}^{\infty}c(U_n). $$ The last inequality holds because $J$ is a finite set of distinct indices and all terms are nonnegative. This bound holds for every admissible $f$ for $U$; taking the supremum over those $f$ proves the result. $\square$

Additivity on Disjoint Open Sets

Theorem: If $U_1,U_2,\ldots$ are pairwise disjoint open subsets of $X$, then $$ c\left(\bigcup_{n=1}^{\infty}U_n\right)=\sum_{n=1}^{\infty}c(U_n). $$ The equality is in $[0,\infty]$.

Proof. Countable subadditivity gives the inequality from left to right. To prove the reverse inequality, first fix a positive integer $N$. For each $1\le n\le N$, take any function $f_n$ admissible for $U_n$. Their supports are pairwise disjoint because each support is contained in its corresponding open set and the open sets are disjoint. Therefore $f=\sum_{n=1}^N f_n$ is continuous with compact support contained in $\bigcup_{n=1}^N U_n$, and $0\le f\le1$. It is admissible for $\bigcup_{n=1}^{\infty}U_n$, so $$ c\left(\bigcup_{n=1}^{\infty}U_n\right)\ge I(f)=\sum_{n=1}^N I(f_n). $$ Taking the supremum independently over the admissible choices of each $f_n$ yields $$ c\left(\bigcup_{n=1}^{\infty}U_n\right)\ge\sum_{n=1}^N c(U_n). $$ This remains valid if one of the finite number of contents is infinite: its admissible values are then unbounded, and the left side must be infinite. Finally, the partial sums on the right increase to $\sum_{n=1}^{\infty}c(U_n)$. Taking their supremum proves the reverse inequality, including the case of an infinite sum. $\square$

Worked Examples

Worked Example: A Weighted Functional on a Finite Discrete Space

Let $X=\{a,b,c\}$ have the discrete topology, and define $I(f)=2f(a)+f(b)+3f(c)$. Every function on $X$ is continuous with compact support. The functional is linear, and it is positive because each coefficient is nonnegative.

For any subset $U\subseteq X$, an admissible function satisfies $f(x)=0$ for $x\notin U$ and $0\le f(x)\le1$ for $x\in U$. Hence $I(f)\le\sum_{x\in U}w_x$, where $w_a=2,w_b=1,w_c=3$. The indicator $\chi_U$ is itself admissible and attains this bound. Therefore $c(U)=\sum_{x\in U}w_x$. For $U=\{a,c\}$, this gives $c(U)=2+3=5$; for $V=\{b\}$, $c(V)=1$. These sets are disjoint, and $c(U\cup V)=2+1+3=6=5+1$. Check: $c(X)=2+1+3=6$, which agrees with the sum of the contents of the disjoint sets $U$ and $V$.

Worked Example: Weighted Counting Content on the Natural Numbers

Give $X=\mathbb N=\{1,2,\ldots\}$ the discrete topology and let $I(f)=\sum_{n\ge1}2^{-n}f(n)$ for $f\in C_c(X)$. Compact subsets of this discrete space are finite, so each such sum has only finitely many nonzero terms. For any open $U\subseteq X$, admissibility implies $I(f)\le\sum_{n\in U}2^{-n}$. Conversely, for every finite $F\subseteq U$, the indicator $\chi_F$ is admissible and $I(\chi_F)=\sum_{n\in F}2^{-n}$. Taking the supremum over finite $F$ gives $c(U)=\sum_{n\in U}2^{-n}$.

Let $O$ be the odd positive integers and $E$ the even positive integers. Both are open and disjoint. The geometric series give $$ c(O)=\sum_{k=0}^{\infty}2^{-(2k+1)} =\frac{1/2}{1-1/4}=\frac23,\qquad c(E)=\sum_{k=1}^{\infty}2^{-2k} =\frac{1/4}{1-1/4}=\frac13. $$ Thus $c(O)+c(E)=2/3+1/3=1$. Since $O\cup E=\mathbb N$, $c(O\cup E)=\sum_{n\ge1}2^{-n}= (1/2)/(1-1/2)=1$ as well. Check: the two disjoint contents sum to $1$, exactly the content of their union.

Worked Example: Content from Evaluation at a Point

Let $X=\mathbb R$ and define $I(f)=f(0)$ for real-valued $f\in C_c(\mathbb R)$. This is linear and positive. If an open set $U$ does not contain $0$, every admissible $f$ vanishes at $0$, so $c(U)=0$. If $0\in U$, the cutoff property supplies an admissible $f$ supported in $U$ with $f(0)=1$. No admissible function can have $I(f)>1$, so $c(U)=1$.

For example, $U=(-1,1)$ and $V=(2,3)$ are disjoint open sets. Since $0\in U$ and $0\notin V$, $c(U)=1$ and $c(V)=0$. Their union contains $0$, so $c(U\cup V)=1=1+0$. The calculation also illustrates why content need not record the size or length of an open set: for this functional, only whether the open set contains the evaluation point matters.

A Common Counting Pitfall

In the subadditivity proof, a finite partition can have several pieces assigned to the same member of the open cover. A tempting but invalid step is to bound each piece by the content of its assigned open set and then count each distinct cover index only once. The piecewise bounds do not justify removing repeated terms.

For a concrete instance, take $X=\{x\}$, $I(f)=f(x)$, and the one-set cover $U_1=X$. Split the function $f(x)=1$ into two pieces $f_1(x)=1/2$ and $f_2(x)=1/2$, both assigned to $U_1$. Each piece has $I(f_j)=1/2$, and $c(U_1)=1$. The sum of the piecewise content bounds is $c(U_1)+c(U_1)=2$; it does not imply $I(f)\le c(U_1)$ by simply deleting one repeated term. The correct grouping is $h_1=f_1+f_2$. Then $h_1(x)=1$, so $0\le h_1\le1$, it is admissible for $U_1$, and $I(f)=I(h_1)=1\le c(U_1)=1$. In general, grouping all pieces assigned to $U_i$ works because their partition functions sum to at most one.

These properties explain why open-set content is useful in constructing a measure from a positive functional. Monotonicity respects containment, subadditivity controls open covers, and disjoint additivity prevents the content from overcounting disjoint pieces. The content is not yet being asserted to be a measure on all subsets; that requires additional construction and verification.

Check Your Understanding

Use the definition of admissibility and the proofs above to answer the following questions.

Why does positivity of $I$ imply that $f\le g$ entails $I(f)\le I(g)$?
In the subadditivity proof, why must the pieces assigned to one cover index be grouped before applying the content bound?
For $I(f)=f(0)$ on $C_c(\mathbb R)$, compute the content of $(-2,-1)\cup(0,3)$.
For the weighted counting functional $I(f)=\sum_{n\ge1}2^{-n}f(n)$, compute the content of $\{2,4,6,\ldots\}$.
Where does compactness enter the proof of countable subadditivity, and what does it allow us to do?